丁香园AV

We begin by the first derivative as an application of the tangent line to approximate $f\text{.}$

Recall that the tangent line to $f(x)$ at a point $\left(a,f(a)\right)$ is given by

\begin{equation*} \begin{split} y -f(a) \amp = f'(a) (x-a) \\ y \amp = f'(a)(x-a) + f(a) \end{split} \end{equation*}

provided that $f$ is differentiable at $x=a\text{.}$ As mentioned earlier, notice that the expression

\begin{equation*} f'(a)(x-a) + f(a) \end{equation*}

is linear in $x\text{.}$ Therefore, the above equation is also called the linear approximation of $f$ at $a\text{.}$ The function defined by

\begin{equation*} L(x) = f'(a)(x-a) + f(a) \end{equation*}

is called the linearization of $f$ at $a\text{.}$

If $f$ is differentiable at $a$ then $L$ is a good approximation of $f$ so long as $x$ is 鈥渘ot too far鈥� from $a\text{.}$ Put another way, if $f$ is differentiable at $a$ then under a microscope $f$ will look very much like a straight line, and thus will look very much like $L\text{;}$ since $L(x)$ is often much easier to compute than $f(x)\text{,}$ then it makes sense to use $L$ as an approximation. 贵颈驳耻谤别听5.3 shows a tangent line to $\ds y=x^2$ at three different magnifications.

Figure 5.3. The linear approximation to $y = x^2\text{.}$

Interactive Demonstration. The graph of $f(x) = x\sin(x) $ is shown below in blue, and the linearization of $f$ at $x=a $ is shown in red. Drag the point $(a,f(a))$ to investigate the linearization of $f $ at $a\text{.}$ Use the controls in the bottom left corner to zoom in. Notice how the linear approximation improves.

Definition 5.14. Linear Approximation.

Suppose we are given a function $y=f(x)\text{.}$

The linearization of $f$ at $x=a$ is given by

\begin{equation*} L(x)=f(a)+f'(a)(x-a) \end{equation*}

provided that $f$ is differentiable at $x=a\text{.}$
The linear approximation of $f$ at $x=a$ is given by

\begin{equation*} f(x) \approx L(x) = f(a) + f'(a)(x-a) \end{equation*}

provided that $f$ is differentiable at $x=a\text{.}$

Thus in practice if we want to approximate a difficult value of $f(b)\text{,}$ then we may be able to approximate this value using a linear approximation, provided that we can compute the tangent line at some point $a$ close to $b\text{.}$ Here are some examples.

Example 5.15. Linear Approximation.

Let $f(x)=\sqrt{x+4}\text{,}$ what is $f(6)\text{?}$

We are asked to calculate $f(6)=\sqrt{6+4}=\sqrt{10}$ which is not easy to do without a calculator. However 9 is (relatively) close to 10 and of course $f(5)=\sqrt{9}$ is easy to compute, and we use this to approximate $\sqrt{10}\text{.}$

To do so we have $f'(x)=1/(2\sqrt{x+4})\text{,}$ and thus the linear approximation to $f$ at $x=5$ is

\begin{equation*} L(x)=\bigg(\frac{1}{2\sqrt{5+4}}\bigg)(x-5)+\sqrt{5+4}=\frac{x-5}{6}+3\text{.} \end{equation*}

Notice that we did not create a common denominator to add teh two terms. This is because calculations are often easier in this form. Now to estimate $\sqrt{10}\text{,}$ we substitute 6 into the linear approximation $L(x)$ instead of $f(x)\text{,}$ to obtain

\begin{equation*} \sqrt{6+4}\approx \frac{6-5}{6}+3=\frac{19}{6}=3\sfrac{1}{6}=3.1\bar{6}\approx 3.17 \end{equation*}

It turns out the exact value of $\sqrt{10}$ is actually 3.16227766鈥� but our estimate of 3.17 was very easy to obtain and is relatively accurate. This estimate is only accurate to two decimal places.

With modern calculators and computing software it may not appear necessary to use linear approximations, but in fact they are quite useful. For example in cases requiring an explicit numerical approximation, they allow us to get a quick estimate which can be used as a 鈥渞eality check鈥� on a more complex calculation. Further in some complex calculations involving functions, the linear approximation makes an otherwise intractable calculation possible without serious loss of accuracy.

Example 5.16. Linear Approximation of Sine.

Find the linear approximation of $\sin x$ at $x=0\text{,}$ and use it to compute small values of $\sin x\text{.}$

If $f(x)=\sin x\text{,}$ then $f'(x)=\cos x\text{,}$ and thus the linear approximation of $\sin x$ at $x=0$ is:

\begin{equation*} L(x)=\cos (0)(x-0)+\sin (0)=x\text{.} \end{equation*}

Thus when $x$ is small this is quite a good approximation and is used frequently by engineers and scientists to simplify some calculations.

For example you can use your calculator (in radian mode since the derivative of $\sin x$ is $\cos x$ only in radian) to see that

\begin{equation*} \sin (0.1)=0.099833416\ldots \end{equation*}

and thus $L(0.1)=0.1$ is a very good and quick approximation without any calculator!

Exercises for Section聽5.3.1.

Exercise 5.3.1.

Determine the linear approximation $L(x)$ at $a$ of each function below. Then use $L(x)$ to approximate the value of each function at the given $x$-value.

$f(x) = \sqrt{x}, \ a=4, \ x=3$
Answer
$f(3) \approx L(3)=\frac{7}{4}$
Solution

We first differentiate:

\begin{equation*} f'(x) = \diff{}{x} \sqrt{x} = \frac{1}{2\sqrt{x}} \end{equation*}

Next, use the formula for $L(x)$ when $a=4\text{:}$

\begin{equation*} \begin{split} L(x) \amp = f(a) + f'(a)(x-a)\\ \amp = \sqrt{4} + \frac{1}{2\sqrt{4}} (x-4)\\ \amp = 2 + \frac{1}{4} (x-4) \end{split} \end{equation*}

Finally, use $L(x)$ to approximate the given quantity:

\begin{equation*} f(3) \approx L(3)= 2 + \frac{1}{4} (3-4) = \frac{7}{4} \end{equation*}
$f(x) = \sqrt[3]{x}, \ a=8, \ x=9$
Answer
$f(9) \approx L(9)=\frac{25}{12}$
Solution

We first differentiate:

\begin{equation*} f'(x) = \diff{}{x} x^{1/3} = \frac{1}{3x^{2/3}} \end{equation*}

Next, use the formula for $L(x)$ when $a=8\text{:}$

\begin{equation*} \begin{split} L(x) \amp = f(a) + f'(a)(x-a)\\ \amp = 8^{1/3} + \frac{1}{3(8^{2/3})} (x-8)\\ \amp = 2 + \frac{1}{12} (x-8) \end{split} \end{equation*}

Finally, use $L(x)$ to approximate the given quantity:

\begin{equation*} f(9) \approx L(9)= 2 + \frac{1}{12} (9-8) = \frac{25}{12} \end{equation*}
$f(x) = \dfrac{1}{x}, \ a=5, \ x=5.3$
Answer
$f(5.3) \approx L(5.3)=\frac{47}{250}$
Solution

We first differentiate:

\begin{equation*} f'(x) = \diff{}{x} \frac{1}{x} = -\frac{1}{x^2} \end{equation*}

Next, use the formula for $L(x)$ when $a=5\text{:}$

\begin{equation*} \begin{split} L(x) \amp = f(a) + f'(a)(x-a)\\ \amp = \frac{1}{5} - \frac{1}{25} (x-5) \end{split} \end{equation*}

Finally, use $L(x)$ to approximate the given quantity:

\begin{equation*} f(5.3) \approx L(5.3)= \frac{1}{5} - \frac{1}{25} (0.3) = \frac{47}{250} \end{equation*}
$f(x) = \dfrac{1}{x^{2}}, \ a=3, \ x=2.8$
Answer
$f(2.8) \approx L(2.8)=\frac{13}{135}$
Solution

We first differentiate:

\begin{equation*} f'(x) = \diff{}{x} \frac{1}{x^2} = -\frac{2}{x^3} \end{equation*}

Next, use the formula for $L(x)$ when $a=3\text{:}$

\begin{equation*} \begin{split} L(x) \amp = f(a) + f'(a)(x-a)\\ \amp = \frac{1}{9} - \frac{2}{27} (x-3) \end{split} \end{equation*}

Finally, use $L(x)$ to approximate the given quantity:

\begin{equation*} f(2.8) \approx L(2.8)= \frac{1}{9} - \frac{2}{27} (0.2) = \frac{13}{135} \end{equation*}
$f(x) = x^{2}+3, \ a=2, \ x=2.2$
Answer
$f(2.2) \approx L(2.2)=7.8$
Solution

We first differentiate:

\begin{equation*} f'(x) = \diff{}{x} (x^2+3) = 2x \end{equation*}

Next, use the formula for $L(x)$ when $a=2\text{:}$

\begin{equation*} \begin{split} L(x) \amp = f(a) + f'(a)(x-a)\\ \amp = (4+3) + 4 (x-2)\\ \amp = 7 + 4 (x-2) \end{split} \end{equation*}

Finally, use $L(x)$ to approximate the given quantity:

\begin{equation*} f(2.2) \approx L(2.2)= 7+4(0.2) = 7+0.8 = 7.8 \end{equation*}
$f(x) = (x-2)^{3}, \ a=3, \ x=3.1$
Answer
$f(3.1) \approx L(3.1)=1.3$
Solution

We first differentiate:

\begin{equation*} f'(x) = \diff{}{x} (x-2)^3 = 3(x-2)^2 \end{equation*}

Next, use the formula for $L(x)$ when $a=3\text{:}$

\begin{equation*} \begin{split} L(x) \amp = f(a) + f'(a)(x-a)\\ \amp = 1 + 3 (x-3) \end{split} \end{equation*}

Finally, use $L(x)$ to approximate the given quantity:

\begin{equation*} f(3.1) \approx L(3.1)= 1+3(0.1) = 1+0.3 =1.3 \end{equation*}

Exercise 5.3.2.

Find the linearization $L(x)$ of $f(x)=\ln (1+x)$ at $a=0\text{.}$ Use this linearization to approximate $f(0.1)\text{.}$

$L(x)=x\text{,}$ $f(0.1)\approx L(0.1)=0.1$

First, compute $f'(x)\text{:}$

\begin{equation*} f'(x)=\diff{}{x} \ln(1+x) = \frac{1}{1+x}\text{.} \end{equation*}

Choose $a=0\text{,}$ since we can readily compute $f(0)$ and $f'(0)\text{.}$ Thus,

\begin{equation*} L(x) = f(0) + f'(0)(x-0) = x\text{.} \end{equation*}

We can now estimate $f(0.1)\text{:}$

\begin{equation*} f(0.1) \approx L(0.1) = 0.1\text{.} \end{equation*}

Note: $|f(0.1)-L(0.1)| \approx 0.005\text{.}$

Exercise 5.3.3.

Use linear approximation to estimate $(1.9)^3\text{.}$

6.8

Let $f(x)=x^3\text{.}$ Then $f'(x) = 3x^2\text{.}$ The nearest integer to $1.9$ is $a=2\text{,}$ which makes our linearization:

\begin{equation*} L(x) = f(2) + f'(2) (x-2) = 8+12(x-2)\text{.} \end{equation*}

Therefore, our linear approximation at $x=1.9$ is

\begin{equation*} f(1.9) \approx L(1.9) = 8+12(0.1) = 6.8\text{.} \end{equation*}

Note: $|f(1.9)-L(1.9)| \approx 0.06\text{.}$

Exercise 5.3.4.

Show in detail that the linear approximation of $\sin x$ at $x=0$ is $L(x)=x$ and the linear approximation of $\cos x$ at $x=0$ is $L(x)=1\text{.}$

Let $f(x) = \sin x\text{.}$ Then

\begin{equation*} f'(x) = \cos x\text{,} \end{equation*}

and the linearization of $f$ at $x=0$ is

\begin{equation*} L(x) = \sin 0 + \cos 0 (x-0) = x\text{.} \end{equation*}

Next, let $f(x) = \cos x\text{.}$ Then

\begin{equation*} f'(x) = -\sin x\text{,} \end{equation*}

and the linearization of $f$ at $x=0$ is

\begin{equation*} L(x) = \cos 0 - \sin 0 (x-0) = 1\text{.} \end{equation*}

Exercise 5.3.5.

Use $f(x)=\sqrt[3]{x+1}$ to approximate $\sqrt[3]{9}$ by choosing an appropriate point $x=a\text{.}$ Are we over- or under-estimating the value of $\sqrt[3]{9}\text{?}$ Explain.

2.08\bar{3}; overestimate.

We wish to estimate $\sqrt[3]{9}$ using a linear approximation to $f(x) = \sqrt[3]{x+1}\text{.}$

The closest $x$-value for which we know the exact evaluation of $f(x)$ is $x=7\text{:}$

\begin{equation*} f(7) = \sqrt[3]{8} = 2\text{.} \end{equation*}

We then choose $a=7$ for our approximation, giving

\begin{equation*} \begin{split} L(x) \amp = f(7) + f'(7)(x-7)\\ \amp = f(7) + \frac{1}{3}\left(x+1\right)^{-2/3}\big\rvert_{x=7} \cdot (x-7) \\ \amp = 2 + \frac{1}{3}8^{-2/3} (x-7) \\ \amp = 2 + \frac{1}{12}(x-7) \end{split} \end{equation*}

So,

\begin{equation*} \sqrt[3]{9} = f(8) \approx L(8) = 2 + \frac{1}{12}(1) = \frac{25}{12}\text{.} \end{equation*}

This is a slight overestimation of the true value, since the slope of the tangent line to $f$ is decreasing from $x = 7$ to $x = 8\text{.}$ To check our calculations, we compute

\begin{equation*} L(8) = 12 \approx 2.0833 \dots \end{equation*}

and the exact value

\begin{equation*} f(8) = 9 \approx 2.080008\dots \end{equation*}

The approximation is therefore accurate to 0.03. See the plots below of $f(x) (x \geq -1)$ , and the corresponding linear approximation around $x = 7\text{.}$

Subsection 5.3.2 Differentials

Very much related to linear approximations are the differentials $dx$ and $dy\text{,}$ used not to approximate values of $f\text{,}$ but instead the change (or rise) in the values of $f\text{.}$

Definition 5.17. Differentials $dx$ and $dy$.

Let $y=f(x)$ be a differentiable function. We define a new independent variable $dx\text{,}$ and a new dependent variable $dy=f'(x)\,dx\text{.}$ Notice that $dy$ is a function both of $x$ (since $f'(x)$ is a function of $x$) and of $dx\text{.}$ We call both $dx$ and $dy$ differentials.

Now fix a point $a$ and let $\Delta x =x-a$ and $\Delta y= f(x)-f(a)\text{.}$ If $x$ is near $a$ then $\Delta x$ is clearly small. If we set $dx=\Delta x$ then we obtain

\begin{equation*} dy = f'(a)\,dx \approx \frac{\Delta y}{\Delta x}\Delta x = \Delta y\text{.} \end{equation*}

Thus, $dy$ can be used to approximate $\Delta y\text{,}$ the actual change in the function $f$ between $a$ and $x\text{.}$ This is exactly the approximation given by the tangent line:

\begin{equation*} dy = f'(a)(x-a) = f'(a)(x-a)+f(a)-f(a)=L(x)-f(a)\text{.} \end{equation*}

While $L(x)$ approximates $f(x)\text{,}$ $dy$ approximates how $f(x)$ has changed from $f(a)\text{.}$ 贵颈驳耻谤别听5.4 illustrates the relationships.

Figure 5.4. Differentials.

Note:

$x\text{,}$ $dx$ and $\Delta x$ are independent variables.
Both $dx$ and $\Delta x$ measure the change as $x$ changes from $x$ to $x+\Delta x\text{.}$
$y$ is a dependent variable of $x\text{,}$ and $\Delta y$ is a dependent variable of both $x$ and $\Delta x\text{.}$
$\Delta y$ measures the actual change in $y$ as $x$ changes from $x$ to $x+\Delta x\text{.}$
$dy$ measures the approximate change in $y$ as $x$ changes from $x$ to $x+\Delta x\text{.}$

Interactive Demonstration. The linearization of $f(x)$ at $x=a$ is shown in red below. Drag the point $(a,f(a))$ and use the slider to investigate the relationship between the differentials at any point $(a, f(a))\text{.}$

Since differentials are used to estimate the change in the dependent variable corresponding to a small change in the independent variable, they are a useful concept to analyze change in cost, revenue, and profit functions, which is summarized below.

Definition 5.18. Differentials in Marginal Analysis.

Suppose we are given the cost function $p=C(q)\text{.}$ If $C$ is differentiable, then

\begin{equation*} dC = C'(q)dq \ \ \text{ and } \ \ \Delta C \approx C'(q)\Delta q\text{.} \end{equation*}
Suppose we are given the revenue function $p=R(q)\text{.}$ If $R$ is differentiable, then

\begin{equation*} dR = R'(q)dq \ \ \text{ and } \ \ \Delta R \approx R'(q)\Delta q\text{.} \end{equation*}
Suppose we are given the profit function $p=P(q)\text{.}$ If $P$ is differentiable, then

\begin{equation*} dP = P'(q)dq \ \ \text{ and } \ \ \Delta P \approx P'(q)\Delta q\text{.} \end{equation*}

Example 5.19. Actual and Approximate Changes in $y$.

Let $y=x^{4}\text{.}$

Calculate $\Delta x$ and $\Delta y$ when $x$ changes from $2$ to $2.1\text{,}$ and from $2$ to $1.9\text{.}$
Calculate the differential $dy$ of $y\text{.}$ Use $dy$ to approximate $\Delta y$ when $x$ changes from $2$ to $2.1\text{,}$ and from $2$ to $1.9\text{.}$
Compare the results of part (b) with those of part (a).

Let $f(x) = x^{3}\text{.}$

When $x$ changes from $2$ to $2.1\text{,}$ $\Delta x = 2.1 - 2 = 0.1\text{.}$ Next,

\begin{equation*} \begin{split} \Delta y \amp = f(x + \Delta x) - f(x) = f(2.1) - f(2)\\ \amp = (2.1)^{3}-2^{3} = 9.261 - 8 = 1.261. \end{split} \end{equation*}

Similarly, when $x$ changes from $2$ to $1.9\text{,}$ we have $\Delta x = 1.9 - 2 = -0.1\text{,}$ and

\begin{equation*} \begin{split} \Delta y \amp = f(x+\Delta x) - f(x) = f(1.9) - f(2) \\ \amp = (1.9)^{3} - 2^{3} = 6.859 - 8 = -1.141. \end{split} \end{equation*}
$dy = f'(x)dx = 3x^{2}dx\text{.}$ First, take $x=2$ and $dx=2.1-2=0.1\text{.}$ Then,

\begin{equation*} dy = 3x^{2}dx = 3(2)^{2}(0.1) = 1.2\text{.} \end{equation*}

Next, we have $x=2$ and $dx =1.9-2 = -0.1\text{.}$ Therefore,

\begin{equation*} dy = 3x^{2}dx = 3(2)^{2}(-0.1) = -1.2\text{.} \end{equation*}
Comparing the actual changes from part (a) to the approximate changes from part (b), we notice that the approximation $dy = 1.2$ is quite close to the actual change $\Delta y = 1.20601\text{,}$ but the approximation $dy=-1.41$ is not very close to the actual change $\Delta y = -1.2\text{.}$

Example 5.20. Rise of Natural Logarithm.

Approximate the rise of $f(x)=\ln x$ from $x=1$ to $x=1.1\text{,}$ using differentials.

Note that $\ln (1.1)$ is not readily calculated (without a calculator) hence why we wish to use linear approximation to approximate $f(1.1)-f(1)\text{.}$

We fix $a=1$ since we know that $f(1) = \ln(1)= 0\text{,}$ and so

\begin{equation*} \Delta x=dx=x-1 \ \ \ \text{ and } \ \ \ \Delta y=f(x)-f(1)=\ln(x)- \ln(1) = \ln(x)\text{.} \end{equation*}

Since

\begin{equation*} f'(x) = \frac{1}{x}\text{,} \end{equation*}

we also have that

\begin{equation*} dy = f'(1)dx = \left(\frac{1}{1}\right) \left(x-1\right) = x-1\text{.} \end{equation*}

Then the actual change in $f$ as $x$ changes from $1$ to $1.1$ is approximated as follows:

\begin{equation*} \begin{split} \Delta y \amp \approx dy \\ f(1.1) - f(1) \amp \approx f'(1)(1.1-1) \\ \ln(1.1) \amp \approx 0.1 \end{split} \end{equation*}

The correct value of $\ln (1.1)$ is 0.0953鈥� and thus we were relatively close.

Example 5.21. Approximating Function Value.

Use differentials to approximate the value of $\sqrt{24.5}\text{.}$ Compare your result using a calculator.

Consider the function $y=f(x)=\sqrt{x}\text{.}$ We want to pick a number $x$ close to $24.5$ for which we know the value of $\sqrt{x}\text{.}$ Appropriately, we take $x=25\text{.}$ Then the change in $y\text{,}$ $\Delta y\text{,}$ as $x$ changes from $x=25$ to $x=24.5$ is

\begin{equation*} \begin{split} \Delta y \approx dy \amp = f'(x)\Delta x \\ \amp = \left(\frac{1}{2\sqrt{x}}\right)\bigg\rvert_{x=25} \cdot(-1.5) \\ \amp =\left(\frac{1}{10}\right)(-1.5) = -0.15 \end{split} \end{equation*}

Therefore,

\begin{equation*} \sqrt{24.5} - \sqrt{25} = \Delta y \approx -0.15 \end{equation*}

\begin{equation*} \sqrt{24.5} \approx \sqrt{25} - 0.15 = 4.75\text{.} \end{equation*}

A calculator tells us that $\sqrt{24.5} \simeq 4.94975\text{,}$ and so the error in our approximation is about 0.2.

Example 5.22. Approximating Operating Cost.

Suppose that the total operating cost of relocating a car 500 km at an average speed of $\nu$ km/h, is

\begin{equation*} C(\nu) = 150 + \nu + \frac{6000}{\nu} \end{equation*}

dollars. Find the approximate change in cost when the average speed is increased from 80 km/h to 85 km/h.

We use $\nu = 80$ to approximate $C(85)\text{.}$

\begin{equation*} \begin{split} \Delta C \approx dC \amp = C'(\nu)d\nu \\ \amp = \left(1-\frac{6000}{\nu^{2}}\right)\bigg\rvert_{\nu=80} \cdot (5) \\ \amp = \left(1-\frac{6000}{6400}\right)(5) = 0.0625, \end{split} \end{equation*}

that is, the total cost would increase by approximately $0.06.

Example 5.23. Approximating Sales.

A company determines that the relationship between the amount of money $q$ spent on advertising and total sales $S(q)$ is

\begin{equation*} S(q) = -0.02q^{3} + q^{2} + 2q + 100 \ \ \ 0 \leq q \leq 60 \end{equation*}

where $q$ is measured in thousands of dollars. Estimate the change in the company's total sales if the amount spent on advertising is increased from $50,000 ($q=50$) to $55,000 ($q=55$).

We approximate

\begin{equation*} \begin{split} \Delta S \approx dS \amp = S'(55)dq \\ \amp = \left(-0.06q^{2} + 2q + 2\right)\big\rvert_{q=50} \cdot (55-50) \\ \amp = (-150 + 100 + 2)(5) = -240. \end{split} \end{equation*}

That is, total sales will decrease by approximately $240,000.

Example 5.24. Approximating Drop in Price.

Suppose the demand for a certain product is given by

\begin{equation*} p = f(q) = \frac{100}{q^{2}+2} \end{equation*}

where $p$ is expressed in dollars/unit and $q$ is the quantity demanded each year. The manufacturer predicts they will be able to produce 6 billion units for the year. If the actual production is 6.2 billion units instead, what would happen to the predicted price, $p\text{?}$

The differential is given by

\begin{equation*} dp = -\frac{100q}{(q^{2}+2)^{2}} dq\text{.} \end{equation*}

So when $q=6$ and $dq = 0.2\text{,}$

\begin{equation*} \Delta p \approx dp = -\frac{100(6)}{(36+2)^{2}} (0.2) = -0.0831\text{,} \end{equation*}

that is, the price will drop by approximately $0.08.

Exercises for Section聽5.3.2.

Exercise 5.3.6.

Find the differential of the given function.

$\ds f(x) = 2x^{2}$
Answer
$\ds dy = 4x dx$
Solution
$dy = f'(x)\,dx = 4x \,dx$
$\ds g(t) = t^{3}-t$
Answer
$\ds dy = (3t^{2}-1)dt$
Solution
$dy = g'(t)\,dt = (3t^2-1)\,dt$
$\ds f(t) = \sqrt{t+1}$
Answer
$\ds dy = \frac{1}{2\sqrt{t+1}} dt$
Solution
$dy = f'(t)\,dt = \dfrac{1}{2\sqrt{t+1}}\,dt$
$\ds p(q) = 2q^{3/2}+q^{1/2}$
Answer
$\ds dy = \frac{6q+1}{2\sqrt{q}}dq$
Solution
$dy = p'(q)\,dq = \dfrac{6q+1}{2\sqrt{q}}\,dq$
$\ds h(s) = s + \dfrac{2}{s}$
Answer
$\ds dy = \frac{s^{2}-2}{s^{2}} ds$
Solution
$dy = h'(s)\,ds= \dfrac{s^2-s}{s^2}\,ds$
$\ds p(q) = \frac{q-1}{q^{2}+1}$
Answer
$\ds dy = \frac{-q^{2}+2q+1}{(q^{2}+1)^{2}} dq$
Solution
$dy = p'(q)\,dq = \dfrac{-q^2+2q+1}{(q^2+1)^2}\,dq$
$\ds f(x) = \sqrt{3x^{2} - x}$
Answer
$\ds dy = \frac{6x-1}{2\sqrt{3x^{2}-x}}dx$
Solution
$dy = f'(x)\,dx = \dfrac{6x-1}{2\sqrt{3x^2-x}}\,dx$

Exercise 5.3.7.

For the following functions $f(x)\text{,}$ determine $\Delta y$ and $dy$ at the given values of $a$ and $\Delta x\text{.}$

$\ds f(x) = x^4\text{,}$ $a=1\text{,}$ $dx= \Delta x =1/2$
Answer
$\Delta y=-15/16\text{,}$ $dy=2$
Solution

The differential is given by

\begin{equation*} dy = f'(x) dx \big\vert_{x=1} = 4(1)^3 \cdot \frac{1}{2} = 2\text{.} \end{equation*}

The actual change in $y$ is

\begin{equation*} \Delta y = f(a+\Delta x) - f(a) = 1-\left(\frac{1}{2}\right)^4 = -\frac{15}{16}\text{.} \end{equation*}
$\ds f(x) = \sqrt{x}\text{.}$ If $a=1$ and $\Delta x=1/10$
Answer
$\ds \Delta y=\sqrt{11/10}-1\text{,}$ $dy=0.05$
$f(x) = \sin (2x)\text{.}$ If $a=\pi$ and $\Delta x=\pi/100$
Answer
$\ds \Delta y=\sin(\pi/50)\text{,}$ $dy=\pi/50$

Exercise 5.3.8.

For the functions (i) $f(x)=x^{2}-1$ with $x$ changing from 1 to 0.9 and (ii) $f(x)=\frac{1}{x}$ with $x$ changing from $-1$ to $-1.01\text{,}$ do the following:

Calculate the differential of $f\text{.}$
Answer
(i) $\ds dy=2x dx\text{,}$ and (ii) $\ds dy=-\frac{1}{x^{2}} dx$
Solution

The differential of $f(x) = x^2-1$ is given by

\begin{equation*} df = f'(x)dx = 2xdx\text{.} \end{equation*}

The differential of $f(x)=1/x$ is given by

\begin{equation*} df = f'(x)dx = \frac{-1}{x^2}dx\text{.} \end{equation*}
Use your results from part (a) to find the approximate change in $y$ for the given change in $x\text{.}$ Answer
(i) $\ds dy \approx -0.02\text{,}$ and (ii) $\ds dy \approx -0.0098$
Solution

Let $f(x) = x^2-1.$ We approximate $\Delta y$ between $x=1$ and $x=0.9$ by

\begin{equation*} \begin{split} \Delta y \approx dy \amp = f'(x)dx \\ \amp =f'(1)(0.9-1) \\ \amp = 2(-0.1)= -0.2. \end{split} \end{equation*}

Let $f(x) = 1/x. $ We approximate $\Delta y$ between $x=-1$ and $x=-1.01$ by

\begin{equation*} \begin{split} \Delta y \approx dy \amp = f'(x)dx \\ \amp =f'(-1)(-1.01+1) \\ \amp = -1(-0.01)= 0.01. \end{split} \end{equation*}
Calculate the actual change in $y$ for the given change in $x$ and compare your results with that obtained in part (b). Answer
(i) $\ds \Delta y = -0.19\text{,}$ and (ii) $\ds \Delta y = -0.0099\text{.}$
Solution

Let $f(x) = x^2-1.$ The exact change in $y$ is given by

\begin{equation*} \Delta y = f(0.9) - f(1) = (0.9^2-1)-(1-1) = -0.19\text{.} \end{equation*}

Our estimate in part (b) was therefore accurate to 0.01.

Let $f(x) = 1/x. $ The exact change in $y$ is given by

\begin{equation*} \Delta y = f(-1.01) - f(-1) = (1/(-1.01))-(-1) = 0.00990..\text{.} \end{equation*}

Our estimate in part (b) was therefore accurate to $0.0001\text{.}$

Exercise 5.3.9.

Use differentials to approximate the given quantity.

$\ds \sqrt{49.5}$
Answer
$\ds \frac{197}{28}$
Solution

We know that $f(49) = \sqrt{49} = 7\text{,}$ and so let $f(x) = \sqrt{x}\text{,}$ $a = 49$ and $dx = 0.5\text{.}$ Therefore, the differential $df$ is

\begin{equation*} df =\frac{1}{2\sqrt{49}} (0.5) = \frac{1}{28}\text{.} \end{equation*}

Therefore, since $df \approx \Delta f\text{,}$ we have that

\begin{equation*} \Delta f = f(49.5) - f(49) = f(49.5) - 7 \approx \frac{1}{28}\text{.} \end{equation*}

Rearranging, we find that

\begin{equation*} f(49.5) \approx \frac{1}{28} + 7 = \frac{197}{28}\text{.} \end{equation*}
$\ds \sqrt[3]{8.2}$
Answer
$\ds 2.042$
Solution

Since we know that $f(8) = \sqrt[3]{8} = 2\text{,}$ let $f(x) = \sqrt[3]{x}\text{,}$ $a=8$ and $dx = 0.2\text{.}$ Then the differential $df$ is

\begin{equation*} df = \frac{1}{3x^{2/3}}\bigg\vert_{x=8} (0.2) = \frac{2}{30\cdot 4} =\frac{1}{60}\text{.} \end{equation*}

Now if we let $df \approx \Delta f\text{,}$ then

\begin{equation*} \Delta f = f(8.2)- f(8) \approx \frac{1}{60} \implies f(8.2) \approx \frac{1}{60} + 2 = \frac{121}{60}\text{.} \end{equation*}

Note: $|\frac{121}{60} - \sqrt[3]{8.2}| \approx 0.0001\text{.}$
$\ds \sqrt{4.05} + \frac{1}{\sqrt{4.05}}\text{.}$
Answer
$\ds 2.509$
Solution

Let $f(x) = \sqrt{x} + \dfrac{1}{\sqrt{x}}\text{.}$ The differential, $df\text{,}$ of $f$ is given by

\begin{equation*} df=f'(x)\,dx = \left(\frac{1}{2\sqrt{x}} - \frac{1}{2x^{3/2}}\right)dx\text{.} \end{equation*}

So we can evaluate exactly both $f(4)$ and $f'(4)\text{.}$ We can now estimate $f(4.05) = \sqrt{4.05} + \dfrac{1}{\sqrt{4.05}}\text{:}$

\begin{equation*} \begin{split} \Delta f = f(4.05) - f(4) \amp \approx f'(4) (0.05) f(4.05) \\\amp \approx f'(4) (0.05) + f(4) \\ \amp \approx \left(\frac{1}{4} - \frac{1}{16}\right) (0.05) + \left(2+ \frac{1}{2}\right) \\ \amp \approx \frac{3}{16}(0.05)+\frac{5}{2} \approx 2.509375. \end{split} \end{equation*}

We can compare this value to the exact value

\begin{equation*} f(4.02)=2.509365 \dots \end{equation*}

We see that this approximation is accurate to around $10^{-6}\text{!}$

Exercise 5.3.10.

Approximate the amount of paint needed to apply a coat of paint 0.02 cm thick to a sphere with diameter $40$ meters. You may use the fact that the volume of a sphere of radius $r$ is $V =(4/3)\pi r^3\text{,}$ where in this example, $dr=0.02\text{.}$

$dV=4\pi(20^2) (0.0002)$

We know that the volume of the sphere is given by $V = \frac{4}{3} \pi r^3\text{,}$ and so

\begin{equation*} dV = 4\pi r^2 \,dr\text{.} \end{equation*}

Here, we have $r=20$ m and $dr = 0.02$ cm. Thus,

\begin{equation*} dV = 4\pi(20^2) (0.0002) \ \text{m} ^3\text{.} \end{equation*}

Exercise 5.3.11.

It is determined that a certain country's gross domestic product (GDP) can be approximated by

\begin{equation*} f(x) = 350 x^{1/4} \end{equation*}

where $f(x)$ is measured in millions of dollars and $x$ is the capital expenditure in billions of dollars. Approximate the change in GDP if the country's capital expenditure changes from $200 million to $210 million.

$16 million

The differential is given by

\begin{equation*} df = \frac{350}{4} x^{-3/4} \,dx\text{.} \end{equation*}

So when $x= 200$ and $dx = 10$ million,

\begin{equation*} df = \frac{350}{4} (200)^{-3/4} (10) \approx 16.46\text{.} \end{equation*}

Hence, the country's GDP will increase by approximately $16.5 million.

Exercise 5.3.12.

A major supermarket determines that their yearly profit $P(q)$ is related to the amount $q$ spent on advertising by

\begin{equation*} P(q)=-\frac{1}{6}q^{2}+12q+15 \ \ \ 0\leq q \leq 73 \end{equation*}

where both $P(q)$ and $q$ are measured in thousands of dollars. Approximate the change in profits when advertising expenditure is increased from $30,000 to $32,000.

Increase of approximately $4,000

The differential, $dP\text{,}$ of the profit function is found to be

\begin{equation*} dP = P'(q)dq = -\left(\frac{1}{3}q+12\right)dq\text{.} \end{equation*}

Notice that $P'(q)$ is easily evaluated when $q=30\text{.}$ Therefore, when the advertising expenditure is increased from $q=30$ to $q=32\text{,}$ we estimate

\begin{equation*} \Delta P \approx dP = \left(-\frac{1}{3}(30)+12\right)(32-30) = 4\text{.} \end{equation*}

That is, we can expect an increase in profits of approximately $4,000.

Exercise 5.3.13.

A bank determines that the number $N(t)$ of loans issued over the course of one year is related to the interest rate $r$ by

\begin{equation*} N(t)=\frac{8}{1+0.02r^{2}} \end{equation*}

where $N$ is measured in millions. Approximate the change in the number of loans the bank issues when the interest rate is increased from 10% to 10.5%.

Approximately 160 fewer loans

The differential is given by

\begin{equation*} dN = \frac{-8(0.04 r)}{(1+0.02r^2)^2}\,dr\text{.} \end{equation*}

So when $r=0.1$ and $dr = 0.005\text{,}$

\begin{equation*} dN \approx -0.000159\text{.} \end{equation*}

That is, the number of loans is expected to decrease by approximately 160.

Exercise 5.3.14.

The supply equation for a certain product is given by

\begin{equation*} p=s(q)=0.5\sqrt{q}+8 \end{equation*}

where $q$ is the quantity supplied and $p$ is the unit price in dollars. Approximate the change in price when the quantity supplied is increased from 10,000 to 10,200 units.

Increase by $0.50.

We first compute the differential:

\begin{equation*} dp = s'(q)\,dq = \frac{1}{4\sqrt{q}} \, dq\text{.} \end{equation*}

Then when $q=10,000$ and $dq = 200\text{,}$

\begin{equation*} dp = \frac{1}{4\sqrt{10,000}} (200) = 0.5\text{.} \end{equation*}

That is, the unit price is expected to increase by $0.50.

Subsection 5.3.3 Error Approximation

When working with differentials, we approximate function values, and therefore an error is introduced compared to the actual function values. Suppose we are given a function $y=f(x)$ with a measured quantity as input. If $a$ is the exact value of the measured quantity, but $a+dx$ is the measured value, then $dx=\Delta x$ represents the so-called measurement error. Furthermore, this measurement error causes an error in the calculation of $f(x)\text{,}$ which is known as propagation error $\Delta y=f(a+dx)-f(a)\text{.}$ Both types of errors are known as absolute errors.

If $y=f(a)$ is calculated with the absolute error $\Delta y\text{,}$ then the relative error in the calculation of $y$ is given by the quantity $\frac{\Delta y}{y}\text{,}$ while the percentage error is given by the quantity $\frac{\Delta y}{y} \times 100\%\text{.}$ Since $\Delta y$ is approximated by $dy\text{,}$ the relative error is approximated by $\frac{dy}{y}$ and the percentage error by $\frac{dy}{y}\times 100\%\text{.}$

\begin{equation*} \begin{array}{lcc} \amp \qquad \textbf{ True Value} \amp \qquad \textbf{ Approximate Value} \\[2ex] \textbf{ Absolute Error} \amp \qquad \Delta y \amp \qquad dy \\[2ex] \textbf{ Relative Error} \amp \qquad \dfrac{\Delta y}{y} \amp \qquad \dfrac{dy}{y} \\[2ex] \textbf{ Percentage Error} \amp \qquad \dfrac{\Delta y}{y} \times 100\% \amp \qquad \dfrac{dy}{y} \times 100\% \end{array} \end{equation*}

Table 5.5. Types of Error when Working with Differentials.

Example 5.25. Approximating Errors in Measurement.

We are given that the radius of a spherical object is measured to be $0.4$ m to within an error of $\pm 0.001$ m. What are the relative and percentage errors?

The relative error in $r$ is

\begin{equation*} \frac{dr}{r} = \pm \frac{0.001}{0.4} = \pm 0.0004\text{.} \end{equation*}

The percentage error is then

\begin{equation*} \frac{dr}{r} \times 100 \% = 0.0004 \times 100 \% = \pm 0.04 \%\text{.} \end{equation*}

Example 5.26. Approximating Errors in Measurement.

The sides of a cubical object are measured with an absolute percentage error of 3%. Approximate the maximum percentage error in the calculated volume of the cube using differentials.

Let $x$ be the side-length of the cube. Then its volume is

\begin{equation*} V = x^{3}\text{,} \end{equation*}

and

\begin{equation*} dV = 3x^{2}dx\text{.} \end{equation*}

Therefore,

\begin{equation*} \frac{dV}{V}=\frac{3x^{2}dx}{x^{3}}=3\frac{dx}{x}\text{.} \end{equation*}

Hence,

\begin{equation*} \left\rvert\frac{dV}{V}\right\rvert = 3\left\rvert\frac{dx}{x}\right\rvert \leq 3(0.03) = 0.09\text{,} \end{equation*}

where we used the fact that $\left\rvert\frac{dx}{x}\right\rvert \leq 0.03\text{.}$

Thus, the maximum percentage error in the measurement of the volume of the cube is 9%.

Exercises for Section聽5.3.3.

Exercise 5.3.15.

The edges of a cube are measured to be 12 cm in length, with a maximum possible error of 0.02 cm. What is the maximum possible error that could occur when calculating the volume of the cube?

$\pm 8.64$ cm$^{3}$

The length of each side of the cube is $x=12$ with maximum measurement error $0.02$ cm. The possible error in the measurement of the volume of the cube, $V=x^3\text{,}$ can be approximated by

\begin{equation*} \begin{split} \big\rvert dV \big\rvert \amp \leq \big\rvert V'(x) \big\rvert \big\rvert dx \big\rvert \amp \leq \big\rvert 3(12)^2 \big\rvert \big\rvert 0.02\big\rvert \amp \leq 8.64 \text{ cm } ^3 \end{split} \end{equation*}

The maximum possible error in the measurement of the volume of the cube is thus $\pm 8.64$ cm$^3\text{.}$

Exercise 5.3.16.

A wooden box with a lid is lacquered to an even thickness of 0.04 cm. If the edges of the box measure 0.5 m, then calculate the approximate amount of lacquer required.

$300$cm$^3$

The volume of the box is given by

\begin{equation*} V = l^3\text{,} \end{equation*}

where $l$ is the side length. Therefore,

\begin{equation*} dV = 3l^2 \, dl\text{.} \end{equation*}

So if $l=0.5$ m and $dl = 0.04$ cm, then

\begin{equation*} dV = 3(0.5)^2(0.0004) =0.0003 \ \text{m} ^3\text{.} \end{equation*}

Therefore, we require 0.3 L.

Exercise 5.3.17.

A dome of radius 20 m is to be coated with a layer of paint. What is the approximate amount of paint needed if the coat is to be 0.05 cm thick? Note that the volume of a dome of radius $r$ is $V = \frac{2}{3}\pi r^{3}\text{.}$

1257 L

We are given that the volume of a dome of radius $r$ is $V = \frac{2}{3}\pi r^3\text{.}$ Therefore,

\begin{equation*} dV = 2\pi r^2 dr\text{.} \end{equation*}

Therefore, if we have $dr = 0.05$ cm when $r=20$ m, then

\begin{equation*} dV = 2\pi (20)^2 \cdot 5 \times 10^{-4} \approx 1.2566 \text{m} ^3\text{.} \end{equation*}

That is, approximately 1257 L of paint is required.

Exercise 5.3.18.

True or false: If $A=f(x)\text{,}$ then the percentage change in $A$ is

\begin{equation*} \frac{100f'(x)}{f(x)}dx\text{.} \end{equation*}

Explain your answer.

True.

True. The percentage change in $A$ is approximately

\begin{equation*} \frac{100\left[f(x+\Delta x) - f(x)\right]}{f(x)} \approx \frac{100 f'(x)\,dx}{f(x)}\text{,} \end{equation*}

for $\Delta x$ small.

Exercise 5.3.19.

The quarterly profit of a certain manufacturer is given by

\begin{equation*} P(q)=-0.000032q^{3} + 6q - 300 \end{equation*}

million dollars, where $q$ is measured in tens of thousands of units. The expected number of units sold over the next quarter is 320,000, with a maximum error of 18%. Determine the maximum error in the expected profit.

$\pm$ 1.06 million dollars

We calculate the differential:

\begin{equation*} dp = f'(q)\,dq = -\frac{150q}{(3q^2+2)^2}\,dq\text{.} \end{equation*}

So the percentage error in the unit price is

\begin{equation*} \frac{dp}{p} = \left(-\frac{150q}{(3q^2+2)^2}\,dq\right)\left(\frac{3q^2+2}{25}\right)= -\frac{6 q}{3 q^2 + 2}\,dq\text{.} \end{equation*}

If the maximum percentage error in $q$ is $10$%, then

\begin{equation*} \left\vert \frac{dp}{p} \right\vert = \frac{6 q^2}{3 q^2 + 2}\left\vert \frac{dq}{q} \right\vert \leq \frac{6 q^2}{3 q^2 + 2} (0.1)\text{.} \end{equation*}

When $q=2\text{,}$ we have

\begin{equation*} \left\vert \frac{dp}{p} \right\vert \leq \frac{6(4)}{3(4) + 2} (0.1) \leq 0.18\text{.} \end{equation*}

Hence, the maximum percentage error in unit price is $\pm 18$ %.

Exercise 5.3.20.

The demand equation for a certain product is given by

\begin{equation*} p=f(q)=\frac{25}{3q^{2}+2} \end{equation*}

where $p$ is the unit price in dollars and $q$ is the quantity demanded each year, measured in thousands of units. It is expected that the demand will be 2000 units for the year, with a maximum error of 10%. What is the maximum error in the predicted price?

$\pm 0.36$ %

We are given that

\begin{equation*} \bigg\rvert \frac{dq}{q} \bigg\rvert \leq 0.10 \implies \big\rvert dq \big\rvert \leq 0.1(2)\text{,} \end{equation*}

when the demand is expected to be 2000 units. This corresponds to an approximate error in the price $p$ of

\begin{equation*} \begin{split} \big\rvert dp \big\rvert \amp \leq \big\rvert p'(q) \big\rvert \big\rvert dq \big\rvert \amp \leq \bigg\rvert 25\diff{}{q} \left(3q^2+2\right)^{-1} \bigg\rvert \big\rvert dq\big\rvert \amp \leq \bigg\rvert -150(2)(3(2)^2+1)^{-2} \bigg\rvert \big\rvert 0.1(2) \big\rvert \amp \leq \bigg\rvert \frac{-300}{(13)^2} (0.2) \bigg\rvert \amp \leq \dots \amp \leq 0.355, \end{split} \end{equation*}

or approximately $0.36 per unit.

Exercise 5.3.21.

Suppose a monthly mortgage payment $P\text{,}$ in dollars, is computed using the formula

\begin{equation*} P = \frac{10,000 r}{1-\left(1+\frac{r}{12}\right)^{-360}} \end{equation*}

where $r$ is the interest rate per year.

Find the differential of $P\text{.}$
Answer
$\ds dP = \dfrac{10,000\left(1-(1+\frac{r}{12})^{-360}-30r(1+\frac{r}{12})^{-361}\right)}{\left(1-\left(1+\frac{r}{12}\right)^{-360}\right)^{2}} dr$
Solution

We first differentiate:

\begin{equation*} P'(r) = \diff{}{r}\left(\frac{10,000 r}{1-\left(1+\frac{r}{12}\right)^{-360}}\right) = \dfrac{10,000\left(1-(1+\frac{r}{12})^{-360}-30r(1+\frac{r}{12})^{-361}\right)}{\left(1-\left(1+\frac{r}{12}\right)^{-360}\right)^{2}} \end{equation*}

Hence,

\begin{equation*} dP = \diff{}{r}\left(\frac{10,000 r}{1-\left(1+\frac{r}{12}\right)^{-360}}\right) = \frac{10,000\left(1-(1+\frac{r}{12})^{-360}-30r(1+\frac{r}{12})^{-361}\right)}{\left(1-\left(1+\frac{r}{12}\right)^{-360}\right)^{2}}\,dr \end{equation*}
Approximately how much more will the monthly mortgage payments be if the interest rate increases from the present rate of 3% per year to 3.2 % per year? From 3% to 3.3% per year? To 3.4% per year? To 3.5% per year?
Answer
$12.90,$19.40,$25.89,$32.36
Solution
Let $r=0.03$ and $dr = 0.002\text{.}$ Then $dP \approx 12.9\text{.}$ When $dr = 0.003\text{,}$ $dP \approx 19.4\text{,}$ and so on. We find that, when the interest rate increases to 3.2% per year, 3.3% per year, 3.4% per year and 3.5% per year, the monthly mortgage payments increase by about $12.90, $19.40, $25.89 and $32.36, respectively.

Exercise 5.3.22.

Suppose $10,000 is deposited into an account that pays interest at the rate $r$/year compounded monthly. Then the account balance at the end of 10 years is given by

\begin{equation*} A = 10,000\left(1+\frac{r}{12}\right)^{120}\text{.} \end{equation*}

Find the differential of $A\text{.}$
Answer
$dA = 100,000\left(1+\frac{r}{12}\right)^{119}dr$
Solution

We first differentiate:

\begin{equation*} A'(r) = 100,000\left(1+\frac{r}{12}\right)^{119} \end{equation*}

Therefore,

\begin{equation*} dA = 100,000\left(1+\frac{r}{12}\right)^{119}\,dr\text{.} \end{equation*}
Approximately how much more would the account be worth at the end of 10 years with an interest rate of 1.1% per year instead of 1%? 1.2% per year instead of 1%? 1.3% per year instead of 1%?
Answer
$110,$220,$331
Solution
When $r = 0.01$ and let $dr = 0.001\text{,}$ we compute $dA = 110.42\text{.}$ Hence, the account will be worth approximately $110.42 more after 10 years when the interest rate is raised to 1.1%. Similarly, we find that when the interest rate is raised to 1.2$ and 1.3%, the account is worth approximately $220 and $331 more, respectively.

Exercise 5.3.23.

Suppose $2000 per month is deposited into an account that pays interest at the rate $r$/year compounded monthly. Then the account balance at the end of 25 years is given by

\begin{equation*} S = \frac{24,000\left(\left(1+\frac{r}{12}\right)^{300}-1\right)}{r} \end{equation*}

dollars.

Find the differential of $S\text{.}$
Answer
$dS =24,000\left(\dfrac{300r(1+\frac{r}{12})^{299}(\frac{1}{12})-(1+\frac{r}{12})^{300}+1}{r^{2}} \right)dr$
Solution

We first differentiate:

\begin{equation*} S'(r) = 24,000\left(\dfrac{300r(1+\frac{r}{12})^{299}(\frac{1}{12})-(1+\frac{r}{12})^{300}+1}{r^{2}} \right) \end{equation*}

Therefore,

\begin{equation*} dS = 24,000\left(\dfrac{300r(1+\frac{r}{12})^{299}(\frac{1}{12})-(1+\frac{r}{12})^{300}+1}{r^{2}} \right)\,dr\text{.} \end{equation*}
Approximately how much more would the account be worth at the end of 25 years with an interest rate of 1.6%/year instead of 1.5%? 1.7%/year instead of 1.5%? 1.8%/year instead of 1.5%?
Answer
$9617; $19,235; $28,852
Solution
When $r=0.015$ and $dr = 0.001\text{,}$ we compute $dS = 9617\text{.}$ That is, the account will be worth approximately $9617 more after 25 years if the interest rate is raised to 1.6% per year. Similarly, the account will be worth about $19,235 and $28,852 more if the interest rate is raised to 1.7% and 1.8% per year, respectively.

Subsection 5.3.4 Newton's Method

A well-known numerical method is Newton's Method (also sometimes referred to as Newton-Raphson's Method), named after Isaac Newton and Joseph Raphson. This method is used to find roots, or $x$-intercepts, of a function. While we may be able to find the roots of a polynomial which we can easily factor, we saw in the previous chapter on Limits, that for example the function $e^x + x = 0$ has a solution ($i.e\text{.}$ root, or $x$-intercept) at $x \approx-0.56714\text{.}$ By the Intermediate Value Theorem we know that the function $e^x + x = 0$ does have a solution. We cannot here simply solve for such a root algebraically, but we can use a numerical method such as $Newton's\text{.}$ Such a process is typically classified as an $iterative$ method, a name given to a technique which involves repeating similar steps until the desired accuracy is obtained. Many computer algorithms are coded with a for-loop, repeating an iterative step to converge to a solution.

The idea is to start with an initial value $x_0$ (approximating the root), and use linear approximation to create values $x_1\text{,}$ $x_2\text{,}$ $\cdots$ getting closer and closer to a root.

The first value $x_1$ corresponds to the intercept of the tangent line of $f(x_0)$ with the $x$-axis, which is:

\begin{equation*} x_1 = x_0 -\frac{f(x_0)}{f'(x_0)} \end{equation*}

Figure 5.6. First iteration of Newton's Method.

We can see in 贵颈驳耻谤别听5.6, that if we compare the point $(x_0,0)$ to $(x_1,0)\text{,}$ we would likely come to the conclusion that $(x_1,0)$ is closer to the actual root of $f(x)$ than our original guess, $(x_0,0)\text{.}$ As will be discussed, the choice of $x_0$ must be done correctly, and it may occur that $x_1$ does not yield a better estimate of the root.

Newton's method is simply to repeat this process again and again in an effort to obtain a more accurate solution. Thus at the next step we obtain:

\begin{equation*} x_2 = x_1 -\frac{f(x_1)}{f'(x_1)} \end{equation*}

Figure 5.7. Second iteration of Newton's Method.

We can now clearly see how $(x_2,0)$ is a better estimate of the root of $f(x)\text{,}$ rather than any of the previous points. Moving forward, we will get:

\begin{equation*} x_3 = x_2 -\frac{f(x_2)}{f'(x_2)} \end{equation*}

Rest assured, $(x_3,0)$ will be an even better estimate of the root! We express the general iterative step as:

\begin{equation*} x_{n+1} = x_n -\frac{f(x_n)}{f'(x_n)} \end{equation*}

Interactive Demonstration. Investigate the first four iterations of Newton's Method below. Drag the initial point $x_1 $ to see how the subsequent iterations change. Credit: Jamie Mulholland.

The idea is to iterate these steps to obtain the desired accuracy.

Newton's Method.

Choose an initial estimate $x_{0}$ of the root $r\text{.}$
Calculate the next estimate using the iterative formula

\begin{equation*} x_{n+1} = x_{n} - \frac{f(x_{n})}{f'(x_{n})} \ \ \ \ (n=0,1,2,...) \end{equation*}
Calculate $\left\rvert x_{n} - x_{n+1}\right\rvert\text{,}$ which determines the number of accurate digits that have been achieved in the estimation of the root $r\text{.}$
Either repeat as of step 2 or terminate the algorithm with $r \approx x_{n+1}$ unless Newton's Method failed (see Key Points at the end of this section).

Here is an example.

Example 5.27. Newton's Method to Approximate a Root.

Approximate the roots of $f(x)=x^3-x+1$ by Newton's Method, accurate to six decimal places.

Since the function is a cubic, solving the equation algebraically is difficult. We therefore use Newton's Method to compute an approximate root.

Our function $f$ has only one real root as a sketch confirms (see 厂别肠迟颈辞苍听5.7 on how to perform curve sketching. We note that $f(-1)=-5$ and $f(0)=1\text{.}$ We apply the Intermediate Value Theorem to determine that $f$ has a root between these two values. We choose to start with the initial value $x_0=-1\text{.}$ We encourage you to try Newton's Method with a different initial value such as -0.5 or -0.7 or any other value between -1 and 0.

We calculate the derivative to be $f'(x)=3x^2-1\text{.}$ Therefore, Newton's formula is given by

\begin{equation*} x_{n+1} = x_n -\frac{f(x_n)}{f'(x_n)} = x+n - \frac{x^3_n-x_n+1}{3x^2_n-1} \cdot \end{equation*}

We compute the following approximations of the root. We also encourage you to verify that these are indeed the correct iterations through Newton's algorithm.

\begin{equation*} \begin{array}{l} x_0 = -1 \\ x_1= -1.5000\\ x_2 = -1.347826... \\ x_3= -1.325200... \\ x_4 = -1.324718... \\ x_5 = -1.324717... \\ x_6 = -1.324717... \\ \cdots \end{array} \end{equation*}

Hence, the root we are seeking is approximately $-1.324717\text{.}$

Subsubsection 5.3.4.1 Applications

Example 5.28. Market Equilibrium.

The monthly demand $q$ (in units of a thousand) for a certain product is related to the unit price $p$ (in dollars) by the demand equation

\begin{equation*} p=25e^{-0.1(q+1)^{2}}\text{.} \end{equation*}

The monthly supply for the same product is given by the supply equation

\begin{equation*} p=5+\frac{5}{2}q^2\text{.} \end{equation*}

Estimate the equilibrium point of this system.

We determine the equilibrium point by finding the point of intersection of the demand curve and the supply curve.

\end{adjustbox} To find the equilibrium point, we equate both equations, giving

\begin{equation*} 5 + \frac{5}{2}q^{2} = 25e^{-0.1(q+1)^{2}} \end{equation*}

\begin{equation*} 5+\frac{5}{2}q^{2}-25e^{-0.1(q+1)^{2}} = 0 \end{equation*}

Instead of attempting to solve this equation exactly using algebraic techniques, we instead employ Newton's Method to find an approximate solution. First, write

\begin{equation*} 10+5q^2-50e^{-0.1(q+1)^2} = 0\text{,} \end{equation*}

since it is easier to work without the fraction. Then,

\begin{equation*} \begin{split} f(q) \amp = 10 + 5q^{2} - 50e^{-0.1(q+1)^{2}}\\ \amp = 5\left(2+q^{2}-10e^{-0.1(q+1)^{2}}\right), \text{ and } \\ f'(q) \amp = 10q - 50e^{-0.1(q+1)^{2}}\cdot\left(-0.2(q+1)\right) \\ \amp = 10\left(q+(q+1)e^{-0.1(q+1)^{2}}\right). \end{split} \end{equation*}

And so we construct the required iterative formula as

\begin{equation*} q_{n+1} = q_{n} - \frac{2+q_{n}^{2}-10e^{-0.1(q_{n}+1)^{2}}}{2\left(q_{n}+(q_{n}+1)e^{-0.1(q_{n}+1)^{2}}\right)} \end{equation*}

From the above sketch, we see that a reasonable estimate of the intersection point is $q=2\text{.}$ We now carry out Newton's Method using an initial guess of $q_0 = 2\text{.}$

\begin{equation*} \begin{split} q_{1} \amp = 2 - \frac{2+2^{2}-10e^{-0.1(3)^{2}}}{2\left(2+(3)e^{-0.1(3)^{2}}\right)} \approx 1.69962 \\[1ex] q_{2} \amp = 1.69962 - \frac{2+(1.69962)^{2}-10e^{-0.1(2.69962)^{2}}}{2\left(2+(2.69962)e^{-0.1(2.69962)^{2}}\right)} \approx 1.68899 \\[1ex] q_{3} \amp = 1.68899 - \frac{2+(1.68899)^{2}-10e^{-0.1(2.68899)^{2}}}{2\left(2+(2.68899)e^{-0.1(2.68899)^{2}}\right)} \approx 1.68898 \end{split} \end{equation*}

Therefore, the equilibrium quantity is approximately 1.689 units, and the equilibrium price is correspondingly

\begin{equation*} p=5+\frac{5}{2}(1.689)^{2} \approx 12.1316 \end{equation*}

or approximately $12.1316 per unit. The equilibrium point is thus $(1.689,12.1316)\text{.}$

Another application of Newton's Method is to the internal rate of return on an investment. Suppose an investment yields returns of $R_1,R_2,\dots,R_n$ dollars at the end of the first, second,$\dots\text{,}$$n$-th periods, respectively with an initial payment of $C$ dollars. Then this investment has a net present value of

\begin{equation*} \frac{R_{1}}{1+r} + \frac{R_{2}}{(1+r)^{2}} + \frac{R_{3}}{(1+r)^{3}} + \dots + \frac{R_{n}}{(1+r)^{n}} - C = 0\text{.} \end{equation*}

By multiplying both sides of the above equation with $(1+r)^n\text{,}$ we obtain

\begin{equation*} C(1+r)^{n} - R_{1}(1+r)^{n-1} - R_{2}(1+r)^{n-2} - R_{3}(1+r)^{n-3} - \dots - R_{n} = 0\text{.} \end{equation*}

Typically, a company's executives use the internal rate of return to determine whether an investment is profitable or not.

Example 5.29. Internal Rate of Return.

A company is deciding on whether or not to purchase new equipment. The upfront cost of the equipment is $50,000, but the company predicts that they will save $15,000-$1000(m-1)$ per year after $m$ years for up to $4$ years, after which the equipment will be useless. Approximate the internal rate of return on this investment.

This investment would yield returns of $R_1 = 15,000-1000(1-1)=15,000$ after the first year, of $R_2 = 15,000-1000(2-1)= 14,000$ after the second year, $R_3= 15,000-1000(3-1)=13,000$ after the third year, and $R_4 = 15,000-1000(4-1) = 12,000$ after the fourth year. We also have that the initial investment is $C=50,000\text{.}$ Therefore, we wish to solve

\begin{equation*} 50,000(1+r)^{4} - 15,000(1+r)^{3} - 14,000(1+r)^{2} - 13,000(1+r) - 12,000 = 0 \end{equation*}

for $r\text{.}$ Let $x=1+r$ for simplicity. Then,

\begin{equation*} f(x)=50,000x^4-15,000x^3-14,000x^2-13,000x-12,000\text{,} \end{equation*}

where we are looking to solve $f(x)=0\text{.}$ We can approximate the root of $f$ using Newton's Method. Since

\begin{equation*} f'(x)=200,000x^3-45,000x^2-28,000x-13,000 \end{equation*}

the required iterative formula is

\begin{equation*} x_{n+1} = x_{n} - \frac{50,000x_n^4 - 15,000x_n^3-14,000x_n^2-13,00x_n-12,000}{200,000x_n^3-45,000x_n^2-28,000x_n-13,000} \cdot \end{equation*}

Choose $x_{0}=1.0\text{.}$ Then our iterates are

\begin{equation*} \begin{split} x_1 \amp = 1.0 - \frac{50,000(1.0)^4 - 15,000(1.0)^3-14,000(1.0)^2-13,000(1.0)-12,000}{200,000(1.0)^3-45,000(1.0)^2-28,00(1.0)-13,000}\\[1ex] \amp \approx 1.03509 \\ x_2 \amp \approx 1.03277 \\ x_3 \amp \approx 1.03276 \end{split} \end{equation*}

So if $x \approx 1.033\text{,}$ then $r \approx 1-x = 0.033\text{.}$ Therefore, we find the rate of return on the investment to be about 3.3%.

As with any numerical method, we need to be aware of the weaknesses of any technique we are using.

Figure 5.8. Function with three distinct solutions.

If we know our root is somewhere near $a\text{,}$ we would make our guess $x_0=a\text{.}$ Generally speaking, a good practice is to make our guess as close to the actual root as possible. In some cases we may have no idea where the root is, so it would be prudent to perform the algorithm several times on several different initial guesses and analyze the results.

For example we can see in 贵颈驳耻谤别听5.8 that $f(x)$ in fact has three roots, and depending on our initial guess, we may get the algorithm to converge to different roots. If we did not know where the roots were, we would try the technique several times. In one instance, if our initial guess was $x_a\text{,}$ we'd likely converge to $(a, 0)\text{.}$ Then if we were to choose another guess, $x_b\text{,}$ then we'd likely converge to $(b, 0)\text{.}$ Eventually, using various initial guesses we'd get one of three roots: $a\text{,}$ $b\text{,}$ or $c\text{.}$ Under these circumstances we can clearly see the effectiveness of this numeric method.

Figure 5.9. Newton's Method applied to $\sin x$ with unstable point $x_0\text{.}$

As another example if we attempt to use $Newton's$ $Method$ on $f(x) = \sin x$ using $x_0=\pi/2\text{,}$ then $f'(x_0)=0$ so $x_1$ is undefined and we cannot proceed. Even in general $x_{n+1}$ is typically nowhere near $x_n\text{,}$ and in general not converging to the root nearest to our initial guess of $x_0\text{.}$ In effect, the algorithm keeps "bouncing around", an example of which is depicted in 贵颈驳耻谤别听5.9. Based on our initial guess for such a function, the algorithm may or may not converge to a root, or it may or may not converge to the root closest to the initial guess. This gives rise to the more common issue: Selection of the initial guess, $x_0\text{.}$

Here is a summary.

Key Points in Using Newton's Method.

We attempt to choose $x_0$ as close as possible to the root we wish to find.
A guess for $x_0$ which makes the algorithm 鈥榖ounce around鈥� is considered unstable.
Even the smallest changes to $x_0$ can have drastic effects: We may converge to another root (see 贵颈驳耻谤别听5.8), we may converge very slowly requiring many more iterations, or we may not converge at all due to an unstable point (see 贵颈驳耻谤别听5.9).
We may encounter a stationary point if we choose $x_0$ such that $f'(x_0)=0$ ($i.e\text{.}$ $x_0$ is a critical point, see 顿别蹿颈苍颈迟颈辞苍听5.48!) in which case the algorithm fails (see 贵颈驳耻谤别听5.10).

This is all to say that your initial guess for $x_0$ can be extremely important.

Figure 5.10. Example of a stationary point, $f'(x_0)=0.$

Exercises for Section聽5.3.4.

Exercise 5.3.24.

Apply Newton's Method using the following steps: First, determine the function given the radical. Second, approximate the given radical with three iterations based on the given initial guess.

$\ds \sqrt{3}\text{,}$ $x_{0}=1.5$
Answer
$f(x)=x^{2}-3\text{,}$ 1.732051
Solution

We need to find a function $f$ such that $f(x) = 0 \iff x = \sqrt{3}\text{.}$ We cannot take $f(x) = x-\sqrt{3}$ (since we are trying to approximate $\sqrt{3}\text{!}$), let $f(x) = x^2-3$ and search for the positive root. Apply Newton's Method using:

\begin{equation*} x_{k+1} = x_k - \frac{x_k^2-3}{2x_k}, k = 0,1,2,3,..\text{.} \end{equation*}

The first three iterations are shown in the table below.

\begin{equation*} \begin{array}{ll} k \amp x_k \\ \hline 0 \amp 1.5 \\ 1 \amp 1.5 - \dfrac{1.5^2-3}{2(1.5)} =1.75 \\ 2 \amp 1.75 - \dfrac{1.75^2-3}{2(1.75)}=1.73214 \\ 3 \amp 1.73214 - \dfrac{1.73214^2-3}{2(1.73214)} = 1.73205 \end{array} \end{equation*}

Hence, $\sqrt{3} \approx 1.73205\text{.}$ Note: $|1.72305-\sqrt{3}| \approx 0.009\text{.}$
$\ds \sqrt{7}\text{,}$ $x_{0}=2.5$
Answer
$f(x) = x^{2}-7\text{,}$ 2.645751
Solution

Suppose $f(x) = x^2-7\text{.}$ Then $f(x) = 0$ if and only if $x = \pm \sqrt{7}\text{.}$ So to approximate $\sqrt{7}\text{,}$ we can use Newton's Method with $f(x) = x^2-7$ and search for the positive root. The iterative formula is thus:

\begin{equation*} x_{n+1} = x_n - \frac{x_n^2-7}{2x_n}, n = 0,1,2,3,..\text{.} \end{equation*}

The first three iterations are shown in the table below, taking $x_0 = 2.5\text{.}$

\begin{equation*} \begin{array}{ll} n \amp x_n \\ \hline 0 \amp 2.5 \\ 1 \amp 2.5 - \dfrac{2.5^2-7}{2(2.5)} = 2.65 \\ 2 \amp 2.65 - \dfrac{2.65^2-7}{2(2.65)}= 2.64575471... \\ 3 \amp 2.645751... \end{array} \end{equation*}

Therefore, $\sqrt{7} \approx 2.64575\text{.}$
$\ds \sqrt[3]{14}\text{,}$ $x_{0}=2.5$
Answer
$f(x)= x^{3}-14\text{,}$ 2.410142
Solution

Suppose $f(x) = x^3-14\text{.}$ Then $f(x) = 0$ if and only if $x = \sqrt[3]{14}\text{.}$ Therefore, we use Newton's Method with the iterative formula:

\begin{equation*} x_{k+1} = x_{k} - \frac{x_k^3-14}{3x_k^2}, k = 0,1,2,3,..\text{.} \end{equation*}

Taking an initial guess $x_0=2.5\text{,}$ the first three iterations are shown in the table below.

\begin{equation*} \begin{array}{ll} k \amp x_k \\ \hline 0 \amp 2.5 \\ 1 \amp 2.5 - \dfrac{2.5^3-14}{3(2.5)^2} =2.4133 \\ 2 \amp 2.41015\\ 3 \amp 2.41014 \end{array} \end{equation*}

Hence, $\sqrt[3]{14} \approx 2.4101\text{.}$

Exercise 5.3.25.

Apply Newton's Method to approximate the root of each function within the given interval, accurate to three decimal places.

$f(x)=e^{-x}-x\text{,}$ $(0,1)\text{.}$
Answer
0.567143
Solution

We wish to approximate a root of $f(x) = e^{-x}-x$ in the interval $(0,1)\text{.}$ Notice that $f(0) = 1 > 0$ and $f(1)=-0.632... \lt 0$ and so by the Intermediate Value Theorem, $f$ must contain at least one root in this interval. Our Newton iterations will be of the form

\begin{equation*} x_{k+1} = x_k - \frac{f(x_k)}{f'(x_k)} = x_k + \frac{e^{-x}-x}{e^{-x}+1}, k = 0,1,2,3,..\text{.} \end{equation*}

For our initial guess, take $x_0 = 0.5\text{.}$

\begin{equation*} \begin{array}{ll} k \amp x_k \\ \hline 0 \amp 0.5 \\ 1 \amp 0.566311\\ 2 \amp 0.567143\\ 3 \amp 0.567143 \end{array} \end{equation*}

We stop after 3 iterations since the method has converged on a root. Thus, $x = 0.567143$ is our approximation to a root of $f\text{.}$
$f(x)=e^{x} - \frac{1}{x}\text{,}$ $(0,1)\text{.}$
Answer
0.567143
Solution

Notice that $x=1$ is a root of $f(x)\text{.}$ We can check if there is a root in the interval $(0,1)$ using Newton's Method with $x_0=0.5\text{:}$

\begin{equation*} x_{k+1} = x_k - \frac{e^{x_k}-\frac{1}{x_k}}{e^{x_k} +\frac{1}{x^2_k}}, k = 0,1,2,3,..\text{.} \end{equation*}

\begin{equation*} \begin{array}{ll} k \amp x_k \\ \hline 0 \amp 0.5 \\ 1 \amp 0.562187\\ 2 \amp 0.56712\\ 3 \amp 0.567143 \end{array} \end{equation*}

Hence, $x=0.567$ is a root of $f$ to three decimal places.
$f(x)=\ln (x^{2})\text{,}$ $(6,7)\text{.}$
Answer
None
Solution

The function $f(x) = \ln(x^2)$ has roots

\begin{equation*} x^2= 1 \implies x = \pm 1 \end{equation*}

and so has no roots in the interval $(6,7)\text{.}$

Exercise 5.3.26.

Sketch the graphs of $f$ and $g$ on the same Cartesian coordinate system. Based on the sketch, choose an initial value to approximate the $x$-coordinate of the intersection point(s) of the two graphs. Then apply Newton's Method to determine the $x$-coordinate of the intersection point accurate to within two decimal places.

$\ds f(x) = \sqrt{x}\text{,}$ $g(x)=-0.5x+2$
Answer
1.528
Solution

From the sketch below, we see that $x_0 = 1.5$ is a reasonable starting guess for the intersection point of $f$ and $g\text{:}$
Let $h(x) = f(x) - g(x)\text{.}$ Then Newton's Method applied to the function $h(x)$ will approximate

\begin{equation*} h(x) = 0 \implies f(x) - g(x) = 0 \implies f(x) = g(x)\text{.} \end{equation*}

Newton's formula is thus

\begin{equation*} x_{k+1} = x_k - \frac{\sqrt{x_k} +0.5x_k - 2}{\frac{1}{2x_k}-0.5} k = 0,1,2,..\text{.} \end{equation*}

The first three iterations are shown below:

\begin{equation*} \begin{array}{ll} k \amp x_k \\ \hline 0 \amp 1.5 \\ 1 \amp 1.52781\\ 2 \amp 1.52786\\ 3 \amp 1.52786 \end{array} \end{equation*}

Hence, the intersection point of $f$ and $g$ is approximately $1.528\text{.}$
$\ds f(x)=e^{-x^{2}}\text{,}$ $g(x)=x^{2}$
Answer
0.753
Solution

From the sketch below, we see that $x_0 = 1$ is a reasonable starting guess for the intersection point of $f$ and $g\text{:}$
Let $h(x) = f(x) - g(x)\text{.}$ Then Newton's Method applied to the function $h(x)$ will approximate

\begin{equation*} h(x) = 0 \implies f(x) - g(x) = 0 \implies f(x) = g(x)\text{.} \end{equation*}

Newton's formula is thus

\begin{equation*} x_{k+1} = x_k - \frac{e^{-x_k^2} - x^2}{-2x_ke^{-x_k^2} - 2x_k} k = 0,1,2,..\text{.} \end{equation*}

The first three iterations are shown below:

\begin{equation*} \begin{array}{ll} k \amp x_k \\ \hline 0 \amp 1 \\ 1 \amp 0.768941\\ 2 \amp 0.753185\\ 3 \amp 0.753089 \end{array} \end{equation*}

Hence, the intersection point of $f$ and $g$ is approximately $0.753\text{.}$
$\ds f(x)=\ln x\text{,}$ $g(x)=2-x$
Answer
1.557
Solution

From the sketch below, we see that $x_0 = 2$ is a reasonable starting guess for the intersection point of $f$ and $g\text{:}$
Let $h(x) = f(x) - g(x)\text{.}$ Then Newton's Method applied to the function $h(x)$ will approximate

\begin{equation*} h(x) = 0 \implies f(x) - g(x) = 0 \implies f(x) = g(x)\text{.} \end{equation*}

Newton's formula is thus

\begin{equation*} x_{k+1} = x_k - \frac{ln(x_k) -2 + x_k}{\frac{1}{x_k} + 1} k = 0,1,2,..\text{.} \end{equation*}

The first three iterations are shown below:

\begin{equation*} \begin{array}{ll} k \amp x_k \\ \hline 0 \amp 2 \\ 1 \amp 1.5379\\ 2 \amp 1.5571\\ 3 \amp 1.55715 \end{array} \end{equation*}

Hence, the intersection point of $f$ and $g$ is approximately $1.557\text{.}$

Exercise 5.3.27.

For each of the given functions $f(x)\text{,}$ show that $f(x)=0$ has a root between the given $x$-values. Use Newton's Method to find the zero(s). Hint

Use the Intermediate Value Theorem.

$f(x)=3x^{2}-9x-11$ between $x=-1$ and $x=0\text{;}$ and between $x=3$ and $x=4\text{.}$
Solution

The function $f(x)$ is a polynomial (and hence continuous everywhere), and so we can apply IVT. We first investigate the root between $x=-1$ and $x=0\text{.}$ Since $f(-1) = 3 +9 - 11 > 0\text{,}$ and $f(0) = -11 \lt 0\text{,}$ by the Intermediate Value Theorem, $f$ must contain at least one root in the interval $(-1,0)\text{.}$ Let $x_0 = -0.5$ and

\begin{equation*} x_{k+1} = x_k - \frac{3x_k^2-9x_k-11}{6x_k+9}, k = 0,1,2,3,..\text{.} \end{equation*}

Then:

\begin{equation*} \begin{array}{ll} k \amp x_k \\ \hline 0 \amp -0.5 \\ 1 \amp -0.979167 \\ 2 \amp -0.932861 \\ 3 \amp -0.93242 \\ 4 \amp -0.93242 \end{array} \end{equation*}

We conclude that $x = -0.93242$ approximates a root of $f(x)\text{.}$ We see that $f(x)$ must also have a root on the interval $(3,4)\text{,}$ since

\begin{equation*} f(3)= 3(9) - 9(3) -11 \lt 0, \ \text{ and } \ f(4) = 3(16)-9(4)-11 > 0\text{.} \end{equation*}

So, take the same iterative formula as above, but now take $x_0 = 3.5\text{:}$

\begin{equation*} \begin{array}{ll} k \amp x_k \\ \hline 0 \amp 3.5 \\ 1 \amp 3.97917 \\ 2 \amp 3.93286 \\ 3 \amp 3.93242\\ 4 \amp 3.93242 \end{array} \end{equation*}

Hence, we conclude that $x = 3.93242$ also approximates a root of $f(x)\text{.}$
$f(x)=x^{3}-x-1$ between $x=1$ and $x=2\text{.}$
Solution

The function $f(x)$ is a polynomial (and hence continuous everywhere), and so we can apply IVT. Since $f(1) = 1-1-1 = -1 \lt 0$ and $f(2) = 8-2-1=5 > 0\text{,}$ by the Intermediate Value Theorem, $f$ must contain at least one root in the interval $(1,2)\text{.}$ To find this root, take $x_0=1.5$ and the Newton iterations

\begin{equation*} x_{k+1} = x_k - \frac{x_k^3-x_k-1}{3x_k^2-1}, k = 0,1,2,3,..\text{.} \end{equation*}

Then:

\begin{equation*} \begin{array}{ll} k \amp x_k \\ \hline 0 \amp 1.5 \\ 1 \amp 1.34783 \\ 2 \amp 1.3252 \\ 3 \amp 1.32472 \\ 4 \amp 1.32472 \end{array} \end{equation*}

We stop after 4 iterations and conclude that $x = 1.32472$ approximates a root of $f\text{.}$
$f(x)=x^{4}-4x^{3} + 10$ between $x=1$ and $x=2\text{.}$
Solution

The function $f(x)$ is a polynomial (and hence continuous everywhere), and so we can apply IVT. We compute

\begin{equation*} f(1) = 1-4+10 > 0, \ \text{ and } \ f(2) = 16 - 4(8) + 10 \lt 0\text{.} \end{equation*}

Therefore, by the Intermediate Value Theorem, $f$ must have at least one root on the interval $(1,2)\text{.}$ Let $x_0 = 1.5$ and

\begin{equation*} x_{k+1} = x_k - \frac{x_k^4-4x_k^3+10}{4x_k^3-12x_k^2}, k = 0,1,2,3,..\text{.} \end{equation*}

Then:

\begin{equation*} \begin{array}{ll} k \amp x_k \\ \hline 0 \amp 1.5 \\ 1 \amp 1.61574 \\ 2 \amp 1.6118 \\ 3 \amp 1.61179 \\ 4 \amp 1.61179 \end{array} \end{equation*}

We conclude that $x=1.61179$ approximates a root of $f\text{.}$

Exercise 5.3.28.

Consider $f(x)=x^3-x^2+x-1\text{.}$

Using initial approximation $x_0=2\text{,}$ find $x_4\text{.}$
Answer
$x_4\approx 1.00022\ldots$
Solution

We are given $f(x) = x^3-x^2+x-1\text{.}$ Then the iterative formula required for Newton's Method is

\begin{equation*} x_{k+1} = x_k - \frac{x_k^3-x_k^2+x_k-1}{3x_k^2-2x_k+1}, k = 0,1,2,3,..\text{.} \end{equation*}

Taking $x_0 = 2\text{,}$ we find:

\begin{equation*} \begin{array}{ll} k \amp x_k \\ \hline 0 \amp 2 \\ 1 \amp 1.44444 \\ 2 \amp 1.13057 \\ 3 \amp 1.01498 \\ 4 \amp 1.00022 \end{array} \end{equation*}

Therefore, $x_4 = 1.00022\text{.}$
What is the exact value of the root of $f\text{?}$ How does this compare to our approximation $x_4$ in part (a)?
Answer
$x=1$ is the root of $f\text{.}$ Our approximation in part (a) was correct to 3 decimal places.
Solution

By inspection, we see that $x=1$ is the exact value of a root of $f\text{.}$ So the error in our approximation is

\begin{equation*} |x_4 - 1| = |1.00022 - 1| = 0.00022\text{.} \end{equation*}

So our approximation $x_4$ is correct to 3 decimal places.
What would happen if we chose $x_0=0$ as our initial approximation?
Answer
$x_1=1\text{.}$ The root is found in one iteration of Newton's Method.
Solution

If we take $x_0 = 0\text{,}$ we find that

\begin{equation*} x_1 = 0 - \frac{-1}{1} = 1\text{,} \end{equation*}

and so the method converges to the exact root after only one iteration.

Exercise 5.3.29.

Consider $f(x)=\sin x\text{.}$ What happens when we choose $x_0=\pi/2\text{?}$ Explain.

$\cos (\pi/2)=0\text{,}$ so $x_1$ is undefined.

Let $f(x) = \sin x\text{.}$ Then we know that the roots of $f(x)$ are $n\pi\text{,}$ for all integers $n\text{.}$

If we tried to apply Newton's Method to $f$ taking $x_0=\frac{\pi}{2}\text{,}$ we can see graphically that this will not converge to a root:

We can also argue algrabraically: the iterative formula required is

\begin{equation*} x_{k+1} = x_{k} - \frac{\sin x_k}{\cos x_k}, k = 0,1,2,3,..\text{.} \end{equation*}

and so if $x_0 = \frac{\pi}{2}\text{,}$ then $\cos(x_0) = 0\text{.}$ Hence, $x_1$ would be undefined.

Exercise 5.3.30.

Suppose a company purchases $6000 worth of new equipment which will be used for the next 3 years. The investment is expected to yield returns of $2,000 at the end of the first year, $3500 at the end of the second year and $1000 at the end of the third year. What is the internal rate of return on this investment?

$r \approx 4.5 \%$ per year.

To find the internal rate of return after 3 years, we solve the following for $r\text{:}$

\begin{equation*} I(r) = 6000(1+r)^3-2000(1+r)^2 - 3500(1+r) - 1000 = 0\text{.} \end{equation*}

Let $x=1+r\text{.}$ Then the required formula for Newton's method (in terms of $x$) is

\begin{equation*} x_{n+1}=x_n - \frac{6000x_n^3-2000x_n^2-3500x_n-1000}{12,000x_n^2-4000x_n-3500} \ \ \ n=0,1,2,..\text{.} \end{equation*}

Before we can start our iterations, we need to find an accetable starting point. Notice that when $x=1\text{,}$ $I \lt 0$ and when $x=2\text{,}$ $I > 0\text{.}$ Therefore, by the Intermediate Value Theorem, we conclude that there must be a root in this interval and we can take, for example, $x_0=1.8\text{.}$ As before, we calculate,

\begin{equation*} \begin{array}{lll} n \amp x_n \amp \rvert x_n-x_{n-1}\rvert \\ \hline 0 \amp 1.8 \amp - \\ 1 \amp 1.354557... \amp 0.445...\\ 2 \amp 1.126354... \amp 0.228...\\ 3 \amp 1.0525702... \amp 0.073...\\ 4 \amp 1.0446339... \amp 0.0079...\\ 5 \amp 1.0445449... \amp 0.0000089\\ 6 \amp 1.0445449... \amp 0.00000001\\ \end{array} \end{equation*}

The internal rate of return on the investment is therefore approximately $1.0445-1 = 0.0445\text{,}$ or $r \approx 4.5 \%$ per year.

Exercise 5.3.31.

Executives of a certain company are contemplating the purchase of $100,000 worth of equipment, which would be in use over the next 4 years. The investment is expected to yield returns of $20,000 at the end of the first year, $30,000 at the end of the second year, $45,000 at the end of the third year and $15,000 at the end of the fourth year. What is the internal rate of return on this investment?

$r \approx -6.67 \%$ per year.

To find the internal rate of return after 3 years, we solve the following for $r\text{:}$

\begin{equation*} I(r) = 100,000(1+r)^4-20,000(1+r)^3 - 30,00(1+r)^2-45,000(1+r) - 15,000 = 0\text{.} \end{equation*}

Let $x=1+r\text{.}$ Then the required formula for Newton's method (in terms of $x$) is

\begin{equation*} x_{n+1}=x_n - \frac{100,000x_n^4-20,000x_n^3 - 30,00x_n^2-45,000x_n - 15,000}{400,000x_n^3-60,000x_n^2-60,000x_n-45,000} \ \ \ n=0,1,2,..\text{.} \end{equation*}

Before we can start our iterations, we need to find an accetable starting point. Notice that when $x=0\text{,}$ $I \lt 0$ and when $x=1\text{,}$ $I > 0\text{.}$ Therefore, by the Intermediate Value Theorem, we conclude that there must be a root in this interval and we can take, for example, $x_0=0.8\text{.}$ As before, we calculate,

\begin{equation*} \begin{array}{ll} n \amp x_n \\ \hline 0 \amp 0.8 \\ 1 \amp 0.990395 \\ 2 \amp 0.939197 \\ 3 \amp 0.933363 \\ 4 \amp 0.933291 \\ 5 \amp 0.933291\\ \end{array} \end{equation*}

The internal rate of return on the investment is therefore approximately $0.933291-1 = -0.066709\text{,}$ or $r \approx -6.67 \%$ per year.

Exercise 5.3.32.

A first time home buyer borrows $250,000 from a bank to finance the purchase of a house. Interest is computed at the end of each month at a rate of 12$r$ per year on the unpaid balance. If the home buyer repays the loan in equal monthly instalments of $2226.10 over the next 10 years, what is the rate of interest charged by the bank?

1.33% per year

Exercise 5.3.33.

A down payment of 10% is made towards a purchase of $10,000. Financing for this purchase is available with monthly payments of $255.50 over 4 years. What is the rate of interest charged?

16.1% per year

Exercise 5.3.34.

Suppose the demand equation for a certain product is given by

\begin{equation*} p=d(q)=\frac{100}{0.02q^{2}+2} \ \ \ \ 1 \leq q \leq 20 \end{equation*}

where $p$ is the unit price in dollars and $q$ is the quantity demanded in units of hundreds. The corresponding supply equation is given by

\begin{equation*} p=s(q)=0.1q+30 \end{equation*}

dollars. What is the equilibrium point?

$(789.88,30.79)\text{.}$

Exercise 5.3.35.

A certain company determines that the demand equation for a product is

\begin{equation*} p=-3q+900 \ \ \ \ 0 \leq q \leq 300 \end{equation*}

where $p$ is the unit price in dollars and $q$ is the quantity demanded in units of a thousand. If the weekly total cost function associated with the production of this product is

\begin{equation*} C(q)=q^2+2q+700\text{,} \end{equation*}

what is the break-even level(s) of operation for the company?

224 units and 1 unit