Related Rates

Section 5.2 Related Rates

When defining the derivative $f^{\prime}\left( x\right)\text{,}$ we define it to be exactly the rate of change of $f\left( x\right)$ with respect to $x\text{.}$ Consequently, any question about rates of change can be rephrased as a question about derivatives. When we calculate derivatives, we are calculating rates of change. Results and answers we obtain for derivatives translate directly into results and answers about rates of change. Let us look at some examples where more than one variable is involved, and where our job is to analyze and exploit relations between the rates of change of these variables. As an aside, this class of problems is known as related rates problems. The mathematical step of relating the rates of change turns out to be largely an exercise in differentiation using the Chain Rule or implicit differentiation. This explains why some textbooks place this section shortly after the sections on the Chain Rule and implicit differentiation.

Let's say we are interested in the relationship between the rate of change of a mortgage rate and the rate of change of the number of houses sold over time. If $x$ represents the mortgage rate and $y$ the number of houses sold at any time $t\text{,}$ then $x$ and $y$ are each functions of this third variable $t\text{.}$ Suppose furthermore that the mortgage rate $x$ is related to the number of houses sold $y\text{,}$ i.e. we also have an equation relating $x$ to $y\text{:}$

\begin{equation*} f(x)=g(y)\text{.} \end{equation*}

Then we can differentiate both sides of this equation implicitly with respect to $t\text{,}$ and get

\begin{equation*} f'(x) \frac{dx}{dt}=g'(y) \frac{dy}{dt}\text{.} \end{equation*}

In other words, we now have an equation that relates $dx/dt$ to $dy/dt\text{.}$ In terms of our problem, this means that the rate of change of the mortgage rate and the rate of change of the number of houses sold are related as a function of time. And so, as $dx/dt$ changes determines how $dy/dt$ changes, i.e. the rate of change of mortgage w.r.t. time controls the rate of change of houses at that instant of time.

Example 5.6. Speed at which a Coordinate is Changing.

Suppose an object is moving along a path described by $\ds y=x^2\text{,}$ that is, it is moving on a parabolic path. At a particular time, say $t=5\text{,}$ the $x$-coordinate is 6 and we measure the speed at which the $x$-coordinate of the object is changing and find that $dx/dt = 3\text{.}$

At the same time, how fast is the $y$-coordinate changing?

Solution

Using the Chain Rule,

\begin{equation*} \ds \frac{dy}{dt} = 2x \frac{dx}{dt}\text{.} \end{equation*}

At $t=5$ we know that $x=6$ and $dx/dt=3\text{,}$ so $dy/dt = (2)(6)(3) = 36\text{.}$

In many cases, particularly interesting ones, $x$ and $y$ will be related in some other way, for example $x=f(y)\text{,}$ or $F(x,y)=k\text{,}$ or perhaps $F(x,y)=G(x,y)\text{,}$ where $F(x,y)$ and $G(x,y)$ are expressions involving both variables. In all cases, you can solve the related rates problem by taking the derivative of both sides, plugging in all the known values (namely, $x\text{,}$ $y\text{,}$ and $\ds dx/dt$), and then solving for $\ds dy/dt\text{.}$

To summarize, here are the steps in doing a related rates problem.

Steps for Solving Related Rates Problems.

Read the problem at least twice.
Sketch and label a diagram of the problem if applicable.
Identify the independent variable (often, but not always, time).
Unless already introduced, use a let statement to introduce dependent variables.
State the known and unknown rate(s) and value(s) using your variable name(s).
Find an equation relating the known and unknown variables.
Differentiate the equation implicitly w.r.t the independent variable.
Use substitution of known values to solve the new equation.
Critically evaluate if your answer makes sense.

Interactive Demonstration. A plane is flying directly away from you at 500 mph at an altitude of 3 miles. Use the slider to time to investigate how the rate at which the distance $s$ between you and the plane changes.

Example 5.7. Receding Airplanes.

A plane is flying directly away from you at 500 mph at an altitude of 3 miles, as shown in the interactive demonstration above. How fast is the plane's distance from you increasing at the moment when the plane is flying over a point on the ground 4 miles from you?

Solution

To see what's going on, we first draw a schematic representation of the situation, as shown below.

Because the plane is in level flight directly away from you, the rate at which $x$ changes is the speed of the plane, $dx/dt=500$ mph. The distance between you and the plane is $s\text{;}$ it is $ds/dt$ that we wish to know. By the Pythagorean Theorem we know that $\ds x^2+9=s^2\text{.}$ Taking the derivative with respect to the independent variable $t\text{,}$ we obtain

\begin{equation*} 2x \frac{dx}{dt} = 2s\frac{ds}{dt} s\text{.} \end{equation*}

We are interested in the time at which $x=4\text{;}$ at this time we know that $\ds 4^2+9=s^2\text{,}$ so $s=5\text{.}$ Putting together all the information we get

\begin{equation*} 2(4)(500)=2(5)\frac{ds}{dt}\text{.} \end{equation*}

Thus, $\ds \frac{ds}{dt}=400$ mph.

Example 5.8. Rate of Change of Housing Starts.

It is estimated that the number of housing starts, $N(t)$ (in units of a million), over the next 5 years is related to the mortgage rate $r(t)$ (percent per year) by the equation

\begin{equation*} 8N^2+r=36\text{.} \end{equation*}

What is the rate of change of the number of housing starts with respect to time when the mortage rate is 4% per year and is increasing at the rate of 0.25% per year?

Solution

We want to find $dN/dt$ when

\begin{equation*} r = 4 \ \ \ \text{ and } \ \ \ \frac{dr}{dt} = 0.25\text{.} \end{equation*}

We are given the relationship

\begin{equation*} 8N^2 + r = 36\text{,} \end{equation*}

so we can find the relationship between the rate of change of $N$ and the rate of change of $r$ by differentiating this equation implicitly with respect to time. This gives

\begin{equation*} \begin{split} \frac{d}{dt}\left(8N^2 + r\right) \amp = \frac{d}{dt} \left(36\right) \\ 2(8)N\frac{dN}{dt} + \frac{dr}{dt} \amp = 0 \\ 16N\frac{dN}{dt} \amp = -\frac{dr}{dt}\\ \frac{dN}{dt} \amp = \frac{-1}{16N}\frac{dr}{dt} \end{split} \end{equation*}

At the instant in time we are considering, the number of housing starts $N$ is unknown. However, we know that $N$ satisfies

\begin{equation*} 8N^2 + r = 36\text{,} \end{equation*}

so when $r=4\text{,}$ we must have

\begin{equation*} \begin{split} 8N^2 + 4 \amp = 36 \\ N^2 \amp = 4 \\ N \amp =2, \end{split} \end{equation*}

where we have rejected the negative root. Therefore, when $r=4$ and $dr/dt = 0.25\text{,}$ the rate of change of housing starts is

\begin{equation*} \frac{dN}{dt} = \frac{-1}{16(2)}(0.25) = -0.0078125\text{,} \end{equation*}

that is, the number of housing starts is decreasing by approximately 7,813 units.

Example 5.9. Supply-Demand.

It is found that a certain manufacturer produces $q$ thousand units per week when the unit price is $$p\text{.}$ Suppose the relationship between $q$ and $p$ is

\begin{equation*} q^{2} - 3qp + p^{2} = 5\text{.} \end{equation*}

What is the rate of change of the supply when the quantity produced is 4000 units and the unit price is $11, increasing at a rate of $0.10 per week?

Solution

We differentiate the supply equation on both sides with respect to $t\text{,}$ obtaining

\begin{equation*} \begin{split} \frac{d}{dt}\left(q^{2}\right) - \frac{d}{dt}\left(3qp\right) + \frac{d}{dt}\left(p^{2}\right) \amp = \frac{d}{dt}\left(5\right) \\[1ex] 2q\frac{dq}{dt} - 3\left(p\frac{dq}{dt} + q\frac{dp}{dt}\right) + 2p\frac{dp}{dt} \amp = 0, \end{split} \end{equation*}

where we used the Product Rule on the second term. So when $p=11\text{,}$ $dp/dt = 0.1$ and $q=4000\text{,}$ we have

\begin{equation*} \begin{split} 2(4)\frac{dq}{dt} - 3\left((11)\frac{dq}{dt} + 4(0.1)\right) + 2(11)(0.1) \amp = 0 \\[1ex] 8\frac{dq}{dt} - 33\frac{dq}{dt} - 1.2 + 2.2 \amp = 0 \\[1ex] 25 \frac{dq}{dt} \amp = 1 \\[1ex] \frac{dq}{dt} \amp = 0.04. \end{split} \end{equation*}

Thus, at the instant of time under consideration, the supply is increasing at the rate of $(0.04)(1000)\text{,}$ or 40, units per week.

Interactive Demonstration. A spherical balloon is being inflated at a rate of 7 cm${}^3$/sec. Use the slider to advance time to investigate how the rate at which the radius $r$ increases depends on the rate the ballon is being inflated.

Example 5.10. Spherical Balloon.

You are inflating a spherical balloon at the rate of 7 cm${}^3$/sec. How fast is its radius increasing when the radius is 4 cm?

Solution

Here the independent variable is time $t$ and the dependent variables are the radius $r$ and the volume $V\text{.}$ We know $dV/dt\text{,}$ and we want $dr/dt\text{.}$ The two variables are related by the equation $\ds V=4\pi r^3/3\text{.}$ Taking the derivative with respect to the independent variable $t\text{,}$ we get

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

We now substitute the values we know at the instant in question:

\begin{equation*} \ds 7=4\pi 4^2\frac{dr}{dt}\text{,} \end{equation*}

so $\ds dr/dt=7/(64\pi)$ cm/sec.

Example 5.11. Conical Container.

Water is poured into a conical container at the rate of 10 cm${}^3$/sec. The cone points directly down, and it has a height of 30 cm and a base radius of 10 cm; see below. How fast is the water level rising when the water is 4 cm deep (at its deepest point)?

Solution

The water forms a conical shape within the big cone; its height and base radius and volume are all increasing as water is poured into the container. This means that we actually have three things varying with time: the water level $h$ (the height of the cone of water), the radius $r$ of the circular top surface of water (the base radius of the cone of water), and the volume of water $V\text{.}$ The volume of a cone is given by

\begin{equation*} \ds V=\pi r^2h/3\text{.} \end{equation*}

Again, the independent variable is time $t\text{.}$ We know $dV/dt\text{,}$ and we want $dh/dt\text{.}$ At first something seems to be wrong: we have a third variable, $r\text{,}$ whose rate we don't know.

However, the dimensions of the cone of water must have the same proportions as those of the container. That is, because of similar triangles,

\begin{equation*} \frac{r}{h}=\frac{10}{30}\text{,} \end{equation*}

so $r=h/3\text{.}$ Now we can eliminate $r$ from the problem entirely:

\begin{equation*} \ds V=\pi(h/3)^2h/3=\pi h^3/27\text{.} \end{equation*}

We take the derivative of both sides and plug in $h=4$ and $dV/dt=10\text{,}$ obtaining

\begin{equation*} \ds 10=3\pi(\frac{4^2}{27})\frac{dh}{dt}\text{.} \end{equation*}

Thus, $dh/dt=90/(16\pi)$ cm/sec.

Example 5.12. Swing Set.

A swing consists of a board at the end of a 10 ft long rope. Think of the board as a point $P$ at the end of the rope, and let $Q$ be the point of attachment at the other end. Suppose that the swing is directly below $Q$ at time $t=0\text{,}$ and is being pushed by someone who walks at 6 ft/sec from left to right. Find

how fast the swing is rising after 1s;
the angular speed of the rope in deg/sec after 1s.

Solution

We start out by asking: What is the geometric quantity whose rate of change we know, and what is the geometric quantity whose rate of change we're being asked about? Again, the independent variable is time $t\text{.}$ Note that the person pushing the swing is moving horizontally at a rate we know. In other words, the horizontal coordinate of $P$ is increasing at 6 ft/sec. In the $x$-$y$-plane let us make the convenient choice of putting the origin at the location of $P$ at time $t=0\text{,}$ i.e., a distance 10 directly below the point of attachment as shown below. Then the rate we know is $dx/dt\text{,}$ and in part (a) the rate we want is $dy/dt$ (the rate at which $P$ is rising). In part (b) the rate we want is $\ds d\theta/dt\text{,}$ where $\theta$ stands for the angle in radians through which the swing has swung from the vertical. Actually, since we want our answer in deg/sec, at the end we must convert $d\theta/dt$ from rad/sec by multiplying by $180/\pi\text{.}$

From the diagram we see that we have a right triangle whose legs are $x$ and $10-y\text{,}$ and whose hypotenuse is 10. Hence

\begin{equation*} \ds x^2+(10-y)^2=100\text{.} \end{equation*}

Taking the derivative of both sides with respect to $t$ we obtain

\begin{equation*} 2x\frac{dx}{dt}+2(10-y)(0-\frac{dy}{dt})=0\text{.} \end{equation*}

We now look at what we know after 1 second, namely $x=6$ (because $x$ started at 0 and has been increasing at the rate of 6 ft/sec for 1 sec), thus $y=2$ (because we get $10-y=8$ from the Pythagorean Theorem applied to the triangle with hypotenuse 10 and leg 6), and $\ds dx/dt=6\text{.}$ Putting in these values gives us

\begin{equation*} 2\cdot 6\cdot 6-2\cdot 8\frac{dy}{dt}=0\text{,} \end{equation*}

from which we can easily solve for $\ds dy/dt\text{:}$ $\ds dy/dt=4.5$ ft/sec.
Here our two variables are $x$ and $\theta\text{,}$ so we want to use the same right triangle as in part (a), but this time relate $\theta$ to $x\text{.}$ Since the hypotenuse is constant (equal to 10), the best way to do this is to use the sine: $\sin\theta=x/10\text{.}$ Taking derivatives we obtain

\begin{equation*} \ds (\cos\theta)\frac{d\theta}{dt}=0.1\frac{dx}{dt}\text{.} \end{equation*}

At the instant in question ($t=1$ sec), when we have a right triangle with sides 6鈥�8鈥�10, $\ds \cos\theta=8/10$ and $\ds dx/dt=6\text{.}$ Thus $(8/10)d\theta/dt=6/10\text{,}$ i.e., $\ds d\theta/dt=6/8=3/4$ rad/sec, or approximately $43$ deg/sec.

We have seen that sometimes there are apparently more than two variables that change with time, but in reality there are just two, as the others can be expressed in terms of just two. However sometimes there really are several variables that change with time; as long as you know the rates of change of all but one of them you can find the rate of change of the remaining one. As in the case when there are just two variables, take the derivative of both sides of the equation relating all of the variables, and then substitute all of the known values and solve for the unknown rate.

Example 5.13. Distance Changing Rate.

A road running north to south crosses a road going east to west at the point $P\text{.}$ Car A is driving north along the first road, and car B is driving east along the second road. At a particular time car A is $10$ kilometres to the north of $P$ and traveling at 80 km/hr, while car B is 15 kilometres to the east of $P$ and traveling at 100 km/hr. How fast is the distance between the two cars changing accurate to one decimal place?

Solution

Let $a(t)$ be the distance of car A north of $P$ at time $t\text{,}$ and $b(t)$ the distance of car B east of $P$ at time $t\text{,}$ and let $c(t)$ be the distance from car A to car B at time $t$ as shown below.

By the Pythagorean Theorem, $\ds c(t)^2=a(t)^2+b(t)^2\text{.}$ Taking derivatives we get

\begin{equation*} \ds 2c\frac{dc}{dt}=2a(t)\frac{da}{dt}+2b\frac{db}{dt}\text{,} \end{equation*}

\begin{equation*} \frac{dc}{dt}=\frac{a\frac{da}{dt}+b\frac{db}{dt}}{c}=\frac{a\frac{da}{dt}+b\frac{db}{dt}}{\sqrt{a^2+b^2}}\text{.} \end{equation*}

Substituting known values we get:

\begin{equation*} \frac{dc}{dt}=\frac{10\cdot 80+15\cdot 100}{\sqrt{10^2+15^2}}=\frac{460}{\sqrt{13}} \approx 127.6 \hbox{ km/hr} \end{equation*}

at the time of interest.

Notice how this problem differs from 贰虫补尘辫濒别听5.7. In both cases we started with the Pythagorean Theorem and took derivatives on both sides. However, in 贰虫补尘辫濒别听5.7 one of the sides was a constant (the altitude of the plane), and so the derivative of the square of that side of the triangle was simply zero. In this Example, on the other hand, all three sides of the right triangle are variables, even though we are interested in a specific value of each side of the triangle (namely, when the sides have lengths 10 and 15). Make sure that you understand at the start of the problem what are the variables and what are the constants.

Exercises for Section聽5.2.

Exercise 5.2.1.

Suppose the quantity demanded weekly of a product is related to its unit price by the equation

\begin{equation*} p+q^{2} = 144 \end{equation*}

where $p$ is measured in dollars and $q$ is measured in units of a thousand. What is the rate of change of the quantity demanded when $q=9\text{,}$ $p=63\text{,}$ and the unit price is increasing at the rate of $2/week?

Answer

111 units/week

Solution

Both $p$ and $q$ are functions of time. Therefore, we differentiate implicitly with respect to time:

\begin{equation*} \begin{split} \diff{}{t} \left(p+q^2\right) \amp = \diff{}{t} (144) \\ \diff{p}{t} + 2 q \diff{q}{t} \amp = 0 \\ \diff{q}{t} \amp = -\frac{1}{2q}\diff{p}{t} \end{split} \end{equation*}

Therefore, when $q=9$ and $\diff{p}{t}=2\text{,}$

\begin{equation*} \diff{q}{t}= -\frac{1}{2(9)} (2) = -\frac{1}{9}\text{.} \end{equation*}

That is, the quantity demanded is decreasing at a rate of about 111 units per week.

Exercise 5.2.2.

The demand equation for a certain product is

\begin{equation*} 100q^{2}+9p^{2}=3600 \end{equation*}

where $q$ is the number (in thousands) of units demanded each week when the unit price is $$p\text{.}$ What is the rate of change of the quantity demanded when the unit price is $14 and the selling price is dropping at the rate of $.15/unit/week?

Answer

44 units/week

Solution

We are given $p=14$ and $\diff{p}{t}=-0.15\text{.}$ To find $q$ when $p=14\text{,}$ we can use the relation:

\begin{equation*} 100q^2+9p^2=3600 \implies q = \pm \frac{3\sqrt{51}}{5}\text{,} \end{equation*}

and so $q \approx 4.29\text{.}$

Let $q=q(t)$ and $p=p(t)\text{.}$ Now to find $\diff{q}{t}\text{,}$ we differentiate the equation implicitly:

\begin{equation*} \begin{gathered} \diff{}{t} \left(100q^2+9p^2\right) = \diff{}{t} 3600\\ 200q \diff{q}{t} + 18p\diff{p}{t} = 0\\ \diff{q}{t} = -\frac{18}{200q} \diff{p}{t} \end{gathered} \end{equation*}

Therefore, when $p=14$ and $q=4.29\text{,}$ we have

\begin{equation*} \diff{q}{t} = -\frac{18}{200(4.29)} (-0.15) \approx 0.044\text{.} \end{equation*}

Hence, the quantity demanded is increasing by approximately 44 units per week.

Exercise 5.2.3.

Suppose the price $p$ (in dollars/unit) of a product is related to the weekly supply $q$ (in units of a thousand) by the equation

\begin{equation*} 625p^{2} - q^{2} = 100\text{.} \end{equation*}

If 25,000 units are produced and the supply is falling at the rate of 1000 units/week, at what rate is the price changing?

Answer

-$.037/unit

Solution

We are given that the unit price $p$ and the quantity demanded $q$ (in units of a thousand) of a product are related by

\begin{equation*} 625p^2 - q^2 = 100\text{.} \end{equation*}

Let $q=q(t)$ and $p=p(t)\text{.}$ When $q=25\text{,}$ $\diff{q}{t} = -1$ thousand units per week. We first find how the rate of change of the unit price is related to the rate of change of the demand. Using the relationship above,

\begin{equation*} \begin{split} \diff{}{t} \left(625p^2-q^2\right) \amp = \diff{}{t}(100) 2(625)p\diff{p}{t} \amp = 2q\diff{q}{t} \diff{p}{t} \amp = \frac{q(t)}{625p(t)} \diff{q}{t} \end{split} \end{equation*}

At the point in time when $q=25\text{,}$ $p$ is given by

\begin{equation*} 625 p^2-(25)^2=100 \implies p=\frac{\sqrt{29}}{5}\text{.} \end{equation*}

Therefore,

\begin{equation*} \diff{p}{t} = \frac{25}{625}\frac{5}{\sqrt{29}} \cdot (-1) = -\frac{1}{5\sqrt{29}}\text{.} \end{equation*}

So the unit price is decreasing by approximately $0.037 per unit at this point in time.

Exercise 5.2.4.

The demand function for a certain product is

\begin{equation*} p=-0.01q^{2}-0.1q+6 \end{equation*}

where $p$ is the unit price in dollars and $q$ is the quantity demanded each week (in units of a thousand). Compute the elasticity of demand and determine whether the demand is inelastic, unitary, or elastic when $q=10\text{.}$

Answer

$E(p) = 4/3\text{;}$ elastic

Solution

We differentiate implicitly:

\begin{equation*} \begin{gathered} \diff{}{p} (p) = \diff{}{p} \left(-0.01q^2-0.1q+6\right)\\ 1 = -0.01(2q)\diff{q}{p}-0.1\diff{q}{p}+0 \\ 1 = \left(-0.02q -0.1\right) \diff{q}{p} \\ -\frac{1}{0.1 + 0.02q} = \diff{q}{p} \end{gathered} \end{equation*}

Therefore,

\begin{equation*} E(p,q) = \frac{p}{q} \frac{1}{0.1+0.02q}\text{.} \end{equation*}

When $q=10\text{,}$ the corresponding unit price is

\begin{equation*} p = -0.01(100)-0.1(10)+6 = -2+6 = 4\text{.} \end{equation*}

Hence, the elasticity of demand is

\begin{equation*} E = \frac{4}{10(0.1+0.02(10))} = \frac{4}{3}\text{,} \end{equation*}

which means that the demand is elastic.

Exercise 5.2.5.

Air is being pumped into a spherical balloon at a constant rate of 3 cm3/s. How fast is the radius of the balloon increasing when the radius reaches 5cm? Answer

0.0095 cm/s

Solution

The volume of the balloon is related to its radius by

\begin{equation*} V = \frac{4}{3}\pi r^3\text{.} \end{equation*}

Let $V=V(t)$ and $r=r(t)\text{.}$ Then

\begin{equation*} \diff{V}{t} = 4\pi r^2 \diff{r}{t}\text{.} \end{equation*}

When $r=5$ cm, we know that $\diff{V}{t} = 3$ cm$^3/s\text{,}$ hence

\begin{equation*} \diff{r}{t} = \frac{3}{4\pi 5^2} \approx 0.0095\text{.} \end{equation*}

That is, the radius of the balloon is increasing by about 0.0095 cm/s.

Exercise 5.2.6.

A cylindrical tank standing upright (with one circular base on the ground) has radius 20 cm. How fast does the water level in the tank drop when the water is being drained at 25 cm${}^3$/sec?

Answer

$1/(16\pi)$ cm/s

Solution

If water is being drained at a rate of 25 cm$^3$/s, this means that

\begin{equation*} \diff{V}{t} = -25\text{.} \end{equation*}

So the volume of the water left in the tank at some time $t$ is

\begin{equation*} V = \pi (20)^2 h\text{,} \end{equation*}

where $h$ is the water level, and $r=20$ is the radius of the tank. We now differentiate implicitly:

\begin{equation*} \diff{V}{t} = 400 \pi \diff{h}{t}\text{.} \end{equation*}

So, when $\diff{V}{t} = -25\text{,}$

\begin{equation*} \diff{h}{t} = \frac{-25}{\pi (400)} (-25) = -\frac{1}{16\pi}\text{.} \end{equation*}

Thus, the water level in the tank is decreasing at a rate of about 0.02 cm/s.

Exercise 5.2.7.

A cylindrical tank standing upright (with one circular base on the ground) has radius 1 metre. How fast does the water level in the tank drop when the water is being drained at 3 litres per second?

Answer

$3/(1000\pi)$ metres/second

Solution

The water is being drained at a rate of 3 L/s. Since 1L = 0.001 m$^3\text{,}$ we have that

\begin{equation*} \diff{V}{t} = -\frac{3}{1000} \end{equation*}

m$^3$/s. So the volume of the water left in the tank at some time $t$ is

\begin{equation*} V = \pi (1)^2 h\text{,} \end{equation*}

where $h$ is the water level in metres, and $r=1$ metre is the radius of the tank. We now differentiate implicitly:

\begin{equation*} \diff{V}{t} = \pi \diff{h}{t}\text{.} \end{equation*}

So, when $\diff{V}{t} = -\frac{3}{1000}\text{,}$

\begin{equation*} \diff{h}{t} = \frac{-3}{1000\pi}\text{.} \end{equation*}

Thus, the water level in the tank is decreasing at a rate of about 0.00096 m/s.

Exercise 5.2.8.

A ladder 13 metres long rests on horizontal ground and leans against a vertical wall. The foot of the ladder is pulled away from the wall at the rate of 0.6 m/sec. How fast is the top sliding down the wall when the foot of the ladder is 5 m from the wall?

Answer

$1/4$ m/s

Solution

We make the following diagram:

Where $h$ and $l$ are both functions of time. Then at this instance, $h=\sqrt{13^2-5^2} = 12$ m and the quantity needed is $\diff{h}{t}\text{.}$

Since

\begin{equation*} 13^2 = \sqrt{h^2 + l^2}\text{,} \end{equation*}

we differentiate implicitly:

\begin{equation*} \begin{split} \diff{}{t} (13^2) \amp = \diff{}{t} \sqrt{h^2+l^2}\\ 0 \amp = \frac{1}{\sqrt{h^2+l^2}} \left(2h \diff{h}{t} + 2l \diff{l}{t}\right)\\ \diff{h}{t} \amp = -\frac{l}{h} \diff{l}{t} \end{split} \end{equation*}

So when $h=12$ m, $l=5$ m, and $\diff{l}{t} = 0.6$ m/s,

\begin{equation*} \diff{h}{t} = -\frac{1}{4} \end{equation*}

m/s. That is, the top of the ladder is moving downwards at a rate of about 0.25 m/s.

Exercise 5.2.9.

A ladder 13 metres long rests on horizontal ground and leans against a vertical wall. The top of the ladder is being pulled up the wall at $0.1$ metres per second. How fast is the foot of the ladder approaching the wall when the foot of the ladder is 5 m from the wall?

Answer

$-6/25$ m/s

Solution

We make the following diagram:

Now, we are given that $\diff{h}{t} = 0.1$ m/s, $h=\sqrt{13^2-5^2} = 12$ m and the quantity needed is $\diff{l}{t}\text{.}$

Since

\begin{equation*} 13^2 = \sqrt{h^2 + l^2}\text{,} \end{equation*}

we differentiate implicitly:

\begin{equation*} \begin{split} \diff{}{t} (13^2) \amp = \diff{}{t} \sqrt{h^2+l^2}\\ 0 \amp = \frac{1}{\sqrt{h^2+l^2}} \left(2h \diff{h}{t} + 2l \diff{l}{t}\right)\\ \diff{l}{t} \amp = -\frac{h}{l} \diff{h}{t} \end{split} \end{equation*}

So when $h=12$ m, $l=5$ m, and $\diff{h}{t} = 0.1$ m/s,

\begin{equation*} \diff{l}{t} = -\frac{6}{25} \end{equation*}

m/s. That is, the foot of the ladder is approaching the wall at a rate of about 0.24 m/s.

Exercise 5.2.10.

A rotating beacon is located 2 miles out in the water. Let $A$ be the point on the shore that is closest to the beacon. As the beacon rotates at 10 rev/min, the beam of light sweeps down the shore once each time it revolves. Assume that the shore is straight. How fast is the point where the beam hits the shore moving at an instant when the beam is lighting up a point 2 miles along the shore from the point $A\text{?}$

Answer

$80\pi$ mi/min

Solution

Let $B$ the the location of the beacon and $A$ be the reference point on the shoreline. Then we can make the following (top-view) diagram:

If the beacon makes 10 revolutions per minute, then

\begin{equation*} \diff{\alpha}{t} = 10 (2\pi) = 20\pi \ \text{ rad/min. } \end{equation*}

Our known rate is $\diff{\alpha}{t}$ and our unknown rate is $\diff{x}{t}\text{.}$ Therefore, we relate $\alpha$ to $x\text{:}$

\begin{equation*} \tan \alpha = \frac{x}{2} \end{equation*}

Next, we differentiate this relation:

\begin{equation*} \begin{split} \diff{}{t} \tan \alpha \amp = \diff{}{t} \frac{x}{2} \\ \sec^2 \alpha \diff{\alpha}{t} \amp = \frac{1}{2}\diff{x}{t}\\ \diff{x}{t} \amp = 2 \sec^2 \alpha \diff{\alpha}{t} \end{split} \end{equation*}

When $x = 2\text{,}$ we have that

\begin{equation*} \sec\alpha = \frac{1}{\cos \alpha} = \frac{\sqrt{8}}{2} = \sqrt{2}\text{.} \end{equation*}

Thus, when $x=2$ miles,

\begin{equation*} \diff{x}{t} = 2 \cdot(\sqrt{2})^2 \cdot (20 \pi) = 80\pi \ \text{ miles/min. } \end{equation*}

Exercise 5.2.11.

A baseball diamond is a square 90 ft on a side. A player runs from first base to second base at 15 ft/sec. At what rate is the player's distance from third base decreasing when she is half way from first to second base?

Answer

$\ds 3\sqrt5$ ft/s

Solution

Let $x$ be the position of the player (where $x=0$ corresponds to first base), and let $d$ be the distance from the player to third base. Then we can make the following diagram:

Our known rate is $\diff{x}{t}$ and our unknown rate is $\diff{d}{t}\text{.}$ From our diagram, we see that we can relate $d$ to $x$ by the equation

\begin{equation*} d = \sqrt{(90-x)^2 + 90^2}\text{.} \end{equation*}

Now differentiate:

\begin{equation*} \diff{d}{t} = \frac{1}{2 \sqrt{(90-x)^2 + 90 ^2}} \left(-2(90-x)\diff{x}{t}\right) = \frac{x-90}{\sqrt{(90-x)^2 + 90 ^2}} \diff{x}{t}\text{.} \end{equation*}

Therefore, when $x=45$ ft and $\diff{x}{t} = 15$ ft/s, we calculate

\begin{equation*} \diff{d}{t} = \frac{-45}{\sqrt{45^2 + 90 ^2}} (15) = -3 \sqrt{5}\text{.} \end{equation*}

Thus, the distance between the player and third base is decreasing at a rate of about 6.7 ft/s.

Exercise 5.2.12.

Sand is poured onto a surface at 15 cm${}^3$/sec, forming a conical pile whose base diameter is always equal to its altitude. How fast is the altitude of the pile increasing when the pile is 3 cm high?

Answer

$20/(3\pi)$ cm/s

Solution

Let $h$ be the altitude of the cone. Then we have the following diagram:

Let $V$ be the volume of sand in the pile. Then the known rate is $\diff{V}{t}$ and the unknown rate is $\diff{h}{t}\text{.}$ We first relate $V$ to $h\text{:}$

\begin{equation*} V = \pi r^2 \frac{h}{3} = \pi \left(\frac{h}{2}\right)^2 \frac{h}{3} = \pi \frac{h^3}{12}\text{.} \end{equation*}

Now differentiate:

\begin{equation*} \diff{V}{t} = \frac{\pi}{12} 3h^2 \diff{h}{t} \implies \diff{h}{t} = \frac{4}{\pi h^2} \diff{V}{t}\text{.} \end{equation*}

Therefore, when $h=3$ cm and $\diff{V}{t} = 15$ cm$^3$/s, the altitude of the pile is increasing at a rate of

\begin{equation*} \diff{h}{t} = \frac{4}{9 \pi} (15) = \frac{20}{3\pi} \ \text{ cm/s }\text{.} \end{equation*}

Exercise 5.2.13.

A boat is pulled in to a dock by a rope with one end attached to the front of the boat and the other end passing through a ring attached to the dock at a point 5 ft higher than the front of the boat. The rope is being pulled through the ring at the rate of 0.6 ft/sec. How fast is the boat approaching the dock when 13 ft of rope are out?

Answer

$13/20$ ft/s

Solution

Let $l$ be the length of rope in the system, and let $x$ be the distance from the boat to the dock. Then we can make the following diagram:

The known rate is $\diff{l}{t}$ and the unknown rate is $\diff{x}{t}\text{.}$ We relate $x$ and $l\text{:}$

\begin{equation*} l^2 = 5^2 + x^2 = 25+ x^2\text{.} \end{equation*}

Differentiating both sides gives

\begin{equation*} 2l \diff{l}{t} = 2x \diff{x}{t} \implies \diff{x}{t} = \frac{l}{x} \diff{l}{t}\text{.} \end{equation*}

When $l=13$ ft, we must have $x=\sqrt{13^2-5^2} = 12$ ft. Furthermore, we know that at this point in time $\diff{l}{t} = 0.6$ ft/s. Thus,

\begin{equation*} \diff{x}{t} = \frac{13}{12} 0.6 = \frac{13}{20}\text{.} \end{equation*}

Therefore, the boat is approaching the dock at approximately 0.65 ft/s.

Exercise 5.2.14.

A balloon is at a height of 50 metres, and is rising at the constant rate of 5 m/sec. A bicyclist passes beneath it, traveling in a straight line at the constant speed of 10 m/sec. How fast is the distance between the bicyclist and the balloon increasing 2 seconds later?

Answer

$\ds 5\sqrt{10}/2$ m/s

Solution

Let $h$ denote the height of the balloon, $x$ denote the position of the cyclist and $d$ denote the distance between them. Assuming that the balloon has no horizontal movement, we can make the following diagram:

We can relate the three quantities by:

\begin{equation*} d^2 = h^2 + x^2\text{.} \end{equation*}

Differentiating, we have

\begin{equation*} \diff{d}{t} = \frac{1}{d(t)} \left(h(t) \diff{h}{t} + x(t) \diff{x}{t}\right)\text{.} \end{equation*}

At $t=2\text{,}$ this means that

\begin{equation*} \diff{d}{t} = \frac{1}{d(2)} \left(h(2) \diff{h}{t} + x(2) \diff{x}{t}\right)\text{.} \end{equation*}

At $t=0\text{,}$ the bicyclist is directly beneath the balloon. Therefore, after 2 seconds travelling at a constant speed of $10$ m/s, the bicyclist will be $x=20$ m from this reference position. At this time, the balloon will be at a height of $h=50+5(2) = 60$ m. Hence,

\begin{equation*} \diff{d}{t} = \frac{1}{\sqrt{20^2+60^2}} \left(60 (5) + 20 (10)\right) = 5\sqrt{\frac{5}{2}}\text{.} \end{equation*}

That is, 2 seconds after the cyclist passes under the balloon, the distance between them is increasing at a rate of approximately 7.9 m/s.

Exercise 5.2.15.

A pyramid-shaped vat has square cross-section and stands on its tip. The dimensions at the top are 2 m 脳 2 m, and the depth is 5 m. If water is flowing into the vat at 3 m${}^3$/min, how fast is the water level rising when the depth of water (at the deepest point) is 4 m? Note: the volume of any 鈥渃onical鈥� shape (including pyramids) is $(1/3)(\hbox{height})(\hbox{area of base})\text{.}$

Answer

$75/64$ m/min

Solution

We show the following cross-section of the vat:

The volume of water in the tank is given by

\begin{equation*} V = \frac{1}{3} \times h \times (b^2)\text{.} \end{equation*}

Since the known rate is $\diff{V}{t}$ and the unknown rate is $\diff{h}{t}\text{,}$ we want to express volume as a function of $h$ only. Using the above diagram and special triangles, we see that

\begin{equation*} \frac{h}{b} = \frac{5}{2} \implies b = \frac{2h}{5}\text{.} \end{equation*}

Hence,

\begin{equation*} V = \frac{4}{75} h^3\text{.} \end{equation*}

Differentiating this equation, we find

\begin{equation*} \diff{V}{t}= \frac{12}{75} h^2 \diff{h}{t} \implies \diff{h}{t} = \frac{75}{12 h^2} \diff{V}{t}\text{.} \end{equation*}

Therefore, when the height of the water is $4$ m, the water level is rising at a rate of

\begin{equation*} \diff{h}{t} = \frac{75}{192} (3) = \frac{75}{64} \ \text{ m/min }\text{.} \end{equation*}

Exercise 5.2.16.

A woman 5 ft tall walks at the rate of 3.5 ft/sec away from a streetlight that is 12 ft above the ground. At what rate is the tip of her shadow moving? At what rate is her shadow lengthening?

Answer

tip: 6 ft/s, length: $5/2$ ft/s

Solution

We make the following diagram to illustrate the problem:

The known rate is therefore $\diff{x}{t}\text{,}$ and the unknwon rate at which the shadow is lengthening is $\diff{a}{t}\text{.}$ By special triangles, we have

\begin{equation*} \frac{12}{5} = \frac{a}{x+a} \implies a = \frac{5}{7}x\text{.} \end{equation*}

Differentiating, we find

\begin{equation*} 17 \diff{a}{t} = 5\diff{x}{t}\text{.} \end{equation*}

Hence, when the woman is walking at a rate of $3.5$ ft/s, the shadow is lengthening at a rate of

\begin{equation*} \diff{a}{t} = \frac{5}{7}\times \frac{7}{2} = \frac{5}{2} \ \text{ ft/s }\text{.} \end{equation*}

Let $L = x+a\text{.}$ Then the tip of the shadow is moving at the rate $\diff{L}{t}\text{.}$ Hence,

\begin{equation*} \diff{L}{t} = \diff{x}{t} + \diff{a}{t} = \frac{5}{2} + \frac{7}{2} = 6 \ \text{ ft/s }\text{.} \end{equation*}

Exercise 5.2.17.

A man 1.8 metres tall walks at the rate of 1 meter per second toward a streetlight that is 4 metres above the ground. At what rate is the tip of his shadow moving? At what rate is his shadow shortening?

Answer

tip: $20/11$ m/s, length: $9/11$ m/s

Solution

We make a sketch of the scenario:

From the above sketch, we can make the following diagram:

Where $\diff{x}{t}=-1$m/s in this coordinate system. By similar triangles, we can relate $a(t)$ to $x(t)\text{:}$

\begin{equation*} \frac{4}{1.8} = \frac{a+x}{a} \implies a = \frac{9}{11} x\text{.} \end{equation*}

Differentiating both sides with respect to time, we see that

\begin{equation*} \diff{a}{t} = \frac{9}{11}\diff{x}{t} = -\frac{9}{11}\text{.} \end{equation*}

Hence, the length of the shadow is shortening at a rate of $\frac{9}{11}$ m/s. Now, to calculate the rate at which the tip of the shadow is moving, we let $L = x+a\text{.}$ From our work above, this means that

\begin{equation*} \frac{4}{1.8} = \frac{L}{a} \implies 4a = 1.8 L\text{.} \end{equation*}

Differentiating implicitly, we find

\begin{equation*} 4\diff{a}{t} = 1.8 \diff{L}{t}\text{.} \end{equation*}

Using our answer above, we have

\begin{equation*} \diff{L}{t} = \frac{4}{1.8} \cdot \left(-\frac{9}{11}\right) = -\frac{20}{11}\text{.} \end{equation*}

Therefore, the tip of the shadow is moving at a rate of $\frac{20}{11}$ m/s in the direction towards the light.

Exercise 5.2.18.

A police helicopter is flying at 150 mph at a constant altitude of 0.5 mile above a straight road. The pilot uses radar to determine that an oncoming car is at a distance of exactly 1 mile from the helicopter, and that this distance is decreasing at 190 mph. Find the speed of the car.

Answer

$\ds 380/\sqrt3-150\approx 69.4$ mph

Solution

Let $x_H$ be the position of the helicopter, $x_C$ be the position of the car, and $d$ be the distance between them. Then we can make the following diagram:

Therefore, the known rates are $\diff{x_H}{t}$ and $\diff{d}{t}\text{,}$ while the unknown rate is $\diff{x_C}{t}\text{.}$ We relate the quantites in question by

\begin{equation*} d^2 = 0.5^2 + l^2\text{,} \end{equation*}

and so differentiating, we find

\begin{equation*} 2d \diff{d}{t} = 2l \diff{l}{t} \implies \diff{l}{t} = \frac{d}{l} \diff{d}{t}\text{.} \end{equation*}

Since $\diff{l}{t} = \diff{x_H}{t} + \diff{x_C}{t}\text{,}$

\begin{equation*} \diff{x_C}{t} = \frac{d}{l} \diff{d}{t} - \diff{x_H}{t}\text{.} \end{equation*}

When $d=1\text{,}$ we know that $l = \sqrt{1-0.5^2}\text{,}$ and so

\begin{equation*} \diff{x_C}{t} = \frac{1}{\sqrt{1-0.5^2}} (190) - 150 \approx 69 \ \text{ mph }\text{.} \end{equation*}

Exercise 5.2.19.

A police helicopter is flying at 200 kilometres per hour at a constant altitude of 1 km above a straight road. The pilot uses radar to determine that an oncoming car is at a distance of exactly 2 kilometres from the helicopter, and that this distance is decreasing at 250 kph. Find the speed of the car.

Answer

$\ds 500/\sqrt3-200\approx 88.7$ km/hr

Solution

Let $x_H$ be the position of the helicopter, $x_C$ be the position of the car, and $d$ be the distance between them. Then we can make the following diagram:

Therefore, the known rates are $\diff{x_H}{t}$ and $\diff{d}{t}\text{,}$ while the unknown rate is $\diff{x_C}{t}\text{.}$ We relate the quantites in question by

\begin{equation*} d^2 = 0.5^2 + l^2\text{,} \end{equation*}

and so differentiating, we find

\begin{equation*} 2d \diff{d}{t} = 2l \diff{l}{t} \implies \diff{l}{t} = \frac{d}{l} \diff{d}{t}\text{.} \end{equation*}

Since $\diff{l}{t} = \diff{x_H}{t} + \diff{x_C}{t}\text{,}$

\begin{equation*} \diff{x_C}{t} = \frac{d}{l} \diff{d}{t} - \diff{x_H}{t}\text{.} \end{equation*}

When $d=2\text{,}$ we know that $l = \sqrt{3}\text{,}$ and so

\begin{equation*} \diff{x_C}{t} = \frac{2}{\sqrt{3}} (250) - 200 \approx 89 \ \text{ kph }\text{.} \end{equation*}

Exercise 5.2.20.

A light shines from the top of a pole 20 m high. An object is dropped from the same height from a point 10 m away, so that its height at time $\ds t$ seconds is $\ds h(t)=20-9.8t^2/2\text{.}$ How fast is the object's shadow moving on the ground one second later?

Answer

$4000/49$ m/s

Solution

We first make a sketch of the scenario described.

Notice we have set up a reference frame with the $(0,0)$ position in the right-hand corner. The position of the shadow is given by $(10+a,0)\text{,}$ where $a \geq 0$ is the distance between the shadow and where the object will hit the ground, and the height of the object is given by $(10,h)\text{.}$ Let $a = a(t)$ and $h=h(t)=20-\frac{9.8}{2}t^2\text{.}$ We therefore must find the relationship between the rate the object is falling, $\diff{h}{t}\text{,}$ and the rate the shadow is moving on the ground, $\diff{a}{t}\text{.}$ This relationship will rely on the angle $\theta$ the object makes with the light source.

At some point in time $0 \lt t \lt \sqrt{20/4.9}$ (notice that $h(\sqrt{20/4.9}) = 0$), we have

Then by similar triangles,

\begin{equation*} \frac{10+a}{20} = \frac{10}{20-h(t)} = \frac{10}{4.9t^2} \cdot \end{equation*}

Differentiating both sides gives

\begin{equation*} \diff{a}{t} = \diff{}{t}\left(\frac{200}{4.9}t^{-2}\right) = -\frac{400}{4.9}t^{-3} \cdot \end{equation*}

$\diff{a}{t}$ gives the rate at which the distance between the shadow and the (stationary) point $x=10$ m on the ground is changing. Therefore, after 1s, the shadow is moving at a speed of $400/4.9 \approx 81.63$ m/s in the direction towards the falling object.

丁香园AV

Calculus Early Transcendentals: Differential & Multi-Variable Calculus for Social Sciences

Section 5.2 Related Rates

Example 5.6. Speed at which a Coordinate is Changing.

Steps for Solving Related Rates Problems.

Example 5.7. Receding Airplanes.

Example 5.8. Rate of Change of Housing Starts.

Example 5.9. Supply-Demand.

Example 5.10. Spherical Balloon.

Example 5.11. Conical Container.

Example 5.12. Swing Set.

Example 5.13. Distance Changing Rate.

Exercises for Section聽5.2.

Exercise 5.2.1.

Exercise 5.2.2.

Exercise 5.2.3.

Exercise 5.2.4.

Exercise 5.2.5.

Exercise 5.2.6.

Exercise 5.2.7.

Exercise 5.2.8.

Exercise 5.2.9.

Exercise 5.2.10.

Exercise 5.2.11.

Exercise 5.2.12.

Exercise 5.2.13.

Exercise 5.2.14.

Exercise 5.2.15.

Exercise 5.2.16.

Exercise 5.2.17.

Exercise 5.2.18.

Exercise 5.2.19.

Exercise 5.2.20.