6.3 Related Rates

In this section, we shall look at ‘related rates.’ In this class of problems, two or more related quantities are changing. The rate of change of one quantity is given, and we seek to determine the rate of change of the other related quantities. To start, we need to define the average and instantaneous rates of change; these are not new concepts but new names for old, familiar ones.

Definition 1. If $y=f(x)$, then the average rate of change of $y$ with respect to $x$ over the interval $\left[x_{0},x_{1}\right]$ is the slope of the secant line, $m_{\text{secant}}$, passing through $(x_{0},f(x_{0}))$ and $\left(x_{1},f\left(x_{1}\right)\right)$ (which are on the graph of $f$) (Figure 1(a)).

If $y=f(x)$, then the instantaneous rate of change of $y$ with respect to $x$ at the point $x_{0}$ is the slope of the tangent line, $m_{\text{tangent}}$, to the graph of $f$ at the point $x_{0}$ (Figure 1(b)).

If a rate is positive, it is often called a rate of increase.
If a rate is negative, its absolute value is often called a rate of decrease.
If we simply say “rate,” we mean instantaneous rate.


(a) The slope of the secant, $\dfrac{f(x_{1})-f(x_{0})}{x_{1}-x_{0}}$, is the average rate of change of $y$ with respect to $x$ over the interval $[x_{0},x_{1}]$.	(b) The slope of the tangent line at the point $x_{0}$, $m_{\text{tangent}}=f^\prime(x_{0})$, is the instantaneous rate of change of $y$ at $x_{0}$.

Figure 1

Example 1

A sunken oil tanker is leaking oil in an expanding slick the radius of whose circle is increasing at the rate of $3\text{ m/s}$. What is the rate at which the area of the circle expands when the radius of the oil slick is $50\ \text{m}$? (That is, how fast is the area of the circle expanding when the radius of the oil slick is $50\ \text{m}$?)

Figure 2: Assume the oil slick is spreading in a circular pattern.

Solution 1

Let

$t=$ time in seconds since the tanker sank
$r=$ the radius of the circular oil slick in meters after $t$ seconds
and
$A=$ the area of the slick in square meters after $t$ seconds

In each moment the instantaneous rate of change in the radius of the circle is $\frac{dr}{dt}$ and the instantaneous rate of change in its area is $\frac{dA}{dt}$. We want to find $\left.\frac{dA}{dt}\right|_{r=50}$, that is, the instantaneous rate of change of area of the oil slick when its radius is $50$ m.

Because the radius of the oil slick is increasing at the rate of 3 m/sec, at each moment $t$: \[\frac{dr}{dt}=3\tag{1}\] Also, we know that the area and radius of a circle are related through \[A=\pi r^{2}\tag{2 }\] Because $A$ and $r$ are both functions of $t$, we can differentiate both sides of (2) with respect to $t$ and obtain: \[\frac{dA}{dt}=2\pi r\frac{dr}{dt}\tag{3}\] Upon substitution of (1) in (3), we obtain: \[\frac{dA}{dt}=2\pi r(3)=6\pi r\] Because $r=60$, then the instantaneous rate of change in the area of the slick equals: \[\frac{dA}{dt}=\left.6\pi r\right|_{r=50}=300\pi\ \frac{\text{m}^{2}}{\text{s}}\approx942.48\ \frac{\text{m}^{2}}{\text{s}}.\]

As René Descartes says in Discourse on Method, “Each problem that I solved became a rule, which served afterward to solve other problems.” Similarly, we can explain the rule to solve problems of this kind as follows.

Strategy for Solving Related Rates Problems.

Assign appropriate letters to the variables and constants that remind you of their actual meaning. For example, represent the volume by $V$, the length by $L,x$, or $y$, temperature by $T$, and time by $t$.
Write down the given information and what you are asked to find.
Find an equation that relates the quantity whose rate you do not know to those other quantities whose rates are already known. You may need to combine two or more equations to get a single equation. Drawing a simple picture sometimes facilitates this step.
Differentiate both sides of the equation with respect to the independent variable (which is usually time $t$) thereby you can express the rate you want in terms of the known rates and variables.
Substitute the given values into the resulting equation and solve for the unknown rate using the known values.

Always substitute the given information for the varying quantities AFTER differentiation. For example, in the above example, if we had substituted $r=50$ before differentiation, we would have gotten $A=50^{2}\pi\Rightarrow dA/dt=0$ (because $50\pi^{2}$ is a constant and the derivative of a number is zero).

We are filming a rocket launch with a camera installed 900 m away from the launching pad (Figure 3). When the rocket lifts vertically, obviously the elevation angle increases. The next two examples deal with this problem, assuming that the camera’s focal point changes automatically in such a way that we can get a clear view of the rocket at all times.

Example 2

In Figure 3, if the rocket is ascending at $400$ m/s when it is $1200$ m above the launching pad, how fast the distance between the rocket and the camera change?

Solution 2

Let

$t=$ time in seconds since the launch
$h=$ height of the rocket in meters after $t$ seconds
$s=$ distance between the rocket and the camera in meters after $t$ seconds

In each moment:

$dh/dt=$ instantaneous rate of change of the height of the rocket
$ds/dt=$ instantaneous rate of change of the distance between the rocket and the camera.

We want to find \[\left.\frac{ds}{dt}\right|_{h=1200}\quad\text{given}\quad\left.\frac{dh}{dt}\right|_{h=1200}=400\text{ m/s}.\]

Figure 4

According to the Pythagorean Theorem, we can write:

\[(900)^{2}+h^{2}=s^{2}\tag{i}\] Since $h$ and $s$ are functions of $t$ (time), we can differentiate both sides of the equation with respect to $t$:

\[0+2h\frac{dh}{dt}=2s\frac{ds}{dt}\] [where we have used the chain rule]

or \[\frac{ds}{dt}=\frac{h}{s}\frac{dh}{dt}.\tag{ii}\] When $h=1200$ m, it follows from (i) that $s=1500$ m. Since the problem states that the change of the height of the rocket is 400 m/s when $h=1200$ m, we can put these numbers in (ii) and obtain

\[\frac{ds}{dt}=\frac{1200}{1500}\cdot400=320\ \frac{\text{m}}{\text{s}}\] The distance between the rocket and the camera is increasing at a rate of 320 m/s.

Example 3

Let us assume that the rocket in the previous example is speeding away from earth at 400 m/s when its altitude is $h=1200$ m. In this situation, what should the instantaneous rate of change of the camera’s elevation angle in this moment be to keep the rocket in view?

Solution 3

Let
$t=$ time in seconds since the launch
$h=$ height of the rocket in meters after $t$ seconds
$\theta=$ camera’s elevation angle in radians after $t$ seconds

We want to find \[\left.\frac{d\theta}{dt}\right|_{h=1200}\quad\text{given}\quad\left.\frac{dh}{dt}\right|_{h=1200}=400\text{ m/s}\] From Figure 5, we see that \[\tan\theta=\frac{h}{900}.\tag{i}\] Here $\theta$ and $h$ are functions of $t$, and differentiation with respect to $t$ yields \[(\sec^{2}\theta)\frac{d\theta}{dt}=\frac{1}{900}\frac{dh}{dt}\tag{ii }\] [Here we have applied the chain rule] or \[\frac{d\theta}{dt}=\frac{1}{900(\sec^{2}\theta)}\frac{dh}{dt}\tag{iii}\] When $h=1200$, by the Pythagorean Theorem $y=1500$ m and \[\sec\theta=\frac{1}{\cos\theta}=\frac{1500}{900}=\frac{5}{3}\quad\text{when} \quad h=1200\text{ m}\] The problem states that $\left.\frac{dh}{dt}\right|_{h=1200}=400\text{ m/s}$ and now we can plug these numbers in (iii) and get \[\left.\frac{d\theta}{dt}\right|_{h=1200}=\frac{1}{900\left(\frac{5}{3}\right)^{2}}400=\frac{4}{25}=0.16\text{ rad/s.}\] That is, when $h=1200$ m and the speed of the rocket is 400 m/s, The elevation angle must be increased at a rate of 0.16 rad/s to keep the rocket in view.

Figure 5

Example 4

A ladder $150$ cm long leaning against the wall slips and its foot moves away from the wall at 60 cm/s. When the foot’s base is 120 cm from the wall, how fast does the top of the ladder approach the ground?

Figure 6

Solution 4

Let
$t=$ time in seconds since the ladder slipped
$x=$ distance of the ladder’s base from the wall in meters after $t$ seconds
$y=$ distance of the ladder’s top from the ground in meters after $t$ seconds

We can see from Figure 7 that
$dx/dt=$ speed at which the foot of the ladder moves away from the wall
$dy/dt=$ speed at which the top of the ladder moves towards the ground

We want to find \[\left.\frac{dy}{dt}\right|_{x=120}\quad\text{given} \quad\left.\frac{dx}{dt}\right|_{x=120}=60\text{ cm/s}\]

Figure 7

The relationship between $x$ and $y$ is given by the Pythagorean Theorem

\[x^{2}+y^{2}=(150)^{2}\tag{i}\] Here $x$ and $y$ are functions of $t$ (they vary with time), and differentiation with respect to $t$ yields

\[2x\frac{dx}{dt}+2y\frac{dy}{dt}=0\] [Here we applied the chain rule]

or \[\frac{dy}{dx}=\frac{-x}{y}\frac{dx}{dt}.\tag{ii}\] When $x=120$ cm, Equation (ii) gives $y=90$ cm, and substituting these numbers and $dx/dt=60$ cm/s in (ii) we obtain \[\left.\frac{dy}{dx}\right|_{x=120}=-\frac{120}{90}(60)=-80\text{ cm/s}\] The minus sign tells that $y$ is decreasing, which, in physical terms, means that the top of the ladder is moving closer to the ground.

Example 5

We want to filter some kind of liquid through a conical funnel. The height of the fennel is 15 cm and the radius of its base is 5 cm. Suppose the liquid is filtering out at the constant rate of $\text{3 cm}{}^{3}/\text{s}$. What is the rate of of the decreasing height of the liquid in the funnel when the height of the liquid in the funner is 4 cm?

Figure 8

Solution 5

Let
$t=$ time in minutes since the first observation
$V=$ volume of the liquid in the funnel in cubic centimeters ($\text{cm}^{3}$) after $t$ minutes
$h=$ height of the liquid in the funnel in centimeters after $t$ minutes
$r=$ radius of the top of the liquid in the funnel in centimeters after $t$ minutes

$dV/dt=$ rate at which the volume is changing
$dh/dt=$ rate at which the height of the liquid is changing

We need to find \[\left.\frac{dh}{dt}\right|_{h=4}\quad\text{given}\quad\frac{dV}{dt}=-3\text{ cm}^{3}/\text{s}\] Here we use the minus sign for $dV/dt$ because the volume is decreasing.

The volume of a cone $V$ and its height $h$ and the radius of its base $r$ are related by \[V=\frac{1}{3}\pi r^{2}h\tag{i }\] The radius of the top of the liquid also varies with time. So if we differentiate both sides of (i) with respect to $t$, in addition to $dh/dt$, the right side of the equation involves $dr/dt$ about which no information is given. Therefore, we need to express $r$ in terms of $h$ before differentiation. This can be done if we notice that the two triangles are similar \[\triangle OAB\sim\triangle OCD\] \[\Rightarrow\frac{r}{h}=\frac{AB}{OA}=\frac{5}{15}\] so \[r=\frac{1}{3}h\] and \[V=\frac{1}{3}\pi r^{2}h=\frac{1}{3}\pi\left(\frac{h}{3}\right)^{2}h=\frac{\pi}{27}h^{3}\tag{ii}\] Now we can easily differentiate each side with respect to $t$: \[\frac{dV}{dt}=\frac{\pi}{27}(3h^{2})\frac{dh}{dt}\] [Here again we used the chain rule]

so \[\frac{dh}{dt}=\frac{9}{\pi h^{2}}\frac{dV}{dt}\] Substituting $h=4$ cm and $dV/dt=-3\text{ cm}^{3}/\text{s}$, we have \[\begin{aligned} \frac{dh}{dt} & =\frac{9}{\pi\times4^{2}\text{ cm}^{2}}\left(-3\frac{\text{cm}^{3}}{\text{s}}\right)\\ & =\frac{-27}{16\pi}\frac{\text{cm}}{\text{s}}\\ & \approx-0.537\text{ cm/s}\end{aligned}\] The minus sign indicates the height of the liquid is decreasing. Therefore, when $h=4$ cm, the height of the liquid is approximately decreasing at the rate of 0.537 $\text{cm}/\text{s}$. The minus sign of $dh/dt$ indicates that $h$ decreases as $t$ increases.

Example 6

A 5-ft person is approaching a 15-ft lamp post at the constant rate of 4 ft/s (Figure 9).
(a) How fast is the length of the person’s shadow decreasing?
(b) How fast is the tip of the person’s shadow moving toward the lamp post?

Figure 9

Solution 6

(a) Let
$x=$ distance of the person from the foot of the lamp post
$s=$ the length of his shadow

$dx/dt=$ rate at which the person is approaching the lamp post
$ds/dt=$ rate at which the length of the person’s shadow decreases

We need to find \[\frac{ds}{dt}\quad\text{given }\quad\frac{dx}{dt}=-4\text{ ft/s}\] Here $dx/dt$ is negative because $x$ is decreasing.

Figure 10

To find an relationship between $x$ and $s$, we notice that the following triangles are similar (because they have the same angle) \[\triangle ABB’\sim\triangle A’B’C\] Therefore, we can write \[\frac{s}{x}=\frac{B’A}{CA’}=\frac{5}{10}\] so \[s=\frac{1}{2}x\] Differentiating each side with respect to $t$, we have \[\frac{ds}{dt}=\frac{1}{2}\frac{dx}{dt}.\] Given $dx/dt=-4$ ft/s, therefore, \[\frac{ds}{dt}=\frac{1}{2}(-4)=-2\text{ ft/s.}\] Here the negative sign shows that the length of the person’s shadow is decreasing.
In fact $|ds/dt|$ is the rate at which $B$ is approaching $A$.
Note that, regardless of where the person is in any moment, $ds/dt$ is always a constant.

(b) From Figure 10, we can see that the position of the tip of the shadow is \[OA+AB=x+s\] Therefore, the rate at which the tip of the shadow $B$ is moving is \[\frac{dx}{dt}+\frac{ds}{dt}=-4+(-2)=-6\text{ ft/s.}\]

Example 7

In the previous example, let $\theta$ be the angle between the ray $CB$ and the lamp post (see Figure 11). What is the rate at which $\theta$ changing when the person is 8 ft away from the lamp post and with the same speed as in the previous example is moving towards it?

Figure 11

Solution 7

Here we need to find \[\left.\frac{d\theta}{dt}\right|_{x=8}\quad\text{given}\quad\frac{dx}{dt}=-4\text{ ft/s.}\]

Figure 12

From Figure 12, we can see that \[\tan\theta=\frac{x}{10},\tag{i }\] where $\theta$ and $x$ are functions of time $t$. Differentiating both sides of (i) with respect to $t$, we obtain \[(1+\tan^{2}\theta)\frac{d\theta}{dt}=\frac{1}{10}\frac{dx}{dt}\] or \[\frac{1}{\cos^{2}\theta}\frac{d\theta}{dt}=\frac{1}{10}\frac{dx}{dt}\] so \[\frac{d\theta}{dt}=\frac{1}{10}\cos^{2}\theta\frac{dx}{dt}.\] Now we need to evaluate $\cos\theta$ in terms of the given data. In $\triangle CA’B’$ \[\cos\theta=\frac{CA’}{CB’}=\frac{CA’}{\sqrt{(A’B’)^{2}+(CA’)^{2}}}=\frac{10}{\sqrt{10^{2}+x^{2}}}\] [Here we have applied the Pythagorean Theorem]

So \[\begin{equation*} \frac{d\theta}{dt} =\frac{1}{10}\left(\frac{10}{\sqrt{10^{2}+x^{2}}}\right)^{2}\frac{dx}{dt}\\ =\frac{10}{10^{2}+x^{2}}\frac{dx}{dt}\tag{ii }\end{equation*}\] We can plug the given information into (ii) and evaluate $d\theta/dx$: \[\left.\frac{d\theta}{dt}\right|_{x=8}=\frac{10}{100+64}\times(-4)=-\frac{10}{41}\text{ rad/s}\] The answer is in units of radians per second, because when we use $\frac{d}{d\theta}\tan\theta=\sec^{2}\theta$, we assume that $\theta$ is in radians. Obviously we can express the answer in units of degrees per second: \[\begin{aligned} \left.\frac{d\theta}{dt}\right|_{x=8} & =-\frac{10}{41}\frac{\cancel{\text{ rad}}}{\text{s}}\times\frac{180\text{ degree}}{\pi\cancel{\text{ rad}}}\\ & =-\frac{1800}{41}\frac{\text{degree}}{\text{s}}\\ & \approx43.9^{\circ}/\text{s}.\end{aligned}\] Equation (ii) clearly shows that $d\theta/dt$ depends on $x$, the distance between the person and the lamp post, while as we saw in the previous example, $ds/dt$, the rate at which the length of the shadow is decreasing, is independent of $x$.

Example 8

A police cruiser is approaching a right-angled intersection from the north and a speeding car is moving toward the intersection from the west (Figure 13). The police determine with radar that when the distance between them and the car is 1 mile, the distance is decreasing at 70 mph (mile per hour). At the instant of measurement, if the cruiser is moving at 30 mph, what is the speed of the car?

Figure 13

Solution 8

Let
$y=$ position of the cruiser at time $t$
$x=$ position of the car at time $t$
$s=$ distance between the cruiser and the car at time $t$

We want to find \[\frac{dx}{dt}\] when $y=0.6$ mile, $s=$ 1 mile, $dy/dt=-30$ mph and $ds/dt=-70$ mph.
Here $dy/dt$ and $ds/dt$ are negative because $y$ and $s$ are decreasing.

Drawing a simple picture can help us understand the problem better.

Figure 14

The distance between the cruiser and the car, $s$, the position of the car, $x$, and the position of the cruiser, $y$, are related by \[\begin{equation*}s^{2}=x^{2}+y^{2}\tag{i}\end{equation*}\] (or $s=\sqrt{x^{2}+y^{2}}$).
Differentiating each side of (i) with respect to time $t$, we have \[\begin{equation*}2s\frac{ds}{dt}=2x\frac{dx}{dt}+2y\frac{dy}{dt}.\tag{ii }\end{equation*}\] Given $s=1$ mile and $y=0.6$ miles, by the Pythagorean Theorem (i), $x^{2}=1-0.36=0.64$. We say $x=-0.8$ miles, because the car is located on the negative $x$-axis.

Now we can substitute the given data in (ii) and solve for $dx/dt$ \[2\times1\times(-70)=2\times(-0.8)\frac{dx}{dt}+2\times0.6\times(-30)\] \[\Rightarrow\frac{dx}{dt}=65\text{ mph}\] The car’s speed at the moment of measurement is 65 mph. $dx/dt$ is positive which shows $x$ is increasing.