6.3 Related rates

$t=$ 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 ${\frac{dA}{dt}|}_{r=50}$, 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 $t$: $$\begin{array}{}\text{(1)}& \frac{dr}{dt}=3\end{array}$$ Also, we know that the area and radius of a circle are related through $$\begin{array}{}\text{(2)}& A=\pi r^2\end{array}$$ Because $A$ and $r$ are both functions of $t$, we can differentiate both sides of (2) with respect to $t$ and obtain: $$\begin{array}{}\text{(3)}& \frac{dA}{dt}=2\pi r\frac{dr}{dt}\end{array}$$ 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: $$\frac{dA}{dt}={6\pi r|}_{r=50}=300\pi \frac{{\text{m}}^2}{\text{s}}\approx 942.48\frac{{\text{m}}^2}{\text{s}}.$$

$t=$ 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 $${\frac{ds}{dt}|}_{h=1200}\text{given}{\frac{dh}{dt}|}_{h=1200}=400\text{ m/s}.$$

According to the Pythagorean Theorem, we can write:

$$\begin{array}{}\text{(i)}& (900{)}^2+h^2=s^2\end{array}$$ 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 $$\begin{array}{}\text{(ii)}& \frac{ds}{dt}=\frac{h}{s}\frac{dh}{dt}.\end{array}$$ 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

$$\frac{ds}{dt}=\frac{1200}{1500}\cdot 400=320\frac{\text{m}}{\text{s}}$$ The distance between the rocket and the camera is increasing at a rate of 320 m/s.

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

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 $${\frac{dy}{dt}|}_{x=120}\text{given}{\frac{dx}{dt}|}_{x=120}=60\text{ cm/s}$$

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

$$\begin{array}{}\text{(i)}& x^2+y^2=(150{)}^2\end{array}$$ 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 $$\begin{array}{}\text{(ii)}& \frac{dy}{dx}=\frac{-x}{y}\frac{dx}{dt}.\end{array}$$ When $x=120$ cm, Equation (ii) gives $y=90$ cm, and substituting these numbers and $dx/dt=60$ cm/s in (ii) we obtain $${\frac{dy}{dx}|}_{x=120}=-\frac{120}{90}(60)=-80\text{ cm/s}$$ 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.

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

We need to find $${\frac{dh}{dt}|}_{h=4}\text{given}\frac{dV}{dt}=-3{\text{ cm}}^3/\text{s}$$ 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 $$\begin{array}{}\text{(i)}& V=\frac{1}{3}\pi r^{2}h\end{array}$$ 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 $$\mathrm{△}OAB\sim \mathrm{△}OCD$$ $$\Rightarrow \frac{r}{h}=\frac{AB}{OA}=\frac{5}{15}$$ so $$r=\frac{1}{3}h$$ and $$\begin{array}{}\text{(ii)}& V=\frac{1}{3}\pi r^{2}h=\frac{1}{3}\pi {\left(\frac{h}{3}\right)}^{2}h=\frac{\pi }{27}{h}^3\end{array}$$ 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 $$\begin{array}{rl}\frac{dh}{dt}& =\frac{9}{\pi \times 4^2{\text{ cm}}^2}(-3\frac{{\text{cm}}^3}{\text{s}})\\ & =\frac{-27}{16\pi }\frac{\text{cm}}{\text{s}}\\ & \approx -0.537\text{ cm/s}\end{array}$$ 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.

$dx/dt=$ 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 $$\frac{ds}{dt}\text{given }\frac{dx}{dt}=-4\text{ ft/s}$$ Here $dx/dt$ is negative because $x$ is decreasing.

To find an relationship between $x$ and $s$, we notice that the following triangles are similar (because they have the same angle) $$\mathrm{△}AB{B}^{\prime }\sim \mathrm{△}{A}^{\prime }{B}^{\prime }C$$ Therefore, we can write $$\frac{s}{x}=\frac{B^{\prime }A}{C{A}^{\prime }}=\frac{5}{10}$$ 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 $$\frac{dx}{dt}+\frac{ds}{dt}=-4+(-2)=-6\text{ ft/s.}$$

From Figure 12, we can see that $$\begin{array}{}\text{(i)}& \tan \theta =\frac{x}{10},\end{array}$$ 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 $\mathrm{△}C{A}^{\prime }{B}^{\prime }$ $$\cos \theta =\frac{C{A}^{\prime }}{C{B}^{\prime }}=\frac{C{A}^{\prime }}{\sqrt{(A^{\prime }{B}^{\prime }{)}^2+(C{A}^{\prime }{)}^2}}=\frac{10}{\sqrt{{10}^2+x^2}}$$ [Here we have applied the Pythagorean Theorem]

So $$\begin{gathered} \begin{array}{}\text{(ii)}& \frac{d\theta }{dt}=\frac{1}{10}{\left(\frac{10}{\sqrt{{10}^2+x^2}}\right)}^2\frac{dx}{dt} \\ =\frac{10}{{10}^2+x^2}\frac{dx}{dt}\end{array} \end{gathered}$$ We can plug the given information into (ii) and evaluate $d\theta /dx$: $${\frac{d\theta }{dt}|}_{x=8}=\frac{10}{100+64}\times (-4)=-\frac{10}{41}\text{ rad/s}$$ 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: $$\begin{array}{rl}{\frac{d\theta }{dt}|}_{x=8}& =-\frac{10}{41}\frac{\cancel{\text{ rad}}}{\text{s}}\times \frac{180\text{ degree}}{\pi \cancel{\text{ rad}}}\\ & =-\frac{1800}{41}\frac{\text{degree}}{\text{s}}\\ & \approx {43.9}^{\circ }/\text{s}.\end{array}$$ 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 $x$.

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.

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 $$\begin{array}{}\text{(i)}& s^2=x^2+y^2\end{array}$$ (or $s=\sqrt{x^2+y^2}$).
Differentiating each side of (i) with respect to time $t$, we have $$\begin{array}{}\text{(ii)}& 2s\frac{ds}{dt}=2x\frac{dx}{dt}+2y\frac{dy}{dt}.\end{array}$$ 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 $dx/dt$ $$2\times 1\times (-70)=2\times (-0.8)\frac{dx}{dt}+2\times 0.6\times (-30)$$ $$\Rightarrow \frac{dx}{dt}=65\text{ mph}$$ The car’s speed at the moment of measurement is 65 mph. $dx/dt$ is positive which shows $x$ is increasing.