3.8 Related Rates

A spherical balloon is being filled with air at the constant rate of $2\mathrm{c}{\mathrm{m}}^3/\sec$ (Figure 3.17). How fast is the radius increasing when the radius is $3\mathrm{c}\mathrm{m}\mathrm{?}$

An airplane is flying overhead at a constant elevation of $4000\mathrm{f}\mathrm{t}$. A man is viewing the plane from a position $3000\mathrm{f}\mathrm{t}$ from the base of a radio tower. The airplane is flying horizontally away from the man. If the plane is flying at the rate of $600\mathrm{f}\mathrm{t}/\sec$, at what rate is the distance between the man and the plane increasing when the plane passes over the radio tower?

As shown, $x$ denotes the distance between the man and the position on the ground directly below the airplane. The variable $s$ denotes the distance between the man and the plane. Note that both $x$ and $s$ are functions of time. We do not introduce a variable for the height of the plane because it remains at a constant elevation of $4000\mathrm{f}\mathrm{t}$. Since an object’s height above the ground is measured as the shortest distance between the object and the ground, the line segment of length 4000 ft is perpendicular to the line segment of length $x$ feet, creating a right triangle.

Step 2. Since $x$ denotes the horizontal distance between the man and the point on the ground below the plane, $dx/dt$ represents the speed of the plane. We are told the speed of the plane is 600 ft/sec. Therefore, ${\displaystyle \frac{dx}{dt}}=600$ ft/sec. Since we are asked to find the rate of change in the distance between the man and the plane when the plane is directly above the radio tower, we need to find $ds/dt$ when $x=3000\mathrm{f}\mathrm{t}$.

Step 3. From the figure, we can use the Pythagorean theorem to write an equation relating $x$ and $s:$

$${[x(t)]}^2+{4000}^2={[s(t)]}^2.$$

Step 4. Differentiating this equation with respect to time and using the fact that the derivative of a constant is zero, we arrive at the equation

$$x\frac{dx}{dt}=s\frac{ds}{dt}.$$

Step 5. Find the rate at which the distance between the man and the plane is increasing when the plane is directly over the radio tower. That is, find ${\displaystyle \frac{ds}{dt}}$ when $x=3000\mathrm{f}\mathrm{t}$. Since the speed of the plane is $600\mathrm{f}\mathrm{t}/\sec$, we know that ${\displaystyle \frac{dx}{dt}}=600\mathrm{f}\mathrm{t}/\sec$. We are not given an explicit value for $s;$ however, since we are trying to find ${\displaystyle \frac{ds}{dt}}$ when $x=3000\mathrm{f}\mathrm{t}$, we can use the Pythagorean theorem to determine the distance $s$ when $x=3000$ and the height is $4000\mathrm{f}\mathrm{t}$. Solving the equation

$${3000}^2+{4000}^2=s^2$$

for $s$, we have $s=5000\mathrm{f}\mathrm{t}$ at the time of interest. Using these values, we conclude that $ds/dt$ is a solution of the equation

$$(3000)(600)=(5000)\cdot \frac{ds}{dt}.$$

Therefore,

$$\frac{ds}{dt}=\frac{3000\cdot 600}{5000}=360\mathrm{f}\mathrm{t}/\sec .$$

Note: When solving related-rates problems, it is important not to substitute values for the variables too soon. For example, in step 3, we related the variable quantities $x(t)$ and $s(t)$ by the equation

$${[x(t)]}^2+{4000}^2={[s(t)]}^2.$$

Since the plane remains at a constant height, it is not necessary to introduce a variable for the height, and we are allowed to use the constant 4000 to denote that quantity. However, the other two quantities are changing. If we mistakenly substituted $x(t)=3000$ into the equation before differentiating, our equation would have been

$${3000}^2+{4000}^2={[s(t)]}^2.$$

After differentiating, our equation would become

$$0=s(t)\frac{ds}{dt}.$$

As a result, we would incorrectly conclude that ${\displaystyle \frac{ds}{dt}}=0$.

A rocket is launched so that it rises vertically. A camera is positioned $5000\mathrm{f}\mathrm{t}$ from the launch pad. When the rocket is $1000\mathrm{f}\mathrm{t}$ above the launch pad, its velocity is $600\mathrm{f}\mathrm{t}/\sec$. Find the necessary rate of change of the camera’s angle as a function of time so that it stays focused on the rocket.

Let $h$ denote the height of the rocket above the launch pad and $\theta$ be the angle between the camera lens and the ground.

Step 2. We are trying to find the rate of change in the angle of the camera with respect to time when the rocket is 1000 ft off the ground. That is, we need to find ${\displaystyle \frac{d\theta }{dt}}$ when $h=1000\mathrm{f}\mathrm{t}$. At that time, we know the velocity of the rocket is ${\displaystyle \frac{dh}{dt}}=600\mathrm{f}\mathrm{t}/\sec$.

Step 3. Now we need to find an equation relating the two quantities that are changing with respect to time: $h$ and $\theta$. How can we create such an equation? Using the fact that we have drawn a right triangle, it is natural to think about trigonometric functions. Recall that $\tan \theta$ is the ratio of the length of the opposite side of the triangle to the length of the adjacent side. Thus, we have

$$\tan \theta =\frac{h}{5000}.$$

This gives us the equation

$$h=5000\tan \theta .$$

Step 4. Differentiating this equation with respect to time $t$, we obtain

$$\frac{dh}{dt}=5000{\sec }^2\theta \frac{d\theta }{dt}.$$

Step 5. We want to find ${\displaystyle \frac{d\theta }{dt}}$ when $h=1000\mathrm{f}\mathrm{t}$. At this time, we know that ${\displaystyle \frac{dh}{dt}}=600\mathrm{f}\mathrm{t}/\sec$. We need to determine ${\sec }^2\theta$. Recall that $\sec \theta$ is the ratio of the length of the hypotenuse to the length of the adjacent side. We know the length of the adjacent side is $5000\mathrm{f}\mathrm{t}$. To determine the length of the hypotenuse, we use the Pythagorean theorem, where the length of one leg is $5000\mathrm{f}\mathrm{t}$, the length of the other leg is $h=1000\mathrm{f}\mathrm{t}$, and the length of the hypotenuse is $c$ feet as shown in the following figure.

We see that

$${1000}^2+{5000}^2=c^2$$

and we conclude that the hypotenuse is

$$c=1000\sqrt{26}\mathrm{ft}.$$

Therefore, when $h=1000$, we have

$${\sec }^2\theta ={\left(\frac{1000\sqrt{26}}{5000}\right)}^2=\frac{26}{25}.$$

Recall from step 4 that the equation relating ${\displaystyle \frac{d\theta }{dt}}$ to our known values is

$$\frac{dh}{dt}=5000{\sec }^2\theta \frac{d\theta }{dt}.$$

When $h=1000\mathrm{f}\mathrm{t}$, we know that ${\displaystyle \frac{dh}{dt}}=600\mathrm{f}\mathrm{t}/\sec$ and ${\sec }^2\theta ={\displaystyle \frac{26}{25}}$. Substituting these values into the previous equation, we arrive at the equation

$$600=5000\left(\frac{26}{25}\right)\frac{d\theta }{dt}.$$

Therefore, ${\displaystyle \frac{d\theta }{dt}}={\displaystyle \frac{3}{26}}\mathrm{rad}/\sec$.

Water is draining from the bottom of a cone-shaped funnel at the rate of $0.03{\mathrm{ft}}^3/\sec$. The height of the funnel is 2 ft and the radius at the top of the funnel is $1\mathrm{f}\mathrm{t}$. At what rate is the height of the water in the funnel changing when the height of the water is ${\displaystyle \frac{1}{2}}\mathrm{ft}\mathrm{?}$

Let $h$ denote the height of the water in the funnel, $r$ denote the radius of the water at its surface, and $V$ denote the volume of the water.

Step 2: We need to determine ${\displaystyle \frac{dh}{dt}}$ when $h={\displaystyle \frac{1}{2}}\mathrm{ft}$. We know that ${\displaystyle \frac{dV}{dt}}=-0.03{\mathrm{ft}}^3/\sec$.

Step 3: The volume of water in the cone is

$$V=\frac{1}{3}\pi r^{2}h.$$

From the figure, we see that we have similar triangles. Therefore, the ratio of the sides in the two triangles is the same. Therefore, ${\displaystyle \frac{r}{h}}={\displaystyle \frac{1}{2}}$ or $r={\displaystyle \frac{h}{2}}$. Using this fact, the equation for volume can be simplified to

$$V=\frac{1}{3}\pi {\left(\frac{h}{2}\right)}^{2}h=\frac{\pi }{12}{h}^3.$$

Step 4: Applying the chain rule while differentiating both sides of this equation with respect to time $t$, we obtain

$$\frac{dV}{dt}=\frac{\pi }{4}{h}^2\frac{dh}{dt}.$$

Step 5: We want to find ${\displaystyle \frac{dh}{dt}}$ when $h={\displaystyle \frac{1}{2}}\mathrm{ft}$. Since water is leaving at the rate of $0.03{\mathrm{ft}}^3/\sec$, we know that ${\displaystyle \frac{dV}{dt}}=-0.03{\mathrm{ft}}^3/\sec$. Therefore,

$$-0.03=\frac{\pi }{4}{\left(\frac{1}{2}\right)}^2\frac{dh}{dt},$$

which implies

$$-0.03=\frac{\pi }{16}\frac{dh}{dt}.$$

It follows that

$$\frac{dh}{dt}=-\frac{0.48}{\pi }=-0.153\mathrm{ft}/\sec .$$