4.1 Related rates  (Page 3/7)

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What is the speed of the plane if the distance between the person and the plane is increasing at the rate of $300\phantom{\rule{0.2em}{0ex}}\text{ft/sec}?$

$500\phantom{\rule{0.2em}{0ex}}\text{ft/sec}$

We now return to the problem involving the rocket launch from the beginning of the chapter.

Chapter opener: a rocket launch

A rocket is launched so that it rises vertically. A camera is positioned $5000\phantom{\rule{0.2em}{0ex}}\text{ft}$ from the launch pad. When the rocket is $1000\phantom{\rule{0.2em}{0ex}}\text{ft}$ above the launch pad, its velocity is $600\phantom{\rule{0.2em}{0ex}}\text{ft/sec}.$ Find the necessary rate of change of the camera’s angle as a function of time so that it stays focused on the rocket.

Step 1. Draw a picture introducing the variables.

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 $\frac{d\theta }{dt}$ when $h=1000\phantom{\rule{0.2em}{0ex}}\text{ft}.$ At that time, we know the velocity of the rocket is $\frac{dh}{dt}=600\phantom{\rule{0.2em}{0ex}}\text{ft/sec}.$

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 $\text{tan}\phantom{\rule{0.1em}{0ex}}\theta$ is the ratio of the length of the opposite side of the triangle to the length of the adjacent side. Thus, we have

$\text{tan}\phantom{\rule{0.1em}{0ex}}\theta =\frac{h}{5000}.$

This gives us the equation

$h=5000\phantom{\rule{0.1em}{0ex}}\text{tan}\phantom{\rule{0.1em}{0ex}}\theta .$

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

$\frac{dh}{dt}=5000\phantom{\rule{0.1em}{0ex}}{\text{sec}}^{2}\theta \phantom{\rule{0.1em}{0ex}}\frac{d\theta }{dt}.$

Step 5. We want to find $\frac{d\theta }{dt}$ when $h=1000\phantom{\rule{0.2em}{0ex}}\text{ft}.$ At this time, we know that $\frac{dh}{dt}=600\phantom{\rule{0.2em}{0ex}}\text{ft/sec}.$ We need to determine ${\text{sec}}^{2}\theta .$ Recall that $\text{sec}\phantom{\rule{0.1em}{0ex}}\theta$ 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 $5000\phantom{\rule{0.2em}{0ex}}\text{ft}.$ To determine the length of the hypotenuse, we use the Pythagorean theorem, where the length of one leg is $5000\phantom{\rule{0.2em}{0ex}}\text{ft},$ the length of the other leg is $h=1000\phantom{\rule{0.2em}{0ex}}\text{ft},$ 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

$c=1000\sqrt{26}\phantom{\rule{0.2em}{0ex}}\text{ft}.$

Therefore, when $h=1000,$ we have

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

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

$\frac{dh}{dt}=5000\phantom{\rule{0.1em}{0ex}}{\text{sec}}^{2}\theta \phantom{\rule{0.1em}{0ex}}\frac{d\theta }{dt}.$

When $h=1000\phantom{\rule{0.2em}{0ex}}\text{ft},$ we know that $\frac{dh}{dt}=600\phantom{\rule{0.2em}{0ex}}\text{ft/sec}$ and ${\text{sec}}^{2}\theta =\frac{26}{25}.$ Substituting these values into the previous equation, we arrive at the equation

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

Therefore, $\frac{d\theta }{dt}=\frac{3}{26}\phantom{\rule{0.2em}{0ex}}\text{rad/sec}.$

What rate of change is necessary for the elevation angle of the camera if the camera is placed on the ground at a distance of $4000\phantom{\rule{0.2em}{0ex}}\text{ft}$ from the launch pad and the velocity of the rocket is 500 ft/sec when the rocket is $2000\phantom{\rule{0.2em}{0ex}}\text{ft}$ off the ground?

$\frac{1}{10}\phantom{\rule{0.2em}{0ex}}\text{rad/sec}$

In the next example, we consider water draining from a cone-shaped funnel. We compare the rate at which the level of water in the cone is decreasing with the rate at which the volume of water is decreasing.

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