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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 ft/sec ?

500 ft/sec

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

Chapter opener: a rocket launch

A photo of a rocket lifting off.
(credit: modification of work by Steve Jurvetson, Wikimedia Commons)

A rocket is launched so that it rises vertically. A camera is positioned 5000 ft from the launch pad. When the rocket is 1000 ft above the launch pad, its velocity is 600 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.

A right triangle is formed with a camera at one of the nonright angles and a rocket at the other nonright angle. The angle with the camera has measure θ. The distance from the rocket to the ground is h; note that this is the side opposite the angle with measure θ. The side adjacent to the angle with measure θ is 5000 ft.
A camera is positioned 5000 ft from the launch pad of the rocket. The height of the rocket and the angle of the camera are changing with respect to time. We denote those quantities with the variables h and θ , respectively.

Let h denote the height of the rocket above the launch pad and θ 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 d θ d t when h = 1000 ft . At that time, we know the velocity of the rocket is d h d t = 600 ft/sec .

Step 3. Now we need to find an equation relating the two quantities that are changing with respect to time: h and θ . 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 θ is the ratio of the length of the opposite side of the triangle to the length of the adjacent side. Thus, we have

tan θ = h 5000 .

This gives us the equation

h = 5000 tan θ .

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

d h d t = 5000 sec 2 θ d θ d t .

Step 5. We want to find d θ d t when h = 1000 ft . At this time, we know that d h d t = 600 ft/sec . We need to determine sec 2 θ . Recall that sec θ 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 ft . To determine the length of the hypotenuse, we use the Pythagorean theorem, where the length of one leg is 5000 ft , the length of the other leg is h = 1000 ft , and the length of the hypotenuse is c feet as shown in the following figure.

A right triangle has one angle with measure θ. The hypotenuse is c, the side length opposite the angle with measure θ is 1000, and the side adjacent to the angle with measure θ is 5000.

We see that

1000 2 + 5000 2 = c 2

and we conclude that the hypotenuse is

c = 1000 26 ft .

Therefore, when h = 1000 , we have

sec 2 θ = ( 1000 26 5000 ) 2 = 26 25 .

Recall from step 4 that the equation relating d θ d t to our known values is

d h d t = 5000 sec 2 θ d θ d t .

When h = 1000 ft , we know that d h d t = 600 ft/sec and sec 2 θ = 26 25 . Substituting these values into the previous equation, we arrive at the equation

600 = 5000 ( 26 25 ) d θ d t .

Therefore, d θ d t = 3 26 rad/sec .

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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 ft from the launch pad and the velocity of the rocket is 500 ft/sec when the rocket is 2000 ft off the ground?

1 10 rad/sec

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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.

Questions & Answers

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Practice Key Terms 1

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Source:  OpenStax, Calculus volume 1. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11964/1.2
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