Chapter 4 | Applications of Derivatives
347
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 dθ dt when h =1000ft. At that time, we know the velocity of the rocket is dh dt = 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 =5000tan θ . Step 4. Differentiating this equation with respect to time t , we obtain dh dt =5000sec 2 θ dθ dt . Step 5. We want to find dθ dt when h =1000ft. At this time, we know that dh dt = 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 5000ft. To determine the length of the hypotenuse, we use the Pythagorean theorem, where the length of one leg is 5000ft, the length of the other leg is h =1000ft, 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 26 ft.
Therefore, when h =1000, we have
⎛ ⎝ 1000 26 5000
⎞ ⎠
2
sec 2 θ =
= 26 25
.
Recall from step 4 that the equation relating dθ dt
to our known values is
dh dt =5000sec
2 θ dθ
dt .
When h =1000ft, we know that dh
and sec 2 θ = 26 25
dt = 600 ft/sec
. Substituting these values into the
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