Question

question 15
a television camera at ground level is 2000 feet away from the launching pad of a space rocket that is set to take off vertically, as seen in the following figure. the angle of elevation of the camera can be found by $\theta=\tan^{-1}left(\frac{x}{2000}
ight)$ where $x$ is the height of the rocket.
find the rate of change of the angle of elevation after launch when the camera and the rocket are 5093 feet apart.
$\theta =$ radians per foot. round to five decimal places.
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Explanation:

Step1: Find the height of the rocket

Using the Pythagorean theorem $x=\sqrt{l^{2}-2000^{2}}$, where $l = 5093$. So $x=\sqrt{5093^{2}-2000^{2}}=\sqrt{(5093 + 2000)(5093 - 2000)}=\sqrt{7093\times3093}=\sqrt{21938649}=4684.93852$.

Step2: Differentiate the angle - elevation formula

We have $\theta=\tan^{- 1}(\frac{x}{2000})$. By the chain - rule, $\theta^\prime=\frac{1}{1 + (\frac{x}{2000})^{2}}\times\frac{1}{2000}=\frac{2000}{4000000+x^{2}}$.

Step3: Substitute the value of x

Substitute $x = 4684.93852$ into the derivative formula: $\theta^\prime=\frac{2000}{4000000+(4684.93852)^{2}}=\frac{2000}{4000000 + 21938649}=\frac{2000}{25938649}\approx0.00008$.

Answer:

$0.00008$