Question

A major network is televising the launching of a rocket. A camera tracking the lift-off of the rocket is located at point A, as shown in the accompanying figure, where theta is the angle of elevation of the camera at A. How fast is theta changing at the instant when the rocket is at a distance of 13,000 feet from the camera and this distance is increasing at the rate of 480 ft/sec.

Answer 1

Ok, so if i get it right, you wanna find the derivative of this by theta
Ok, first let's make a function do calculate the tangens of the angle of the camera by the location of the rocket at a given time x:
\[
f(x) = tan(\theta) = \frac{\text{speed} \cdot \text{time}} { \text{distance_base_from_camera} } = \\
= \frac{ 480_{ft/sec} \cdot x_{sec}}{12000_{ft}} = \frac{x}{25}
\]
Now we can say this:
\[
\theta = arctan[\;tan(\theta)\;] = arctan[\;f(x)\;]
\]
And now to calculate the derivative we use chain rule, since we got one function's value given to another function:
(The derivative of arctan I took from here: http://www.analyzemath.com/calculus/Differentiation/inverse_trigonometric.html )
\[
\text{chain rule}: \quad {\huge [}\;f[\;g(x)\;] \; {\huge ]}' = f'[\;g(x)\;] \cdot g'(x) \\
arctan'(x) = \frac{1}{1+x^2} \\
f'(x) = 1 \cdot \frac{1}{25} \cdot x^0 = \frac{1}{25} \\
{\huge [}\;arctan[\;f(x)\;] \; {\huge ]}' = arctan'[\;f(x)\;] \cdot f'(x) = \\
= \frac{1}{1 + f(x)^2 } \cdot \frac{1}{25} = \frac{1}{1 + \frac{x^2}{25^2} } \cdot \frac{1}{25} = \\
= \frac{1}{25 + \frac{x^2}{25} } = \frac{1}{ \frac{25^2 + x^2}{25}} = \frac{25}{625 + x^2}
\]
Hope I got it right hehe..

Answer 2

I just noticed that in the picture you wrote 12000ft but in question 13000ft.

Answer 3

The camera is 13000 feet from the rocket, but 12000 feet from the liftoff point.

Answer 4

So we wanna find the derivative at that point I see. forgive me.
Let's start by finding the rocket's height when its distance is 13000ft from the camera
\[
{\small \text{distance_from_camera}^2 = \text{distance_base_from_camera}^2 +\text{rocket_height}^2 } \\
(13000_{ft})^2 = (12000_{ft})^2 +h^2 \\
h^2 = (13000_{ft})^2 - (12000_{ft})^2 \\
h = \pm \sqrt{(13000_{ft})^2 - (12000_{ft})^2} \\
h = \pm \sqrt{169000000_{ft^2} - 144000000_{ft^2}} \\
h = \pm \sqrt{25000000_{ft^2}} = \pm 5000_{ft}
\]
Since the height can't be negative..
\[ h = 5000_{ft} \]
So we know that the rocket is at height of 5000ft in the air at that moment. let's find how many seconds passed so we can use our derivative by x:
\[
\text{seconds_passed} = \frac{height}{speed} = \frac{5000_{ft}}{480_{ft/sec}} = 10.41_{sec} \\
{\huge [} arctan[\;f(10.41_{sec})\;] {\huge ]}' = \frac{25}{625 - 10.41^2}\\
= \frac{25}{625 - 108.37} = \frac{25}{516.63} = 0.05
\]
By this the result should be 0.05
The result is ugly, and i'm pretty tired, perhaps I made mistakes. But I couldn't find any.