# Math Help - Related Rates 4

1. ## Related Rates 4

4) You are videotaping a race from a stand 123 feet away from the track, following a car that is moviing at 180 mi/h (264 ft/sec). How fast will your camera angle theta be changing when the car is right in fron of you? A half second later?

Huge thanks to anyone in advance for any help given

2. Hello, Super Mallow!

You are videotaping a race from a stand 123 feet away from the track,
following a car that is moviing at 180 mi/h (264 ft/sec).
How fast will your camera angle theta be changing
(a) when the car is right in fron of you?
(b) a half second later?
Code:
      B     x     C
* → → → → → *
|         /
|       /
123 |     /
|   /
|θ/
*
A

You are at $A$, 123 feet from the track: $AB = 123$

The car is moving to the right from $B$ to $C$
. . $x = BC$ .and . $\frac{dx}{dt} \,=\,264$ ft/sec
Let $\theta \:=\:\angle BAC$

From right triangle $ABC$, we have: . $\tan\theta \:=\:\frac{x}{123}$

Differentiate with respect to time: . $\sec^2\!\theta\left(\frac{d\theta}{dt}\right)\:=\: \frac{1}{123}\left(\frac{dx}{dt}\right)$
. . and we have: . $\frac{d\theta}{dt} \:=\:\frac{\cos^2\!\theta}{123}\left(\frac{dx}{dt} \right)$ .[1]

(a) When $t=0\!:\;x = 0,\:\theta\,=\,0,\:\cos^2\!0 \,=\,1$

Substitute into [1]: . $\frac{d\theta}{dt}\:=\:\frac{1^2}{123}(264) \:\approx\:\boxed{2.15\text{ radians/second}}$

(b) When $t = \frac{1}{2}\!:\;x = 132$

Right triangle ABC has sides $x = 132$ and $AB =123$.
. . Its hypotenuse is: . $AB \:=\:\sqrt{132^2+123^2} \:=\:\sqrt{32,553} \:\approx\:180.42$
Hence: . $\cos\theta \:=\:\frac{123}{180.42} \:=\:0.681742601\quad\Rightarrow\quad \cos^2\!\theta \:\approx\:0.4648$

Substitute into [1]: . $\frac{d\theta}{dt}\:=\:\frac{0.4648}{123}(264) \:=\:0.997619512$

Therefore: . $\frac{d\theta}{dt}\:\approx\:\boxed{1\text{ radian/second}}$

3. I'm back after 2 days, sorry for not getting to this earlier (Comcast...).

I get the solution to A. I do not understand B - How do we know X=132 when T=1/2?