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 $\displaystyle A$, 123 feet from the track: $\displaystyle AB = 123$

The car is moving to the right from $\displaystyle B$ to $\displaystyle C$

. . $\displaystyle x = BC$ .and .$\displaystyle \frac{dx}{dt} \,=\,264$ ft/sec

Let $\displaystyle \theta \:=\:\angle BAC$

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

Differentiate with respect to time: .$\displaystyle \sec^2\!\theta\left(\frac{d\theta}{dt}\right)\:=\: \frac{1}{123}\left(\frac{dx}{dt}\right) $

. . and we have: .$\displaystyle \frac{d\theta}{dt} \:=\:\frac{\cos^2\!\theta}{123}\left(\frac{dx}{dt} \right)$ .**[1]**

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

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

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

Right triangle ABC has sides $\displaystyle x = 132$ and $\displaystyle AB =123$.

. . Its hypotenuse is: .$\displaystyle AB \:=\:\sqrt{132^2+123^2} \:=\:\sqrt{32,553} \:\approx\:180.42$

Hence: .$\displaystyle \cos\theta \:=\:\frac{123}{180.42} \:=\:0.681742601\quad\Rightarrow\quad \cos^2\!\theta \:\approx\:0.4648$

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

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