Related rate of change problem. Help please

Hello, cedwards08!

Quote:

Chris, who is 6 feet tall, is walking away from a street light pole 30 feet high
at a rate of 2 ft/sec. . ${\color{blue}\frac{dx}{dt} = 2 \text{ ft/sec}}$

a. How fast is his shadow increasing in length when Chris is 24 ft from the pole? 30 feet?

Code:

* | * | * 30 | | * | 6| * | | * *-------*---------------* : - x - : - - - s - - - :

From the similar right triangles: . $\frac{s}{6} \:=\:\frac{x+s}{30} \quad\Rightarrow\quad s \:=\:\frac{1}{4}x$

Differentiate with respect to time: . $\frac{ds}{dt} \;=\;\frac{1}{4}\left(\frac{dx}{dt}\right)$

Therefore: . $\frac{ds}{dt} \:=\:\frac{1}{4}(2) \:=\:\frac{1}{2}\text{ ft/sec}$ always.

Quote:

b. How fast is the tip of his shadow moving?

Code:

* | * | * 30 | | * | 6| * | x | y-x * *-------*---------------* : - - - - - y - - - - - :

We have: . $\frac{y}{30} \:=\:\frac{y-x}{6} \quad\Rightarrow\quad y \:=\:\frac{5}{4}x$

Differentiate with respect to time: . $\frac{dy}{dt} \;=\;\frac{5}{4}\left(\frac{dx}{dt}\right)$

Therefore: . $\frac{dy}{dt} \;=\;\frac{5}{4}(2) \;=\;\frac{5}{2}\text{ ft/sec}$ always.

Quote:

c. To follow the tip of his shadow, at what angular rate must Chris
be lifting his eyes when his shadow is 6ft long?

Code:

* | * | * - - - - - 30 | | * θ | 6| * | | θ * *-------*---------------* : - x - : - - - s - - - :

We have: . $\tan\theta \:=\:\frac{6}{s} \quad\Rightarrow\quad \tan\theta \:=\:6s^{-1}$

Differentiate with respect to time: . $\sec^2\!\theta\left(\frac{d\theta}{dt}\right) \:=\:-6s^{-2}\left(\frac{ds}{dt}\right)$

. . and we have: . $\frac{d\theta}{dt} \;=\;-\frac{6}{s^2}\cos^2\theta\left(\frac{ds}{dt}\right )$ .[1]

We are given: . $s = 6$

From (a), we have: . $\frac{ds}{dt} = \frac{1}{2}$

When $s = 6$ , the right triangle is isosceles.

. . Hence: . $\theta = \frac{\pi}{4}\;\;\text{ and }\cos\frac{\pi}{4} = \frac{1}{\sqrt{2}}$

Substitute into [1]: . $\frac{d\theta}{dt} \;=\;-\frac{6}{6^2}\left(\frac{1}{\sqrt{2}}\right)^2\lef t(\frac{1}{2}\right) \;=\;-\frac{1}{24}$

Therefore, Chris should raise his eyes at the rate of $\frac{1}{24}$ radians/sec.