Hello, cedwards08!

Chris, who is 6 feet tall, is walking away from a street light pole 30 feet high

at a rate of 2 ft/sec. .$\displaystyle {\color{blue}\frac{dx}{dt} = 2 \text{ ft/sec}}$

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: .$\displaystyle \frac{s}{6} \:=\:\frac{x+s}{30} \quad\Rightarrow\quad s \:=\:\frac{1}{4}x$

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

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

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

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

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

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

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

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

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

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

We are given: .$\displaystyle s = 6$

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

When $\displaystyle s = 6$, the right triangle is __isosceles__.

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

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

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