# Math Help - Related rate of change problem. Help please

1. ## Related rate of change problem. Help please

Chris, who is 6 feet tall, is walking away from a street light pole 30 deet high at a rate of 2 feet per second.

a. How fast is his shadow increasing in length when chris is 24ft from the pole? 30 feet? -- I already found that to be 1/2 ft per sec.

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

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

2. 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. . ${\color{blue}\frac{dx}{dt} = 2 \text{ ft/sec}}$

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.

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.

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.