Question

A camera is mounted at a point 4000 feet away from a geyser. If the water is rising vertically at 900 ft/s when it is 3000 feet above the ground, how fast must the camera angle of elevation change (in degrees per second) at that instant to keep the top of the water in the frame?
A. 8

B. 8.25

C. 9

D. 9.5

Answer 1

Answer:

$\dot \theta = 0.144\,\frac{rad}{s}$, $\dot \theta = 8.251\,\frac{deg}{s}$ (Option B)

Step-by-step explanation:

The trigonometric diagram is included herein as attachment. The expression is presented below:

$\tan \theta = \frac{y}{x}$

Where:

$x$ - Horizontal distance between the geyser and the camera.

$y$ - Vertical distance between the geyser and the camera.

The rate of change in terms of time is:

$\dot \theta \cdot \sec^{2}\theta = \frac{\dot y\cdot x-y\cdot \dot x}{x^{2}}$

$\dot \theta \cdot \left(\frac{1}{\cos^{2}\theta} \right) = \frac{\dot y \cdot x - y\cdot \dot x}{x^{2}}$

$\dot \theta = \left(\frac{\dot y \cdot x - y \cdot \dot x}{x^{2}} \right)\cdot \cos^{2}\theta$

$\dot \theta = \left(\frac{\dot y \cdot x - y\cdot \dot x}{x^{2}} \right)\cdot \left(\frac{x^{2}}{x^{2}+y^{2}} \right)$

$\dot \theta = \frac{\dot y \cdot x - y\cdot \dot x}{x^{2}+y^{2}}$

Finally,

$\dot \theta = \frac{\left(900\,\frac{ft}{s} \right)\cdot (4000\,ft)-(3000\,ft)\cdot \left(0\,\frac{ft}{s} \right)}{(4000\,ft)^{2}+(3000\,ft)^{2}}$

$\dot \theta = 0.144\,\frac{rad}{s}$

$\dot \theta = 8.251\,\frac{deg}{s}$