Question

a rotating light is located 14 feet from a wall. the light completes one rotation every 2 seconds. find the rate at which the light projected onto the wall is moving along the wall when the light’s angle is 20 degrees from perpendicular to the wall. wall light feet per second basic funcs trig enter a mathematical expression more..

Explanation:

Step1: Define Variables and Relationship

Let \( \theta \) be the angle between the light's beam and the perpendicular to the wall, and \( x \) be the distance of the light's projection on the wall from the point directly opposite the light. We have \( \tan\theta=\frac{x}{14} \), so \( x = 14\tan\theta \).

Step2: Find \( \frac{d\theta}{dt} \)

The light completes 1 rotation ( \( 2\pi \) radians) every 2 seconds, so \( \frac{d\theta}{dt}=\frac{2\pi}{2}=\pi \) radians per second.

Step3: Differentiate \( x \) with Respect to \( t \)

Differentiate \( x = 14\tan\theta \) with respect to \( t \): \( \frac{dx}{dt}=14\sec^{2}\theta\cdot\frac{d\theta}{dt} \).

Step4: Substitute \( \theta = 20^\circ \) and \( \frac{d\theta}{dt} \)

First, convert \( 20^\circ \) to radians (though \( \sec^{2}\theta \) can be calculated in degrees). \( \sec\theta=\frac{1}{\cos\theta} \), so \( \sec^{2}(20^\circ)=\frac{1}{\cos^{2}(20^\circ)} \). Using a calculator, \( \cos(20^\circ)\approx0.9397 \), so \( \sec^{2}(20^\circ)\approx\frac{1}{0.9397^{2}}\approx1.134 \). Then \( \frac{dx}{dt}=14\times1.134\times\pi \approx14\times1.134\times3.1416 \approx14\times3.563 \approx49.88 \) (more accurately, using exact trigonometric identities: \( \sec^{2}\theta = 1+\tan^{2}\theta \), \( \tan(20^\circ)\approx0.3640 \), so \( \sec^{2}(20^\circ)=1 + 0.3640^{2}\approx1.1325 \), then \( \frac{dx}{dt}=14\times1.1325\times\pi\approx14\times1.1325\times3.1416\approx14\times3.557\approx49.80 \), and with more precise calculation: \( \frac{d\theta}{dt}=\pi \), \( \sec(20^\circ)=\frac{1}{\cos(20^\circ)}\approx1.0642 \), \( \sec^{2}(20^\circ)\approx1.1325 \), so \( 14\times1.1325\times\pi = 14\pi\times1.1325\approx14\times3.1416\times1.1325\approx43.9824\times1.1325\approx49.81 \)).

Answer:

Approximately \( 49.8 \) (or more precisely, \( 14\pi\sec^{2}(20^\circ)\approx49.8 \)) feet per second.