A ladder 10ft long rests against a vertical wall. If the bottom of the ladder slides away from the wall at a speed of 2ft/s, how fast is the angle between the top of the ladder and the wall changing when the angle is #pi/4# rad?

Question

Samantha Adkison · Accepted Answer

#sqrt2/5# #rad#/#s# Let #x# be the distance between the base of the wall and the bottom of the ladder. And let #theta# be the angle between the top of the ladder and the wall, Then #x/10 = sin theta#, so #x = 10sintheta#. Differentiating with respect to time #t# gets us #dx/dt = 10 cos theta (d theta)/dt# We were told that #dx/dt = 2# #ft#/#s# and we seek #(d theta)/dt# when #theta = pi/4# #2 = 10 ( cos(pi/4)) (d theta)/dt# #2 = 10 (sqrt2/2) (d theta)/dt = 5sqrt2 (d theta)/dt# #(d theta)/dt = 2/(5sqrt2) = sqrt2/5# #rad#/#s#

Isabella Ackerman · Answer

undefined We can use related rates to solve this problem. Let $ x $ represent the distance between the bottom of the ladder and the wall, and $ \theta $ represent the angle between the ladder and the wall. We are given $ \frac{{dx}}{{dt}} = 2 $ ft/s, and we want to find $ \frac{{d\theta}}{{dt}} $ when $ \theta = \frac{{\pi}}{{4}} $ rad.

Using the Pythagorean theorem, $ x^2 + 10^2 = (10\cos\theta)^2 $. Differentiating both sides with respect to time $ t $, we get $ 2x\frac{{dx}}{{dt}} = 20\cos\theta\frac{{d\theta}}{{dt}} $.

Substitute the given values: $ 2(10)\cdot2 = 20\cos\frac{{\pi}}{{4}}\frac{{d\theta}}{{dt}} $. Solve for $ \frac{{d\theta}}{{dt}} $:

$$ \frac{{d\theta}}{{dt}} = \frac{{40}}{{20\cdot\frac{{\sqrt{2}}}{{2}}} $$

$$ \frac{{d\theta}}{{dt}} = \frac{{40}}{{10\sqrt{2}}} $$

$$ \frac{{d\theta}}{{dt}} = \frac{{4}}{{\sqrt{2}}} $$

$$ \frac{{d\theta}}{{dt}} = 2\sqrt{2} $$

So, $ \frac{{d\theta}}{{dt}} = 2\sqrt{2} $ rad/s when $ \theta = \frac{{\pi}}{{4}} $ rad.