A 13ft ladder slides down a wall at 2ft/sec when it is 12 ft above the ground. Find the rate at which the angle it makes with the ground changes at this moment?

Answer 1

let the height of the ladder against the wall be y ft
and let the base of the ladder be x ft from the wall
we know x^2 + y^2= 13^

2x dx/dt + 2y dy/dt = 0
when y = 12, x = 5 by above Pythagoras, and dy/dt = -2 ft/s

2(5)(dx/dt) + 2(12)(-2) = 0
dx/dt = 24/10 = 2.4 ft/s

Answer 2

Hi there, wouldn't we need to use right triangle trig for this problem since it's asking for the rate of the angle? theta? Not completely sure on how to relate that exactly.

Answer 3

yes - so, try it

sinθ = y/13
cosθ dθ/dt = 1/13 dy/dt

Now just plug in your numbers. Use the Pythagorean Theorem to figure cosθ when y=12.

Answer 4

ohh ok this helped me too