Question
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?
Answers
Reiny
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
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
Ray
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.
Steve
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.
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.
Anthony
ohh ok this helped me too