Question
A 13ft ladder is placed against a verticle wall. Suppose the bottom of the ladder slides away from the wall at a constant rate of 2 feet per second. How fast is the top of the ladder sliding down the wall (negitive rate) when the bottom is 5ft from the wall? The ladder is sliding down the wall at a rate of _ ft/sec
Answers
Steve
You have
x^2 + y^2 = 13^2
when x=5, y=12
2x dx/dt + 2y dy/dt = 0
Now plug in your numbers and solve for dy/dt.
x^2 + y^2 = 13^2
when x=5, y=12
2x dx/dt + 2y dy/dt = 0
Now plug in your numbers and solve for dy/dt.