Asked by Meghan
A 18 ft ladder is leaning against a wall. If the top of the ladder slides down the wall at a rate of 3 ft/s, how fast (in ft/s) is the bottom moving along the ground when the bottom of the ladder is 9 ft from the wall?
Also, this one. I have done the pathag theorem and then dy/dt = -x/y * dy/dt
Then, plugged stuff in to solve for that, but the answer I got which was 26sqrt3 which is the wrong answer, so I am just confused
Also, this one. I have done the pathag theorem and then dy/dt = -x/y * dy/dt
Then, plugged stuff in to solve for that, but the answer I got which was 26sqrt3 which is the wrong answer, so I am just confused
Answers
Answered by
Steve
no, dy/dx = -x/y
x^2+y^2=18^2
so, when x=9, y=9√3
2x dx/dt + 2y dy/dt = 0
Now plug in your values and you have
2*9 dx/dt + 2*9√3*(-3) = 0
dx/dt = 54√3/18 = 3√3 ft/s
x^2+y^2=18^2
so, when x=9, y=9√3
2x dx/dt + 2y dy/dt = 0
Now plug in your values and you have
2*9 dx/dt + 2*9√3*(-3) = 0
dx/dt = 54√3/18 = 3√3 ft/s
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