Calculus problem?

Question

A winch at the top of 12 meter high building pulls a pipe of the same length to a vertical position. The winch pulls in rope at a rate of -0.2 meters per second. Find the rate of the vertical change and the rate of the horizontal change at the end of the pipe when y=6.

if anyone could even tell me where to start that would be very helpful. i've been trying to figure this problem out for a while now. thanks in advance.

Answer 1

I am assuming that one end of the pipe remains in contact with the foot of the building.

Let:
x be the horizontal distance from the building of the upper end of the pipe,
y be its vertical height above ground,
z be the length of the rope.

From the right angled triangles:
x^2 + y^2 = 12^2 ...(1)
(12 - y)^2 + x^2 = z^2 ...(2)
and
z' = - 0.2 m/s ...(3)

Differentiating (1) and (2) implicitly with respect to time:
2xx' + 2yy' = 0
xx' = -yy' ...(3)

2(12 - y)(-y') + 2xx' = 2zz'
xx' - (12 - y)y' = zz' ...(4)

Eliminate xx' from (3) and (4):
- 12y' = zz'
y' = - zz' / 12 ...(5)

When y = 6:
(1) gives x = sqrt(108) = 6sqrt(3), and
(2) gives z = sqrt(36 + 108) = sqrt(144) = 12.
From (5):
y' = 12 * 0.2
= 2.4 m/s.

From (3):
x' = - yy' / x
= - 6 * 2.4 / 6 sqrt(3)
= - 0.8 sqrt(3) m/s.