Calculus: Related Rates

Question

A 75ft ladder is placed against a large building. The base of the ladder is resting on an oil spill, and it slips at the rate of 5 ft. per minute. Find the rate of change of the height of the top of the ladder above the ground at the instant when the base of the ladder is 30 ft. from the base of the building.

Answer 1

We have a right triangle whose legs are x := distance from base of ladder to wall and y:= distance from top of ladder to ground. The hypotenuse is 75'.

Therefore x² + y² = 75². Note for later: when x = 30', y = √75²-30² ≈ 68.74

Differentiate to get 2xdx/dt + 2ydy/dt = 0 and cancel the 2s. Solve for dy/dt (this is just algebra):

dy/dt = (-x/y)dx/dt.

dx/dt is constant at 5 ft/sec. At the time in question, dy/dt = (-30/68.74) * 5 ≈ -2.18 ft/sec. It makes sense that this is negative: as the bottom of the ladder slides away from the wall, the top of the ladder slides down the wall.

Answer 2

Set up a couple of equations.

Let "x" be how far the base of the ladder is from the wall.

Let "y" be how high up on the wall the ladder is resting.

x²+y²= 75²

So when x = 30, y = sqrt(4725) which is about 68.74.

Take the derivative of our ladder equation with respect to time:

2x (dx/dt) + 2y (dy/dt) = 0

Since the ladder is 30 feet from the wall, x = 30. since it's sliding away at 5 ft/minute, dx/dt = 5. Plugging in all the values we have, we get:

2(30)(5) + 2 (68.74) (dy/dt) = 0.

Solving for dy/dt now, we get dy/dt = -2.18 (this is approximate). The top of the ladder is moving down at the rate of 2.18 feet per minute. The exact answer is dy/dt = (-150/sqrt(4725))