Relation of Rates

Question

A 25-ft ladder rests against a vertical wall. If the bottom of the ladder slides away from the wall at a rate of 0.18 ft/sec, how fast, in ft/sec, is the top of the ladder sliding down the wall, at the instant when the bottom of the ladder is 20 ft from the wall? Answer with 2 decimal places. Type your answer in the space below. If your answer is a number less than 1, place a leading "0" before the decimal point (ex: 0.35).

Answer 1

Draw a picture, a triangle, with base of 20, hypotenuse=25, then the height of the other side is 15. It's a 5,4,3 right triangle. 5 squared = 4 squared plus 3 squared. 25=16+9

As the slide comes to an end, the bottom of the ladder will move 5 feet to the left, and the top of the ladder will move downwards 15 feet, or faster down, as to the left. More of the motion is downward, as the ladder gets closer to fully horizontal. Answer should be a little larger than 0.18 ft/sec.

This is a calculus problem. Let x=horizontal distance, y=vertical distance of the ladder. x²+y²=25

Take the derivative to get 2x(dx/dt) + 2y(dy/dt) =0

plug in the values for x, y and dx/dt, to solve for dy/dt

2(20)(.18)+2(15)(dy/dt)=0

divide through by 2

20(.18)+15(dy/dt)=0

divide through by 5

4(.18)+3(dy/dt)=0

dy/dt = -(4/3)(.18)=-0.18 - (1/3)(0.18) = -0.18 -0.06

=-0.24 negative sign means downward movement

Answer 2

First draw a picture so that you can label what you know and what you need to know. BE SURE TO DRAW A PICTURE.

Let s be the length of the ladder in feet. s = 25 as given. It doesn't change, so ds/dt = 0. Also it's the hypotenuse of a right triangle. That will allow us to use the Pythagorean Theorem.

At a particular point in time, x, the distance from the wall, is 20 ft. We are told that it is sliding away from the wall at .18 ft/sec. In my picture, it is moving to the right. That quantity is dx/dt = 0.18 ft/sec

Hmmm...what do we know about y? The ladder is sliding down. We don't know how fast. But we understand that at the moment that we are concerned with we have a y-20-25 right triangle. I drew the diagram in Quadrant I. So the sides must be in the ratio of 3:4:5, so y = 15.

We know that x^2 + y^2 = s^2; we can differentiate that and see that:

2x(dx/dt) + 2y(dy/dt) = 2s(ds/dt)

We are using the chain rule here and assuming that x, y and s are all functions of t. We coul have written them as x(t), y(t) and s(t)!

We already listed every quantity except dy/dt in this equation in the discussion above, so plugging in:

2*20*0.18 +2*15*(dy/dt) = 2*25*0 or 7.2 + 30(dy/dt) = 0, so dy/dt = -7.2/30 = -0.24 ft/sec.

So the top of the ladder is sliding DOWN the wall at 0.24 ft/sec.