# Related Rates Example

Q: A five foot ladder is leaning against the wall. If the top slides down at a rate of 5 feet per second, then how fast is the base of the ladder moving away from the wall at the instant when the top of the ladder is 3 feet above the ground.

A: Let’s start defining our variables. Let x be the horizontal distance (on the floor) that the ladder is away from the wall. Let y be the vertical distance that the ladder is located while leaning on the wall.

x and y are variables because as the ladder moves, x and y are always changing. x, y and the length of the ladder form a right triangle and ever instant. Since the length of the ladder is never changing, we know (Pythagorean Theorem) that:

x2 + y2 = 52 –>  x2 + y2 =25

We are looking for the rate of change of x with respect to time.  In math-speak, we are trying to find dx/dt.

What do we know?  We know that dy/dt (the rate of change of y with respect to time) is -5.  We also know that we are curious about dx/dt when y = 3.  And, from our equation, x2 + y2 =25, when y = 3 we know x = 4.

Now, to find dx/dt:

Take the derivative of x2 + y2 =25 with respect to time:

d/dt(x2 + y2 =25)

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

We are trying to find dx/dt when x=4, y=3, and dy/dt = -5:

2(4)(dx/dt) + 2(3)(-5) = 0

8(dx/dt) – 30 = 0

dx/dt = 30/8.

So, the base is moving at a rate of 30/8 feet per second!