**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:

x^{2} + y^{2} = 5^{2} –> **x ^{2} + y^{2} =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, **x ^{2} + y^{2} =25**, when y = 3 we know x = 4.

Now, to find dx/dt:

Take the derivative of **x ^{2} + y^{2} =25 **with respect to time:

d/dt(x^{2} + y^{2} =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!

Thanks!

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