Question

A wooden log 10 meters long is leaning against a vertical wall with its other end on the ground. The top end of the log is sliding down the wall. When the top end is 6 meters from the ground, it slides down at 2m/sec. How fast is the bottom moving away from the wall at this instant?

Answer 1

Explanation:

This is a related rates problem from calculus using implicit differentiation. The main equation is Pythagorean's Theorem. Basically, what we are looking for is
`(dx)/(dt)` when y = 6 and
`(dy)/(dt)=-2`.

The equation for Pythagorean's Theorem is

`x^2+y^2=c^2` where x and y are the legs and c is the hypotenuse. The length of the hypotenuse is 10, so when we find the derivative of this function with respect to time, and using implicit differentiation, we get:

`2x(dx)/(dt)+2y(dy)/(dt)=0` and divide everything by 2 to simplify:

`x(dx)/(dt)+y(dy)/(dt)=0`. Looking at that equation, it looks like we need a value for x, y,
`(dx)/(dt)` and
`(dy)/(dt)`.

Since we are looking for
`(dx)/(dt)`, that can be our only unknown and everything else has to have a value. So what do we know?

If we construct a right triangle with 10 as the hypotenuse and use 6 for y, we can solve for x (which is the only unknown we have, actually). Using Pythagorean's Theorem to solve for x:

`x^2+6^2=10^2` and

`x^2+36=100` and

`x^2=64` so

x = 8.

NOW we can fill in the derivative and solve for
`(dx)/(dt)`.

Remember the derivative is

`x(dx)/(dt)+y(dy)/(dt)=0` so

`8(dx)/(dt)+6(-2)=0` and

`8(dx)/(dt)-12=0` and

`8(dx)/(dt)=12` so

`(dx)/(dt)=(12)/(8)=(6)/(4)=(3)/(2)=1.5 m/sec`