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

Answer:

Step-by-step 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 $\frac{dx}{dt}$ when y = 6 and $\frac{dy}{dt}=-2$.

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:

$2x\frac{dx}{dt}+2y\frac{dy}{dt}=0$ and divide everything by 2 to simplify:

$x\frac{dx}{dt}+y\frac{dy}{dt}=0$. Looking at that equation, it looks like we need a value for x, y, $\frac{dx}{dt}$ and $\frac{dy}{dt}$.

Since we are looking for $\frac{dx}{dt}$, 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 $\frac{dx}{dt}$.

Remember the derivative is

$x\frac{dx}{dt}+y\frac{dy}{dt}=0$ so

$8\frac{dx}{dt}+6(-2)=0$ and

$8\frac{dx}{dt}-12=0$ and

$8\frac{dx}{dt}=12$ so

$\frac{dx}{dt}=\frac{12}{8}=\frac{6}{4}=\frac{3}{2}=1.5 m/sec$