Question

A 25-meter ladder is sliding down a vertical wall so the distance between the bottom of the ladder and the wall is increasing at 3.5 meters per minute. At a certain instant, the top of the ladder is 7 meters from the ground. What is the rate of change of the distance between the top of the ladder and the ground at that instant (in meters per minute)?

Answer 1

Answer:

Step-by-step explanation:

This situation models a right triangle.  The hypotenuse is the ladder which has a given length of 25 meters.  Let's call the ground (the base of the triangle) x, and the vertical wall (the height of the triangle) y.  Pythagorean's Theorem tells us that

$x^2+y^2=25^2$

This is our main formula.

We are given that the rate the distance between the bottom of the ladder and the wall is increasing at 3.5 m/min.  This is the rate of change of x, or dx/dt.  We are wanting to find the rate of change of the distance between the top of the ladder and the ground, dy/dt, when y = 7.  Let's find the derivative of our equation first, then fill in what we are given:

$2x\frac{dx}{dt}+2y\frac{dy}{dt}=0$

From our given, we can fill in the dx/dt, and the y.  We are looking for dy/dt, but we are missing what the x value is.  Let's find that by going back to Pythagoreans Theorem and filling in y to find x:

$x^2+7^2=25^2$ and

$x^2=576$ so

x = 24

Now let's fill in our derivative:

$2(24)(3.5)+2(7)\frac{dy}{dt}=0$ and

$168+14\frac{dy}{dt}=0$ and

$14\frac{dy}{dt}=-168$ so

$\frac{dy}{dt}=-12m/min$

Within the context of our problem, this means that the height of the ladder is decreasing at a rate of 12 m/min.  The negative value lets us know that the height is decreasing, not that the height is negative.