Question

A 13 ft ladder slides down a vertical wall at a rate of 3 ft/minute. When the top of the ladder is 5 feet from the ground, what is the rate of change of distance between bottom of the ladder and the wall?

Answer 1

Answer:

\frac{5}{4}[/tex] ft/minute.

Step-by-step explanation:

Let's draw diagram from this given situation.

Let "x" be the distance between the base of the wall and  the foot of the ladder.

Let "y" be the distance between top of the ladder and to the ground.

It is a right triangle, let's form a statement using the Pythagorean theorem.

$x^2 + y^2 = 13^2$

$x^2 + y^2 = 169$ ------(1)

Now let's take the derivative with respect to the time (t), we get

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

When you find the derivative of the constant 160, we get 0.

Given: Ladder slides down a vertical wall at the rate of 3ft. So $\frac{dy}{dt} = -3 ft$

Since it is down, so we took it as negative.

We need to find $\frac{dx}{dt}$ when y = 5 feet.

To do that we need to find the value of "x" when y = 5 feet.

To find the value of x, we need to plug in y = 5 in the equation (1)

$x^2 + 5^2 = 169\\x^2 + 25 = 169\\x^2 = 169 - 25\\x^2 = 144$

Taking square root on both sides, we get

x = 12

Now let's plug in x = 12, y = 5 and $\frac{dy}{dt} = -3$ in the derivative equation (2) and find $\frac{dx}{dt}$

$2*(12)\frac{dx}{dt} + 2(5) (-3) = 0\\24\frac{dx}{dt} - 30 = 0\\24\frac{dx}{dt}  = 30$

$\frac{dx}{dt} = \frac{30}{24}$

Simplifying this fraction, we get

$\frac{dx}{dt} = \frac{5}{4}$ ft/minute

So the rate of change of distance between the bottom of the ladder and the wall is \frac{5}{4}[/tex] ft/minute.