Question

The top of a ladder slides down a vertical wall at a rate of 2 feet per second. At the moment when the bottom of the ladder is 4 feet from the wall, it slides away from the wall at a rate of 3 feet per second. How long is the ladder

Answer 1

Answer:

The length of the ladder is 7.21 feet

Step-by-step explanation:

Let

y = Height of the wall

x = Distance between the wall and the ladder (bottom)

L = Length of the ladder

So, the given parameters are:

$\frac{dy}{dt} = -2ft/s$

When $x = 4; \frac{dx}{dt} = 3ft/s$

The ladder, the wall and the ground forms a right-angled triangle where the hypotenuse is the length of the ladder; So:

$L^2 = x^2 + y^2$

Differentiate with respect to time

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

The length of the ladder does not change with time. So:

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

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

Rewrite as:

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

Divide both sides by 2

$x\frac{dx}{dt} =- y\frac{dy}{dt}$

Recall that:

$\frac{dy}{dt} = -2ft/s$ and  $x = 4; \frac{dx}{dt} = 3ft/s$

So:

$4 * 3 = -y * -2$

$12 = 2y$

$6 = y$

$y = 6$

Substitute $y = 6$ and $x =4$ in: $L^2 = x^2 + y^2$

$L^2 =4^2 + 6^2$

$L^2 =52$

$L = \sqrt{52$

$L = 7.21ft$