Question

The top of a ladder slides down a vertical wall at a rate of 0.5m/s. At the moment when the bottom of the ladder is 3m from the wall, it slides away from the wall at a rate of 0.2m/s. How long is the ladder?

Answer 1

Answer:

The length of the ladder is 3.23 m

Explanation:

Let's call $y$ the distance from the top of the ladder to the ground, $x$ the distance from the bottom of the ladder to the wall and $l$ the the length of the ladder.

So, we can write the following equation:

$x^{2} +y^{2} =l^{2}$

So, if we derive both sides of the equation with respect to t, we get:

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

Solving for y, and replacing the values of x by 3, $\frac{dx}{dt}$ by 0.2m/s and $\frac{dy}{dt}$ by -0.5m/s, we get:

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

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

$y=\frac{-3(0.2)}{-0.5}=1.2m$

Then, if $x$ is equal to 3 m and $y$ is equal to 1.2m, $l$ is equal to:

$x^{2} +y^{2} =l^{2}$

$3^{2} +1.2^{2} =l^{2}$

$10.44 =l^{2}$

$\sqrt{10.44} =l\\3.23=l$