The top of a ladder slides down a vertical wall at a rate of 0.15 m/s. At the moment when the bottom of the ladder is 3 m from the wall, it slides away from the wall at a rate of 0.2 m/s. How long is the ladder?

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Answer 1 · 2016-10-15T18:32:29

$5$ m

The following assumes that (a) the ground is horizontal (i.e is perpendicular to the wall) and (b) the ladder is of constant length.

Variables
Let $y$ be the height up the wall to the top of the ladder.

Let $x$ be the distance between the bottom of the wall and the bottom of the ladder.

Let $L$ be the length of the ladder. (Note that $L$ is the constant we seek.)

Rates of Change

The ladder is sliding down the wall, so $y$ is decreasing at $0.15$ m/s.

$dy/dt = -0.15$ m/s

The bottom of the ladder is sliding away from the wall, so $x$ is increasing at $0.2$ m/s when $x = 3$ m.

$dx/dt = 0.2$ when $x = 3$

Equation relating the variables

$x^2+y^2=L^2$ $" "$ (Here we are using perpendicularity.)

Note that if we knew both $x$ and $y$ at some instant, then we could find $L$

Differentiate to find an equation relating the rates of change.

$2x dx/dt + 2y dy/dt = 0$

We know $dx/dx when$ x=3 $and we know$ dy/dt $, so we can find$ y $when$ x = 3#

$x dx/dt +y dy/dt = 0$

$(3)(0.2)+y(-0.15) = 0$

$0.6 = 0.15y$

$y = 0.6/0.15 = 60/15 = 4$

When $x=3$ m, we find that $y=4$ m.

We also have $x^2+y^2=L^2.$

So, $L=5$ m.