Q. 49 - Calculus

Question

In Exercises 49–51, Alina props a $12$-foot ladder against the side of her house so that she can sneak into her upstairs bedroom window. Unfortunately, the ground is muddy because of a recent rainstorm, and the base of the ladder slides away from the house at a rate of half a foot per second.

How fast is the top of the ladder moving down the side of the house when the base of the ladder is $4$ feet from the house?

Step-by-Step Solution

Answer

The top of the ladder is moving down the side of the house at rate of $0.177$ foot per second when the base of the ladder is $4$ feet from the house.

Step 1 : Given information

Alina props a $12$-foot ladder against the side of her house so that she can sneak into her upstairs bedroom window. Unfortunately, the ground is muddy because of a recent rainstorm, and the base of the ladder slides away from the house at a rate of half a foot per second.

Also given that, the base of the ladder is $4$ feet from the house.

Step 2 : Find the height of house up to top of ladder

Let $h,l$ and $b$ represent height of the house up to top of the ladder, length of ladder and distance of base of ladder from house.

By given data we can write,

$$\begin{gathered} l=12 ft \\ b=4 ft \end{gathered}$$

$\frac{db}{dt}=\frac{1}{2}$

Considering the triangle by using Pythagoras theorem,

$l^2=h^2+b^2$ …(1)

$h^2=l^2-b^2$

$h=\sqrt{l^2-b^2}$

Substitute all values,

$h=\sqrt{{12}^2-4^2}$

$h=\sqrt{144-16}$

$h=\sqrt{128}$

$h=8\sqrt{2}$

Step 3 : Find the rate of moving down of the top of ladder

By using the relation (1),

$l^2=h^2+b^2$

By differentiating both sides with respect to $t$ ,

$2l\frac{dl}{dt}=2h\frac{dh}{dt}+2b\frac{db}{dt}$

$l\frac{dl}{dt}=h\frac{dh}{dt}+b\frac{db}{dt}$ ; by simplifying …(2)

Since length of ladder is not changing,

So, $\frac{dl}{dt}=0$

Substituting this on the equation (2),

$0=h\frac{dh}{dt}+b\frac{db}{dt}$

So, $h\frac{dh}{dt}=-b\frac{db}{dt}$

Substituting all values,

$\left(8\sqrt{2}\right)\frac{dh}{dt}=-\left(4\right)\left(\frac{1}{2}\right)$

$\frac{dh}{dt}=-\left(4\right)\left(\frac{1}{2}\right)\left(\frac{1}{8\sqrt{2}}\right)$

$\frac{dh}{dt}=-0.177$

So, the top of the ladder is moving down at rate of $0.177$ foot per second.

Step 4 : Conclusion

Thus, the top of the ladder is moving down the side of the house at rate of $0.177$ foot per second when the base of the ladder is $4$ feet from the house.