Q. 50 - 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 $10$ 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 $0.754$ foot per second when the base of the ladder is $10$ 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 $10$  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,

$l=12 ft$

$b=10ft$

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

Consider the triangle by using Pythagoras theorem,

$l^2=h^2+b^2$ …(1)

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

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

Substituting all values,

$h=\sqrt{{12}^2-{10}^2}$

 $h=\sqrt{144-100}$

$h=\sqrt{44}$

$h=2\sqrt{11}$

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 the ladder is not changing,

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

Substituting this in the equation(2),

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

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

Substituting all values,

$\left(2\sqrt{11}\right)\frac{dh}{dt}=-\left(10\right)\left(\frac{1}{2}\right)$

$\frac{dh}{dt}=-\left(10\right)\left(\frac{1}{2}\right)\left(\frac{1}{2\sqrt{11}}\right)$

$\frac{dh}{dt}=0.754$

So, the top of the ladder is moving at rate of $0.754$ foot per second.

Step 4 : Conclusion

Thus, the top of the ladder is moving down the side of the house at rate of $0.754$ foot per second when the base of the ladder is $10$ feet from the house.