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 area of the triangle formed by the ladder, the house, and the ground changing when the top of the ladder is $6$ feet from the ground?

Short answer

The area of the triangle is formed by the ladder, the house, and the ground changing at rate of$3$foot per second when the top of the ladder is $6$feet from the ground.

Step-by-step solution

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 top of the ladder is$6$feet from the ground.

Calculation

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.

The diagram will be

From the form triangle,

$h^2+b^2={12}^2$

Differentiating both sides with respect to localid="1650511776722" $t$,

$2h\frac{dh}{dt}+2b\frac{db}{dt}=0$

$h\frac{dh}{dt}+b\frac{db}{dt}=0...\left(1\right)$

When $h=6$,

$6^2+b^2={12}^2$

role="math" localid="1650512766225" $36+b^2=144$

$b^2=144-36$

$b^2=108$

$b=\sqrt{108}$

$b=6\sqrt{3}$

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

From equation (1),

$6·\frac{dh}{dt}+6\sqrt{3}·\left(\frac{1}{2}\right)=0$

$6·\frac{dh}{dt}+3\sqrt{3}=0$

$6·\frac{dh}{dt}=-3\sqrt{3}$

$\frac{dh}{dt}=-\frac{3\sqrt{3}}{6}$

$\frac{dh}{dt}=-\frac{\sqrt{3}}{2}$

So,$h$is decreasing.

Find the changing rate

The area of the formed triangle is

$A=\frac{1}{2}hb$

Differentiating with respect to $t$,

$\frac{dA}{dt}=\frac{1}{2}h\frac{db}{dt}+\frac{1}{2}b\frac{dh}{dt}$

Plugging all values,

$\frac{dA}{dt}=\frac{1}{2}\left(6\right)\left(\frac{1}{2}\right)+\frac{1}{2}\left(6\sqrt{3}\right)\left(-\frac{\sqrt{3}}{2}\right)$

$\frac{dA}{dt}=\frac{3}{2}-\frac{9}{2}$

role="math" localid="1650513894648" $\frac{dA}{dt}=-\frac{6}{2}$

role="math" localid="1650513919174" $\frac{dA}{dt}=-3$

It is decreasing.

The triangle is changing at rate of $3$foot per second.

Conclusion

The area of the triangle is formed by the ladder, the house, and the ground changing at rate of$3$ foot per second when the top of the ladder is$6$ feet from the ground