Q. 48 - Calculus

Question

In Exercises 46–48, suppose that Stuart is 6 feet tall and is walking towards a 20-foot streetlight at a rate of 4 feet per second. As he walks towards the streetlight, his shadow gets

shorter.

How fast is the area of the triangle made up of Stuart’s legs and his shadow changing? Is it increasing or decreasing as Stuart walks towards the streetlight.

Step-by-Step Solution

Answer

Thus, the area of the triangle made up of Stuart’s legs and his shadow is changing at rate of $5.13$ feet square per second.

Yes, it is decreasing as Stuart walks towards the streetlight.

Step 1 : Given information

It is given that, Stuart is 6 feet tall and is walking towards a 20-foot streetlight at a rate of 4 feet per second .

Step 2 : Analyze all information

It is given the rate at which Stuart  walks towards the streetlight .

Now we have to find the rate of change of the length of Stuart  shadow. 

To find a relationship between these two rates we have to find a relationship between their underlying variables: the distances between Stuart and the streetlight and the length $l$ of Stuart's shadow.

By the law of similar triangles,

$s$ and $l$ are the related by the equation $\frac{20}{s+l}=\frac{6}{l}$ , as shown in the following diagram

where $s\left(t\right)=dis\tan ce$ , $l\left(t\right)=lenght of shadow$

Step 3 : Calculation

From above diagram it is given that, 

$\frac{ds}{dt}=4$

We have to find $\frac{dl}{dt}$.

Using above relation,

$\frac{20}{s+l}=\frac{6}{l}$

$20l=6s+6l$ [Simplifying]

$14l=6s$

Differentiation both sides with respect to $t$,

$14\frac{dl}{dt}=6\frac{ds}{dt}$

$14\frac{dl}{dt}=6\left(4\right)$ , since $\frac{ds}{dt}=4$

$\frac{dl}{dt}=\frac{6\left(4\right)}{14}$

$\frac{dl}{dt}=1.71$

So, the length of his shadow is decreasing at rate of $1.71$ feet per second.

Step 4 : Find final answer

The area of the triangle made up of Stuart’s legs and his shadow is

$A=\frac{1}{2}\times 6\times l$ 

By differentiating with respect to $t$,

$\frac{dA}{dt}=\frac{1}{2}\times 6\times \frac{dl}{dt}$

$\frac{dA}{dt}=\frac{1}{2}\times 6\times 1.71$ , since $\frac{dl}{dt}=1.71$

So, $\frac{dA}{dt}=5.13$

Since length of the shadow is decreasing , so the area is also decreasing.

Thus, the area of triangle is decreasing at rate of $5.13$ feet square per second.

Step 5 : Conclusion

Thus, the area of the triangle made up of Stuart’s legs and his shadow is changing at rate of $5.13$ feet square per second.

Yes, it is decreasing as Stuart walks towards the streetlight.