MHB Beverly's question at Yahoo Answers regarding related rates

[MarkFL]

Here is the question:

Calculus Homework Question - related rates help?

Amanda is driving her car south on Interstate 95. A Virginia State Trooper is parked 90 feet west of the interstate, and aims his radar at the car after it passes him. He finds the distance to Amanda's car from his position is 150 feet and the distance separating them is increasing at the rate of 72 feet per second. Find the speed of the car in feet per second.

Here is a link to the question:

Calculus Homework Question - related rates help? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.

 

[MarkFL]

Hello Beverly,

The first thing I would do is draw a diagram representing the scenario:

$T$ represents the trooper's position, $C$ represents the position of Amanda's car on I-95, $x$ represents the distance Amanda has traveled since passing the trooper, and $h$ represents the distance between Amanda and the trooper. All distances are in feet.

Now, we are ultimately being asked to find $$\frac{dx}{dt}$$, and we are given $$\left.\frac{dh}{dt}\right|_{h=150}=72\,\frac{\text{ft}}{\text{s}}$$.

What we need then, is a relationship between $x$ and $h$, and fortunately, we have what we need via Pythagoras:

$$x^2+90^2=h^2$$

Now, implicitly differentiating with respect to time $t$, we find:

$$2x\frac{dx}{dt}=2h\frac{dh}{dt}$$

and solving for $$\frac{dx}{dt}$$, we find:

$$\frac{dx}{dt}=\frac{h}{x}\frac{dh}{dt}$$

Using the Pythagorean relation, we find:

$$x(h)=\sqrt{h^2-90^2}$$

and so we have:

$$\frac{dx}{dt}=\frac{h}{\sqrt{h^2-90^2}}\frac{dh}{dt}$$

and finally, we may compute:

$$\left.\frac{dx}{dt} \right|_{h=150}=\frac{150}{\sqrt{150^2-90^2}}\left.\frac{dh}{dt} \right|_{h=150}=\frac{5}{4}\cdot72\, \frac{\text{ft}}{\text{s}}=90\, \frac{\text{ft}}{\text{s}}$$

To Beverly and any other guests viewing this topic, I invite and encourage you to post other related rates problems in our http://www.mathhelpboards.com/f10/ forum.

Best Regards,

Mark.