Max/Min Application of Derivatives Problem

[mathnerd11]

A North-South highway intersects an East-West highway at a point P. An automobile crosses P at 10:00 am, traveling east at a constant speed of 20 mph. At the same instant another automobile is 2 miles north of P, traveling south at 50 mph. Find the time at which they are closest to each other and approximate the minimum distance between the automobiles.

Okay, so I don't really know where to start. I drew a picture, and thought I should maybe start with pythagorean theorem but... all in all I have no idea. Please help? Thanks!

 

[mmm4444bot]

I don't really know where to start. I drew a picture, and thought I should maybe start with pythagorean theorem

That's a good start.

This scenario involves an infinite number of right triangles; you need to find only one, the one with the shortest hypotenuse.

To facilitate discussion, I'll position these triangles with their right angle at the origin and their hypotenuse in Quadrant I.

At time = 0, there is no triangle. The eastbound car is at the origin, and the southbound car is two miles away, at the point (0, 2).

The instant time begins to elapse (t > 0), right triangles begin forming.

As time elapses, the vertical side of the triangle is getting smaller, by the distance that the 50-mph car goes during time t.

So, to express the length of the vertical side of the triangle, write 2 - d, where you need to express d using the rate and the time variable, as in the classic relationship d = rt.

Likewise, the base length is changing, too. It's growing over time. Since the 20-mph car is already at zero when t = 0, the length of the horizontal side is simply the distance the car goes during t hours, yes? Again, use d = rt.

Now you have expressions for each leg, in terms of t.

Putting these two expressions into the Pythagorean relationship yields a quadratic polynomial which equals the square of the hypotenuse, for any given time t.

When the square of the hypotenuse is smallest, the distance between the cars is the shortest. So, you're looking for the vertex point on a parabola that opens upward.

In other words, the t-coordinate of the vertex point is the amount of elapsed time required for the two cars to be positioned closest.

Oh, wait. You're in an introductory calculus course?

Then they probably want to see you write a function for the hypotenuse square, followed by determining it's first derivative. You've learned that the derivative is the slope of tangent lines on the parabola? So, the tangent line at the parabola's vertex has a slope of zero.

You need to find the value of t which causes the first derivative to evaluate to zero.

Is this enough description for you to proceed?

We welcome specific questions, especially if I wrote something that you do not understand.

 

[BigGlenntheHeavy]

\(\displaystyle Using \ P \ (origin) \ as \ a \ reference, \ t \ in \ hrs., \ t(0) \ = \ 10AM, \ d \ in \ miles \ we \ have;\)

\(\displaystyle d(t) \ = \ [(20t)^{2}+(50t-2)^{2}]^{1/2} \ = \ [2900t^{2}-200t+4]^{1/2}\)

\(\displaystyle Hence, \ d(1) \ = \ 52 \ miles, \ in \ other \ words \ after \ 1 \ hr. (11AM), \ they \ will \ be \ 52 \ miles \ apart.\)

\(\displaystyle But \ we \ want \ to \ know \ at \ what \ time \ they \ will \ be \ the \ closest.\)

\(\displaystyle Hence, \ \frac{d[d(t)]}{dt} \ = \ \frac{1}{2}[2900t^{2}-200t+4}]^{-1/2}(5800t-200)\)

\(\displaystyle Setting \ the \ slope \ = \ to \ zero, \ we \ get \ t \ = \ \frac{1}{29}hrs \ = \ 2 \ min. \ 4 \ sec.\)

\(\displaystyle And \ d\bigg(\frac{1}{29}\bigg) \ = \ \frac{4\sqrt{29}}{29} \ = \ about \ .74278 \ miles.\)

\(\displaystyle Therefore \ at \ 10:02:04 \ AM, \ they \ will \ be \ about \ .74278 \ miles \ apart.\)