Word prob: A boat is on an island that is 3 miles from shore

tarynt1

New member
Joined
Sep 30, 2006
Messages
23
I posted this problem before, but it had several crucial errors in the wording since I was trying to write it from memory. Now I have the actual problem and need some help. Sorry for the confusion before!

A boat is on an island that is 3 miles from shore. A person on the island is severely injured and must be transported by boat to a waiting helicopter on the beach. The hospital is 40 miles inland and 10 miles down the shore from the island. The boat's rate is 40mph and the helicopter's rate is 120 mph. How far down the beach should the helicopter wait in order to minimize the time to get to the hospital?

It's been determined that the total time (T) = (sqrt(x^2+9))/(40) + (sqrt(10-x)^2+(40^2))/(120). I wish I could show the diagram as well, but I'm not sure how.

I took the derivative of that equation and found that the critical number was about 0.238. But since it was a pretty complicated derivative (for me, at least), I'd appreciate it if someone could check my answer. My final derivative was T' = (1/120)((3x)/sqrt(x^2+9) + (x-10)/sqrt(x^2-20x+1700)). It was then that I plugged the equation into my graphing calculator and saw the only zero, 0.238.

I also have to take the second derivative of T', but I'd appreciate it if someone could first tell me if my current answer is right or wrong before I start doing more work.

Thanks!!
 
Re: Word prob: A boat is on an island that is 3 miles from s

Hello, tarynt1!

A boat is on an island that is 3 miles from shore.
A person on the island is severely injured and must be transported by boat
to a waiting helicopter on the beach.
The hospital is 40 miles inland and 10 miles down the shore from the island.
The boat's rate is 40 mph and the helicopter's rate is 120 mph.
How far down the beach should the helicopter wait in order to minimize the time
to get to the hospital?

It's been determined that the total time is:
. . \(\displaystyle \L\:T \;= \;\frac{\sqrt{x^2\,+\,9}}{40} \,+\,\frac{\sqrt{(10-x)^2\,+\,40^2}}{120}\;\;\) . . . Correct!

I took the derivative of that equation.
I'd appreciate it if someone could check my answer.

My final derivative was: \(\displaystyle \L\:T' \:= \:\frac{1}{120}\left[\frac{3x}{\sqrt{x^2\,+\,9}} \,+\,\frac{x\,-\,10}{\sqrt{x^2\,-\,20x\,+\,1700}}\right]\;\;\) . . . Right!

It was then that I plugged the equation into my graphing calculator
and saw the only zero, 0.238.

Right on target . . . lovely work!

 
Top