Optomization problem using integrals

  • Thread starter Quisquis
  • Start date
  • #1

Homework Statement

A boat leaves a dock at 2:00 pm and travels due south at a speed of 20 km/h. Another
boat has been heading due east at 15 km/h and reaches the same dock at 3:00 pm. At
what time were the two boats closest together?

3. The Attempt at a Solution [/b

We actually don't need to use integrals for this(and we haven't really learned how), but I want to.

So a couple of quick questions...
first, integrating 20 equals 20t+c correct?

and second, either the 15km/h or the 20km/h should be negative, and it doesn't matter which one, right?

If those two things are true, I should be all set, if not, I might have more questions. :p

Thanks a lot for the help guys.

Answers and Replies

  • #2
You're overthinking this problem. Both boats are travelling at constant rates, so the distance they travel in t hours is going to be 15t km and 20t km, using the formula d = rt. That's all your integration has bought you, plus you have two constants of integration to worry about.

If you haven't drawn a picture, you should, and maybe two of them, one for the positions of the two boats at 2:00 and another for their positions at 3:00. You need an expression that represents the distance between the two boats, as a function of t, and that's what you have to minimize, using differentiation.
  • #3
right, I have that position with the Pythagorean theorem. I have a^2+b^2=c^2. is a the position function of one boat and b the position function of the other boat? If that's the case, then I have c as a function of time if I integrate the velocities.

The integration constants aren't a problem, because I know the position of one boat at t=0 and the other at t=1 so I can solve for c pretty easily.

Then all I have to do is find the minimum of the Pythagorean function I created which looked to be 9/25.

I'd add more of the math I've done to make what I'm saying more clear, but with latex down... :/