Related Rate Prob - Ships sailing in different directions

• November 5th 2010, 09:15 PM
dbakeg00
Related Rate Prob - Ships sailing in different directions
Ship A is 15 miles east of point O and moving west at 20mi/h. Ship B is 60 miles south of O and moving north at 15mi/h. a)Are they approaching or seperating after 1 hour and at what rate? b)after 3hrs?

Let D=distance between the ships at time t
$D^2=(60-15t)^2+(15-20t)^2$

$2D*\frac{dx}{dt}=(2)(-15)(60-15t)+(2)(-20)(15-20t)$

$2D*\frac{dx}{dt}=(-30)(60-15t)+(-40)(15-20t)$

$2D*\frac{dx}{dt}= 1250t-2400$

$\frac{dx}{dt}=\frac{1250t-2400}{2D}$

$D=\sqrt{5^2+45^2}=5\sqrt{82}$

$\frac{dx}{dt}=\frac{1250t-2400}{10\sqrt{82}}$

a) $\frac{dx}{dt}=\frac{1250*1-2400}{10\sqrt{82}}=approaching\ at \frac{-115}{\sqrt{82}}$ mi/h

b) $\frac{dx}{dt}=\frac{1250*3-2400}{10\sqrt{82}}=seperating\ at \frac{135}{\sqrt{82}}$ mi/h

The book agrees with me on part A, but on part B it says the answer should be $seperating\ at\ \frac{9\sqrt{10}}{2}$ mi/h. Am I missing something obvious or is the book wrong here?
• November 6th 2010, 02:24 AM
Ackbeet

1. On a point of notation, your D = x. I would consolidate to one symbol.
2. The reason your answer to b) is incorrect is because you used the same distance between the two ships as in part a). They're probably not going to be the same distance apart, are they? Find out where the ships are, and re-compute your D. See what that gives you.
• November 6th 2010, 02:43 AM
dbakeg00
1. Thanks for pointing that out. It just leads to sloppiness that causes errors.

2. Thanks for this as well. I recalculated D to $15\sqrt{10}$ which made my final answer match the book's $\frac{9\sqrt{10}}{2}$

3. Its kind of odd that the author of this book was willing to leave a radical in the denominator for the answer to part a but made sure to fix it in part b.
• November 6th 2010, 02:50 AM
Ackbeet
Re: 3.

Well, 82 is 2 * 41, and 41 is a large prime number. That makes it a bit awkward to resolve, though not overmuch. You can always take $1/\sqrt{82}=\sqrt{82}/82.$ I don't know why the author didn't do that, except that, since the radical is not capable of being simplified in such a way as to reduce the total number of symbols, maybe they thought it was easier to leave it in the denominator. Who knows?

In the end, you've got yourself two kinds of people to whom you might report this answer. The first is the mathematician. He's going to want an exact answer, since one is available. He probably won't care which one you pick, as long as they're correct. The second is the engineer. He's going to chew you out unless you give him a decimal number approximation, because he doesn't want to have to think to himself that $\sqrt{82}$ is just slightly greater than 9, and then he has to divide by 9. It's a lot easier to compare decimal approximations.