At noon, ship A is 50 nautical miles due west of ship B. Ship A is sailing west at 18 knots and ship B is sailing north at 19 knots. How fast (in knots) is the distance between the ships changing at 5 PM?

Okay, cool. I got this:

Let's setup our variables:

$A(t) = 50 + A'(t)$

$A'(t) = 18t$

$B(t) = 0 + B'(t)$

$B'(t) = 19t$

$C = ?$

$C' = what\; we\;want$

Let's think about the related rate, and the relationship governing the two of them, one is going north and one is going west. When they start they are 50 nm apart from each other. This forms a right triangle. We're looking at the Pythagorean Theorem:

$c^2 = a^2 + b^2$

Let's differentiate the equation relating ship a to ship b and the distance between them:

$2c(c') = 2a(a') + 2b(b')$

Awesome. Now let's solve for c.

$c^2 = a^2 + b^2\\ c^2 = A(5)^2 + B(5)^2\\ c^2 = 140^2 + 95^2 c^2 = 19600 + 9025\\ c^2 = 28625\\ c = \sqrt{28625} \approx 169.1892\\ $

Now, let's substitute:

$ 2(\sqrt{28625})(c') = 2(140)(18) + 2(95)(19)\\ 2\sqrt{28625}(c') = 3240 + 3610\\ 2\sqrt{28625}(c') = 6850\\ c' = \frac{6850}{2\sqrt{28625}} \approx \frac{6850}{338.3784} \approx 20.2436 $

Right? Nope. Wrong.

Anyway, I tried several other different versions and they didn't work either. Could some enlightened one please help?


3 Answers 3


Everything is fine up through

$$2(\sqrt{28625})(c') = 2(140)(18) + 2(95)(19)\;,$$

though I’d have divided through by $2$ before this point. At that point your arithmetic went astray: $2(140)(18)=5040$, not $3240$, so the righthand side should be $8650$ instead of $6850$. Solving the corrected equation gives you

$$c'\approx 25.56309\;.$$

  • $\begingroup$ Thank you so much for the explanation. My wonderful -- and when I say wonderful, I mean it -- calculus professor makes all his tests and quizzes with pretty numbers and beautiful solutions. Consequently, we're not permitted to use calculators on exams (and for good reason, too). WebAssign, and the homework problems in the book, rarely adopt this philosophy. Paying $200 for a textbook is not enough compensation for the author to make problems with pretty numbers. It's been so long since I've done base 10 arithmetic by hand that, I've discovered, I'm prone to errors. $\endgroup$
    – alvonellos
    Oct 9, 2013 at 19:09
  • 1
    $\begingroup$ @alvonellos: You’re welcome. (Much as I deplore the cost of math textbooks, I suspect that the ugly numbers are at least partly a deliberate attempt to inject a note of reality into the exercises.) $\endgroup$ Oct 9, 2013 at 19:47

I'm having a hard time wading through your work, but I can tell you that I get

$$d(t) = \sqrt{2500+1800 t+ 685 t^2}$$

$$d'(t) = \frac{1800+1370 t}{2 \sqrt{2500+1800 t + 685 t^2}} $$

$$d'(5) = \frac12 \frac{8650}{\sqrt{28625}}$$


I would recommend drawing a picture of where the ships are at noon and where they are and at 5:00 PM. Your second picture should be a right triangle with side lengths:

$$a= 50 \ mi + 18 \ knots * 5 \ h = 140 \ mi,$$ $$b= 19 \ knots * 5 \ h = 95 \ mi.$$

Then use Pythagorean Theorem to find the hypotenuse.

$$c^2=a^2+b^2$$ $$c^2=140^2+95^2$$ $$c^2=28,625$$ $$c=\sqrt{28,625} \ mi$$

You also know that $\frac{da}{dt}=18 \ knots$ because ship A is moving $18 \ knots$ and making side $a$ larger as it moves. Similarly, $\frac{db}{dt}=19 \ knots$ because side $b$ of your triangle is growing at that rate.

Now all you need to do is differentiate the Pythagorean Theorem with respect to time, solve for $\frac{dc}{dt}$, and plug in the other values.

$$c^2=a^2+b^2$$ $$2c\frac{dc}{dt}=2a\frac{da}{dt}+2b\frac{db}{dt}$$ $$c\frac{dc}{dt}=a\frac{da}{dt}+b\frac{db}{dt}$$ $$\frac{dc}{dt}=\frac{a\frac{da}{dt}+b\frac{db}{dt}}{c}$$ $$\frac{dc}{dt}=\frac{140*18+95*19}{\sqrt{28,625}}\approx 25.563 \ knots$$

Here's a link to a site with a much more detailed explanation of how to solve a problem like this. This site has a good explanation of how to draw your picture and go through the problem.



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