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?
Best Answer
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
$$c'\approx 25.56309\;.$$