Related rates (two ships)

[jwpaine]

At noon, ship A is 10 nautical miles due west of ship B. Ship A is sailing west at 21 knots
(nautical miles per hour) and ship B is sailing north at 25 knots. How fast is the distance
between the ships changing at 7 p.m.?

This is what I have:

let h = hour

I create a right-angle triangle of sides \(\displaystyle a = 21(h) + 10\,\,\,\, b = 25(h)\,\,\) with hypotenuse length d

By Pythagorean theorem: \(\displaystyle d = \sqrt{ (21(h) + 10)^2 + (25(h))^2}\)

\(\displaystyle \frac{dd}{dh} = \frac{\sqrt{2}(533h + 105)}{\sqrt{533h^2 + 210h + 50}}\)

Therefore at h = 7 hours: \(\displaystyle \frac{dd}{dh} = \frac{\sqrt{2}(533(7) + 105)}{\sqrt{533(7)^2 + 210(7) + 50}}\)

Did I do that right?

 

[galactus]

Yes, you're correct. Good job. We need to differentiate implicitly.

\(\displaystyle D^{2}=x^{2}+y^{2}\)

\(\displaystyle D\frac{dD}{dt}=x\frac{dx}{dt}+y\frac{dy}{dt}\)

Now, After 7 hours, A has traveled 157 km and B has travelled 175 km.

Therefore, by Pythagoras, \(\displaystyle D=\sqrt{55274}\approx{235.1}\)

\(\displaystyle \sqrt{55274}\frac{dD}{dt}=(157)(21)+(175)(25)\)

\(\displaystyle \frac{dD}{dt}\approx{32.63} \;\ km/hr\)

You can certainly do it the way you done, but it's a little harder. But you have the correct answer.