
Sam A.
asked 11/10/23Calculus I Related Rates (Distance between two sailing ships)
At noon, ship A is 50 nautical miles due west of ship B. Ship A is sailing west at 25 knots and ship B is sailing north at 22 knots. How fast (in knots) is the distance between the ships changing at 4 PM?
(Note: 1 knot is a speed of 1 nautical mile per hour.)
1 Expert Answer

Amanda S. answered 11/11/23
Experienced College-Level Math Tutor
If you were to draw a picture of the movement of these ships, the drawing would create a right triangle, with A heading west (left) and B heading north (up). The line connecting the two ships is like the hypotenuse. The question is asking us to find the rate at which ship A and ship B are moving apart at 4PM, which is essentially the change in the hypotenuse.
At 4 pm, we know that ship B will have moved 22 knots or 22 nautical miles per hour. In 4 hours, ship B will have traveled 22*4 = 88 nautical miles. We'll call this leg b. We also know that ship A will have moved 25 knots, or 25 *4, plus that initial 50 nm it was already west of ship b. So leg a of the triangle is 150 nautical miles.
We have a right triangle with two sides known, so lets find the hypotenuse c using the Pythagorean theorem. This should give us c = 173.91.
Now for the calculus part, we can take the derivative of a2 + b2 = c2. This is because it is an equation that represents and relates the position of the two ships, so once we take the derivative of it, it will represent the change in the position of ships with respect to time.
So take the derivative with respect to t, which uses implicit differentiation :
2a *a' + 2b * b'= 2c * c'
We're trying to find c', or the change in the distance between ship A and B over time. All the other things in the equation can be substituted with information we were given or found We found the a, b, and c distances, and we know a' and b' because that is the speed at which the ships are traveling west and north. So substitute them in!
2(150)*(25) + 2(88)*(22) = 2 (173.91) * c'
This should give us c' = 32.7. So this means, at 4PM, the distance between ship A and ship B is changing by 32.7 knots.
Sam A.
Beautiful explanation! Thank you so much for going into detail; it truly helps a ton :)11/11/23
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ABDOULAYE M.
11/10/23