Two boats leave the same port at the same time, with boat A traveling north at 20 knots (nautical miles per hour) and boat B traveling east at 40 knots. How fast is the distance between them changing when boat A is 4 nautical miles from port? Round your answer to two decimal places.
A)The distance between the boats is changing at the rate of 35.78 knots.
B)The distance between the boats is changing at the rate of 13.42 knots.
C)The distance between the boats is changing at the rate of 44.72 knots.
D)The distance between the boats is changing at the rate of 38.01 knots.
E)The distance between the boats is changing at the rate of 100.00 knots.

Question

Two boats leave the same port at the same time, with boat A traveling north at 20 knots (nautical miles per hour) and boat B traveling east at 40 knots. How fast is the distance between them changing when boat A is 4 nautical miles from port? Round your answer to two decimal places. ​
A)The distance between the boats is changing at the rate of 35.78 knots.
B)The distance between the boats is changing at the rate of 13.42 knots.
C)The distance between the boats is changing at the rate of 44.72 knots.
D)The distance between the boats is changing at the rate of 38.01 knots.
E)The distance between the boats is changing at the rate of 100.00 knots.

Quizplus · Accepted Answer

The distance between the boats can be found using the Pythagorean theorem, $d = \sqrt{x^2 + y^2}$, where $x$ and $y$ are the distances of the boats from the port along the east and north directions, respectively. Given $y = 4$ nautical miles (for boat A) and the rates $dx/dt = 40$ knots (for boat B) and $dy/dt = 20$ knots (for boat A), we can find $dd/dt$ using differentiation: $dd/dt = (1/2)(2x dx/dt + 2y dy/dt) / \sqrt{x^2 + y^2}$. At the time when $y = 4$, assuming $x = 8$ (since boat B travels twice as fast as boat A), we find $dd/dt = (1/2)(2*8*40 + 2*4*20) / \sqrt{8^2 + 4^2} = 44.72$ knots.

Two Boats Leave the Same Port at the Same Time