Question

(1 point) At noon, ship A is 10 nautical miles due west of ship B. Ship A is sailing west at 22 knots and ship B is sailing north at 19 knots. How fast (in knots) is the distance between the ships changing at 6 PM?

Answer 1

Answer:28.8 knots

Explanation:

The ships are moving as the sides of a right triangle. Thus, Pyhogorean theorem will be useful in the following steps. Next, we have to know that the rate of change in distance, which is called velocity, can be described in terms of derivatives.

First, we have to calculate the distances covered by the ships from noon to 6 PM. In 6 hours, ship A moved 22*6=132 nautical mile. However, their first distance was 10 nautical miles, so 132+10=142 miles is the equivalent of A's displacement. For B, the distance travelled is 19*6=114 miles. From now on, A=142 miles and B=114 miles.

The distance between them is described with Pythogorean theorem, which is $D=\sqrt{A^{2} +B^{2} }$ and when we replace the values A and D, we find Distance (D) to be 182 miles.

Now, let's make the notations clear. The velocity of A and B is notated as $\frac{dA}{dt}$ and $\frac{dB}{dt}$. The rate of change of distance is also notated as $\frac{dD}{dt}$. Now, we have to find $\frac{dD}{dt}$ from the Pythogorean theorem. If we derive the Pythogorean expression $D=\sqrt{A^{2} +B^{2} }$ , we would have:

$\frac{dD}{dt} =\frac{1}{2} *(A^{2} +B^{2} )^{-1/2} *(2*A*\frac{dA}{dt} + 2*B*\frac{dB}{dt} )$

The derivation here includes chain rule and derives the interior parts of the parenthesis. When we insert distances for A and B and velocities for derivation notations, the formula becomes:

$\frac{dC}{dt} =\frac{1}{2}*(142^{2} +114^{2})^{-\frac{1}{2} }*(2*142*22 + 2*114*19)$ and the answer is 28.6 knots.