Science:Math Exam Resources/Courses/MATH100 A/December 2023/Question 12

MATH100 A December 2023

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Question 12
At noon, ship ${\textstyle A}$ is 4km west of ship ${\textstyle B}$. Ship ${\textstyle A}$ is sailing south at 2km/h and ship ${\textstyle B}$ is sailing north at 4km/h. The scenario some time after noon is pictured below. How fast is the distance between the ships changing 30 minutes after noon? Diagram of boats.

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If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint

We need to come up with an expression that represents the distance between Ship A and Ship B. Let ${\textstyle d}$ be the distance between Ship A and Ship B, and say that “up” is the positive direction, and at noon when the ships are 4km apart, the ${\textstyle y}$-coordinate of their position is 0. Looking at the upper figure, we could re-arrange the paths of ships A and B to create a right-angle triangle (lower figure). distance between ships distance between ships rearranged

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. We need to come up with an expression that represents the distance between Ship A and Ship B. Let ${\textstyle d}$ be the distance between Ship A and Ship B, and say that “up” is the positive direction, and at noon when the ships are 4km apart, the ${\textstyle y}$-coordinate of their position is 0. Looking at the upper figure, we could re-arrange the paths of ships A and B to create a right-angle triangle (lower figure). distance between ships distance between ships rearranged Using this, we can write an expression for ${\textstyle d}$ that depends on the ${\textstyle y}$-coordinate of the positions of ships A and B: $$d={\sqrt {4^{2}+(p_{B}-p_{A})^{2}}}$$ To determine how fast the distance between the ships is changing, we take the derivative of ${\textstyle d}$ with respect to time. $${\begin{aligned}{\frac {d}{dt}}d&={\frac {d}{dt}}{\sqrt {4^{2}+(p_{B}-p_{A})^{2}}}\\&={\frac {1}{2}}(16+(p_{B}-p_{A})^{2})^{-1/2}\left(2(p_{B}-p_{A})({\frac {dp_{B}}{dt}}-{\frac {dp_{A}}{dt}})\right)\\&={\frac {(p_{B}-p_{A})({\frac {dp_{B}}{dt}}-{\frac {dp_{A}}{dt}})}{\sqrt {16+(p_{B}-p_{A})^{2}}}}\end{aligned}}$$ So, now we need to determine ${\textstyle p_{A}}$, ${\textstyle p_{B}}$, ${\textstyle {\frac {dp_{A}}{dt}}}$, and ${\textstyle {\frac {dp_{B}}{dt}}}$ 30mins after noon. Say at noon, time ${\textstyle t=0}$ and measure time in hours. Then, 30mins later ${\textstyle t=0.5}$. We know ship A and ship B are travelling at a constant speed, so ${\textstyle {\frac {dp_{A}}{dt}}=-2}$ km/h and ${\textstyle {\frac {dp_{B}}{dt}}=4}$ km/h. The ${\textstyle y}$-coordinate of their positions can be determined using ${\textstyle speed={\frac {distance}{time}}}$, so: $${\begin{aligned}p_{A}(0.5)&=(-2km/h)(0.5h)=-1km\\p_{B}(0.5)&=(4km/h)(0.5h)=2km\end{aligned}}$$ Now we can calculate how fast the distance between the ships is changing: $${\begin{aligned}d(0.5h)&={\frac {(2-(-1))(4-(-2))}{\sqrt {16+(2-(-1))^{2}}}}\\&={\frac {18}{5}}\end{aligned}}$$ Therefore, the distance between the ships is changing at ${\textstyle {\frac {18}{5}}}$ km/h.