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

MATH100 A December 2023

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[hide] 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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[show] 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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[show] Solution

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.