Science:Math Exam Resources/Courses/MATH110/April 2012/Question 08

MATH110 April 2012

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[hide] Question 08
Oceanographers on research ships use a device called a CTD probe to make underwater measurements at extreme depths. The probe drops vertically through the water at 4 m/s. Simultaneously, the research ship moves away from the drop site at 5 m/s. What is the rate of change of the distance between the probe and the ship 10 seconds after the probe is dropped?

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If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

[show] Hint 1
Draw a picture.

[show] Hint 2
Can you find an equation that relates the distance between the probe and ship with the distances of the probe and ship from the drop point?

[show] Hint 3
Use Pythagoras.

[show] Hint 4
In order to find the relationship between the time derivatives of these variables, implicitly differentiate your equation with respect to time.

[show] Hint 5
Note that we are trying to find the rate of change of the distance between the probe and ship, at the particular time of 10 s after the drop.

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[show] 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. Let $x$ be the distance of the ship from the drop point, $y$ be the distance of the probe from the drop point, and $r$ be the distance between the ship and the probe. Drawing a labelled picture looks like this: We want to find ${\frac {dr}{dt}}$ , at time 10s after the probe is dropped. We know that ${\frac {dx}{dt}}=5m/s$ and ${\frac {dy}{dt}}=4m/s$ . By the Pythagorean Theorem, $x^{2}+y^{2}=r^{2}.$ In order to find the relationship between the time rates of change of these variables, we differentiate both sides of this equation with respect to time, in order to obtain $2x{\frac {dx}{dt}}+2y{\frac {dy}{dt}}=2r{\frac {dr}{dt}}$ Dividing both sides by $2r$ in order to isolate ${\frac {dr}{dt}}$ , we find ${\frac {dr}{dt}}={\frac {x{\frac {dx}{dt}}+y{\frac {dy}{dt}}}{r}}.$ We are interested in the value of ${\frac {dr}{dt}}$ , at time $10s$ after the probe is dropped. At this point in time, $x=5m/s\cdot 10s=50m,$ $y=4m/s\cdot 10s=40m,$ and ${\begin{aligned}r&={\sqrt {x^{2}+y^{2}}}={\sqrt {(50m)^{2}+(40m)^{2}}}\\&=10{\sqrt {5^{2}+4^{2}}}\ m=10{\sqrt {41}}\ m.\end{aligned}}$ Hence, ${\begin{aligned}{\frac {dr}{dt}}&={\frac {(50m)(5m/s)+(40m)(4m/s)}{10{\sqrt {41}}m}}\\&={\frac {25m/s+16m/s}{\sqrt {41}}}\\&={\sqrt {41}}\ m/s.\end{aligned}}$ That is, $10s$ after the probe is dropped, the distance between the probe and ship is increasing at a rate of ${\sqrt {41}}\ m/s$ .