Science:Math Exam Resources/Courses/MATH110/April 2017/Question 02 (c)

MATH110 April 2017

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Question 02 (c)
The side length of a cube is increasing at a rate of $2{\text{cm/s}}$ , how fast is the cube’s volume changing when the side length is $2{\text{cm}}$? Include units.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

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
You know the rate of change of the side length of the cube, and want to compute the rate of change of the volume of the cube. Because you are looking for one rate of change based on the value of another rate of change, this is a related rates problem. Try writing $V=s^{3}$ and taking the derivative of both sides with respect to time.

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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 begin with the equation $V=s^{3}$ for the volume of a cube. In order to compute the rate of change of V, we differentiate both sides with respect to the time variable t. This derivative must be taken before plugging in the value of s. ${\frac {dV}{dt}}=3s^{2}{\frac {ds}{dt}}$ We know that the value of s is 2 centimeters, and ${\frac {ds}{dt}}$ is 2 centimeters per second. We now plug in these values: ${\begin{aligned}{\frac {dV}{dt}}&=3(2{\text{cm}})^{2}(2{\frac {\text{cm}}{\text{sec}}})\\{\frac {dV}{dt}}&=24{\frac {{\text{cm}}^{3}}{\text{sec}}}\end{aligned}}$ Answer: <math>\color{blue} 24 \frac{\text{cm}^3}{\text{sec}}