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Calculus/Calc Related Rates
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Question
The volume of a cube is expanding at a constant rate of 2cm^3/sec. How fast is the length of an edge changing when t=1? When t=5? Why the difference?
The volume V(t) is equal to the side s(t) to the third.
That is, V(t) = s³(t).
Given this, the change in volume is dV/dt = 3s²(t)(ds/dt).
We are given that dV/dt = 2, so 2 = 3s²(t)(ds/dt).
That means ds/dt = 2/(3s²(t)).
As t increases, our volume is increasing, so s(t) is increasing as well.
As V(t) increases, so does s(t). This means that since s(t) is in the denominator,
ds/dt would be increasing even slower.
When t=1, ds/dt=2/3. When t=5, there are 10 cm³.
This means the side length is the cube root(10).
This says ds/dt = 2/(3*cuberoot²(10)), or (2/3)(10^(-2/3)).
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Scotto
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