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A cylinder is being flattened so that its volume does not change. Find the rate of change of radius when r = 2 inches and h = 5 inches, if the height is decreasing at 0.7 in/sec. Hint: what is the rate of change of volume?

User Cnherald
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1 Answer

6 votes

Answer:


(dr)/(dt) = 0.14 in/s

Step-by-step explanation:

As the volume of the cylinder is constant here so we can say that its rate of change in volume must be zero

so here we can say


(dV)/(dt) = 0

now we have


V = \pi r^2 h

now find its rate of change in volume with respect to time


(dV)/(dt) = 2\pi rh(dr)/(dt) + \pi r^2(dh)/(dt)

now we know that


(dV)/(dt) = 0 = \pi r(2h (dr)/(dt) + r(dh)/(dt))

given that

h = 5 inch

r = 2 inch


(dh)/(dt) = - 0.7 in/s

now we have


0 = 2(5) (dr)/(dt) + 2(-0.7)


(dr)/(dt) = 0.14 in/s

User Henrywright
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