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atakach99

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A rectangular block of ice with a square base has a height half the side of the base. As it melts, the volume of the block of ice is decreasing at the rate of 12 cm^3s^-1. Find the rate at which its surface area will be decreasing when the side of its base is 2.1 cm.

plz help my teacher didn't even know how to solve this
the correct solution at the back of the book is-30.48 i.e decreasing at the rate of 30.48 cm^2s^-1

thnks
 

lyounamu

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atakach99 said:
A rectangular block of ice with a square base has a height half the side of the base. As it melts, the volume of the block of ice is decreasing at the rate of 12 cm^3s^-1. Find the rate at which its surface area will be decreasing when the side of its base is 2.1 cm.

plz help my teacher didn't even know how to solve this
the correct solution at the back of the book is-30.48 i.e decreasing at the rate of 30.48 cm^2s^-1

thnks
Let the base = 2x and height = x (as given by the question)

Therefore, V = 2x . 2x . x = 4x^3
dV/dx = 12x^2

A = 2x . 2x . x + 2x . x . 4 = 16x^2
dA/dx = 32x

dV/dt = -12

dA/dt = dV/dt . dx/dV . dA/dx
= -12 . 1/12x^2 . 32x
= -384x/12x^2
When 2x = 2.1 (i.e. x = 1.05)
dA/dt = -30.461904...

Therefore, the rate at which the area changes is 30.461904... = 30.46 cm/s.

This question is unique in a way that you have to use 3 derivatives to achieve the answer. It is not dA/dt = dV/dt . dA/dt, it is rather dV/dt . dx/dV . dA/dx because you cannot find the entitry that exists between area and volume (as long as you can find it, try but you won't).

EDIT: This does not belong here. It should be in the MX1. I guess you could have got better explanation than mine if you posted up in MX1 section. But, meh.
 
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