Can someone please help me with the 3 calculus one word problems below, I've done all three of them several times but can never seem to arrive at the correct answer. This is greatly frustrating because these questions seem simple enough but I've wasted so much time on them. Thank you.
1)Fred the spherical cow is happily grazing on cubical grass pellets. He grows in volume at a rate of 20 cubic feet per day.
When Fred's radius is 14 feet, at what rate is his surface area growing?
2) Fred the spherical cow is happily grazing on cubical grass pellets. He grows in volume at a rate of 8 cubic feet per day.
When Fred's radius is 11 feet, at what rate is his surface area growing?
3) Fred the spherical cow is happily grazing on cubical grass pellets. He grows in volume at a rate of 11 cubic feet per day.
When Fred's radius is 6 feet, at what rate is his surface area growing?
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$\begingroup$Its volume is given by $$V=\frac{4\pi r^3}{3}$$ Differentiate with respect to $r$: $$\frac{dV}{dr}=4\pi r^2$$ By the chain rule you have $$\frac{dV}{dt}=\frac{dV}{dr}\frac{dr}{dt}=4\pi r^2\frac{dr}{dt}=20$$ Solve for $\frac{dr}{dt}$: $$\frac{dr}{dt}=\frac{20}{4\pi r^2}=\frac{5}{\pi r^2}$$
Its surface area is given by $$A=4\pi r^2$$ Differentiate with respect to $r$: $$\frac{dA}{dr}=8\pi r$$ Now use chain rule to get the rate at which the area changes with respect to time: $$\frac{dA}{dt}=\frac{dA}{dr}\frac{dr}{dt}=8\pi r\frac{dr}{dt}=8\pi r\cdot\frac{5}{\pi r^2}=\frac{40}{r}.$$
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