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the surface area of a cylinder is increasing by 2 pi square inches per hour and the height is decreasing by 0.1 inches per hour when the radius is 16 inches and the height is 7 inches. how fast is the radius of the cylinder changing?

MIT 18.01 Single Variable Calculus (OCW)
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Just try using the chain rule, keeping in mind that Area = f(Height, Radius). In this case you know dA/dt and dH/dt at some particular moment. So, express dR/dt in terms of the stuff you know. The key word is - chain rule for partial derivatives.
If you'll have problems with that I can try to solve it for you, but I'll do it only after you demonstrate that you tried it on your own.
If I got everything right, the answer should be approximately 0.046 inches/hour.

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Other answers:

Write the equation/formula for the surface area. Take the derivative with respect to time, and do the substitutions so that you can solve for dr/dt.
I got .07"/hr. Is the cylinder a solid?

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