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I am attempting a question which asks me to: "Use partial derivatives to find the percentage change in the volume enclosed by the cylinder". I have been told that the cylinder is of length $L$ and circular cross section of radius $r$, its length has increased by 3% and its radius decreased by 2%.

So I tried solving by using $dV = dL (\partial V/\partial L) + dr (\partial V/\partial r)$. I calculated the two partial derivatives and substituted them in to obtain: $dV = \pi r^2dL + 2\pi LrdR$. I have taken $dL = 0.03 $ and $dR = -0.02$. However this is where I become a little lost; When I substitute these values in I get the following: $dV = \pi r(0.03r - 0.04L)$ - Do I need to assume $r$ and $L$ are both 1 (that will give the percentage change of $-\pi$ which is not what I calculate if I use a calculator) or am I missing something in my logic?

Many thanks!

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You must substitute $dL=0.03L$ and $dr=-0.02r$. The percentage change of volume is then $dV/V$.

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