percentage decrease of the edge of an icecube

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I have a question that is asking to find an approximation for the percentage the the edge length of an ice cube will decrease if the cube loses six percent of its volume. The question instructs us to use differential (linear) approximation

My maths background is a bit patchy I'm afraid and I have no real idea what differential approximation is. I have watched a few video on the topic but cant seem to figure out the right way to start this question.

Any hints will be much appreciated.

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Think of volume, $V=x^3$ where $x$ is the length of each side. Differential calculus says, $dV=3x^2.dx$.

Dividing both sides, $\dfrac{dV}{V}=\dfrac{3x^2.dx}{x^3}$

Or, $-6=3\dfrac{dx}{x}$

Hence, $\dfrac{dx}{x}=-2$, that is, percentage decrease in edge length is 2%.

Note that, 1) percentage decrease in any quantity is given by dividing the differential of the quantity by the quantity itself 2) If you are comfortable with logs, taking logs at the first step before taking differentials, makes it slightly neater.

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$V_0=L_0^3$ is the original volume. The final volume is $V=(L_0-\Delta L)^3$ $\Delta V /V_0 =0.06$ You explicitly expand $1-V/V_0$, and obtain an expression from where you can calculate $\Delta L$. If the change is small, only the first term is necessary. You can get the same expression by differentiating $V(L)=L^3$ with respect to $L$. $$\Delta V /V=\partial V/\partial L *\Delta L/V=3 L^2 \Delta L/L^3= 3\Delta L /L $$ So $\Delta L /L=0.02$