Differentials and tolerances

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SUMMARY

The discussion focuses on calculating the required accuracy for measuring the interior diameter of a 10-meter high cylindrical storage tank to ensure the volume is calculated within 1% of its true value. The volume formula used is V = (5/2)πl², where V represents volume and l denotes diameter. The analysis reveals that the measurement error (dl) must not exceed 0.5% of the true diameter to maintain the desired accuracy. An alternative approach suggests that the fractional error in volume is approximately double that of the linear measurement.

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Homework Statement



About how accurately must the interior diameter of a 10-m high cylindrical storage tank be measured to calculate the tank's volume to within 1% of its true value?

Homework Equations



[itex]V=\frac{5}{2}\pi l^{2}[/itex], where [itex]V[/itex] is volume and [itex]l[/itex] is diameter.
[itex]dV=5\pi l \ dl[/itex]

The Attempt at a Solution



I'm really uncertain as to how to go about this problem. What follows is the textbook's method for a similar problem translated into this problem's terms.

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We want any inaccuracy in our measurement to be small enough to make the corresponding increment [itex]\Delta V[/itex] in the volume satisfy the inequality

[itex]|\Delta V|\leq\frac{1}{100}V=\dfrac{\pi l^{2}}{40}[/itex].

We replace [itex]\Delta V[/itex] in this inequality by its approximation

[itex]dV=\left(\dfrac{dV}{dl}\right)dl=5\pi l \ dl[/itex].

This gives

[itex]|5\pi l\ dl|\leq\dfrac{\pi l^{2}}{40}[/itex], or [itex]|dl|\leq\dfrac{1}{5\pi l}\cdot\dfrac{\pi l^{2}}{40}=\dfrac{1}{5}\cdot\dfrac{l}{40}=0.005l[/itex].

We should measure [itex]l[/itex] with an error [itex]dl[/itex] that is no more than 0.5% of its true value.

--

I need some clarification for this solution. Could somebody annotate it, or perhaps write up a more intuitive one?
 
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Your answer is correct. I would tend to start from the other end. V = πl2h/4. If there's a fractional error δ in l, the computed volume will be π(l(1+δ))2h/4 = V(1+δ)2 = V + 2Vδ + Vδ2 ≈ V + 2Vδ. So the fractional error in the volume will be about double that in the linear measurement.
 

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