Linear Approximation and Volume of Cylinder

Question

I was given the question:

The volume V of a cylinder is computed using the values 6m for the diameter and 9.8m for the height. Use the linear approximation to estimate the maximum error in V if each of these values has a possible error of at most 7%.

I am unsure how to begin this problem. All I can assume is to use the volume formula. Any ideas?

Start by understanding the problem statement. What, for instance, does it mean by “the linear approximation?” — amd, Jun 23 '18 at 03:21
I am aware of the formula. In simple terms I know with a function it is used to approximate points at (a,b) with points close to such. — Neil H, Jun 23 '18 at 03:29

score 2 · Accepted Answer · answered Jun 23 '18 at 05:14

2

$$ V= \pi r^2 h $$

$$ dV= \frac {dV}{dr} dr + \frac {dV}{dh} dh$$ $$ = (2\pi rh) dr + (\pi r^2)dh$$

$$ \frac {dV}{V} = 2(dr/r)+ (dh/h) = 2(.07)+ (.07) = 0.21 $$

Thus the relative error in $V$ is $21$ percent.

answered Jun 23 '18 at 05:14

Mohammad Riazi-Kermani

68,728

Perfect explanation. Thank you!!!! – Neil H Jun 23 '18 at 05:20

Key Flex · Answer 2 · 2018-06-23T05:28:51.330

1

Volume of a cylinder $V=\dfrac{\pi}{4}d^2\cdot h$

Possible error in volume $$=\frac{\pi}{4}[(1\pm0.007)d]^2[(1\pm0.007)h]$$ $$=\frac{\pi}{4}[(1\pm0.007)^3]d^2\cdot h$$ $$=v(1\pm0.07)^3$$ Now the percentage error in v $=1.225042$v $=>+22.5043$%

edited Jun 23 '18 at 05:28

answered Jun 23 '18 at 04:23

Key Flex

9,475
7
17
34

Hello, that means V lies between v1.225043 and v0.804357. Now how to compute percentage of error? – Win_odd Dhamnekar Jun 23 '18 at 05:04
The error in v is either of the both. – Key Flex Jun 23 '18 at 05:09
If I calculated v correctly, I got 277.09. So multiplying it by v(1±0.07)^3 will give me the percentage error of V? – Neil H Jun 23 '18 at 05:12
The possible percentage error is in v is either $+22.5043$% or $-19.5643$% – Key Flex Jun 23 '18 at 05:14
What would be used as the experimental and actual value? The computer system is saying it is incorrect – Neil H Jun 23 '18 at 05:17

score 1 · Answer 3 · answered Jun 23 '18 at 05:21

For this kind of problems, logarithmic dfifferentiation is quite useful$$V=\dfrac{\pi}{4}d^2\, h\implies \log(V)=\log\left(\dfrac{\pi}{4}\right)+2\log(d)+\log(h)$$ Take the partial derivatives $$\frac 1 V\frac{\partial V}{\partial d}=\frac 2 d\qquad \text{and}\qquad \frac 1 V\frac{\partial V}{\partial h}=\frac 1 h$$ Going from $\partial$ to $\Delta$ and adding the contributions, we then have $$\frac{\Delta V} V=2\frac{\Delta d } d+\frac{\Delta h } h$$ Since the relative errors are the same, then $\frac{\Delta V} V$ is three times the relative error on each dimension.

Linear Approximation and Volume of Cylinder

3 Answers3