Approximate small change problem

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Jade1886

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Jul 20, 2019
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Hi, i'll get straight to the point, I am studying for an exam and I came across this question which has me stumped:
"A pipe with diameter 4m is covered by 74mm of insulation, what is the approximate increase in the cross sectional area of the pipe?"

So far this is what i've came up with, (small change formula gives: dA ~ dy/dr x dr)
dy/dx = 2pi(2) and (dr = 2.37 - 2) <-- this is the radius when d=4.74, - the radius when d=4.

The answer is 3/4 m/s but I cannot seem to find the method used to obtain this answer, if anyone has a solution to this question i'd be so thankful as I cannot find any information on this!
 
Hello, and welcome to FMH! :)

Yes, we can use:

[MATH]\frac{\Delta A}{\Delta r}\approx\frac{dA}{dr}[/MATH]
Or:

[MATH]\Delta A\approx\frac{dA}{dr}\Delta r[/MATH]
We have:

[MATH]\frac{dA}{dr}=2\pi r[/MATH]
[MATH]r=2\text{ m}[/MATH] (recall the radius is half the diameter)

[MATH]\Delta r=0.074\text{ m}[/MATH]
Hence:

[MATH]\Delta A\approx2\pi(2\text{ m})(0.074\text{ m})=\frac{37\pi}{125}\,\text{m}^2[/MATH]
 
Jade1886 said:
Hi, i'll get straight to the point, I am studying for an exam and I came across this question which has me stumped:
"A pipe with diameter 4m is covered by 74mm of insulation, what is the approximate increase in the cross sectional area of the pipe?"

So far this is what i've came up with, (small change formula gives: dA ~ dy/dr x dr)
dy/dx = 2pi(2) and (dr = 2.37 - 2) <-- this is the radius when d=4.74, - the radius when d=4.

The answer is 3/4 m/s but I cannot seem to find the method used to obtain this answer, if anyone has a solution to this question i'd be so thankful as I cannot find any information on this!
Click to expand...
I think you looked up the answer to the wrong problem. The units m/s are entirely wrong!

As a check of MarkFL's answer (which does have the right units), consider that the exact change in area is

[MATH]\pi(2.074)^2 - \pi(2)^2 = 13.513 - 12.566 = 0.947[/MATH],​

while the approximation works out to 0.9299, which is reasonably close. It is not close to 3/4.
 
Dr Peterson you are correct, I referred back to my book straight after the reply was posted and realised my grave mistake, now I feel like such a doofus for posting before rechecking the answer but thanks very much for the helpful reply
 
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