Using differentials to solve propagated error problem?

May 2010
5
0
I have a practice final and about 10 people in my Calculus class rely on me to help them study for tests, but there is one question I'm stuck on... (might have something with me being absent the day the prof. went over propagated error)

"The instrument used for measuring the height of a cylinder is accurate to the nearest 0.02 cm. The instrument ised for measuring the radius of a cylinder is accurate to the nearest 0.01 cm. A cylinder is measured to have a radius of 6 cm and a height of 20 cm. Use differentials to find (A) the percent error in calculating the surface area and (B) the percent error in calculating the volume of the cylinder."

I know...
(A)
Surface Area = \(\displaystyle S=2\)\(\displaystyle pi\)\(\displaystyle r^2\)\(\displaystyle +2\)\(\displaystyle pi\)\(\displaystyle rh\)

and

(B)
Volume = \(\displaystyle V=\)\(\displaystyle pi\)\(\displaystyle r^2\)\(\displaystyle h\)

Not really sure where to go from here... I assume solve for one variable (like r for h or h for r then plug in and do the derive... but I'm not 100% and before I tell 10 people the wrong way I want to check!!! Any help would be greatly apperciated!!!

P.S. The final is Wednesday, and we have another study group meeting tonight.
 
Last edited:

dwsmith

MHF Hall of Honor
Mar 2010
3,093
582
Florida
I have a practice final and about 10 people in my Calculus class rely on me to help them study for tests, but there is one question I'm stuck on... (might have something with me being absent the day the prof. went over propagated error)

"The instrument used for measuring the height of a cylinder is accurate to the nearest 0.02 cm. The instrument ised for measuring the radius of a cylinder is accurate to the nearest 0.01 cm. A cylinder is measured to have a radius of 6 cm and a height of 20 cm. Use differentials to find (A) the percent error in calculating the surface area and (B) the percent error in calculating the volume of the cylinder."

I know...
(A)
Surface Area = \(\displaystyle S=2\)\(\displaystyle pi\)\(\displaystyle r^2\)\(\displaystyle +2\)\(\displaystyle pi\)\(\displaystyle rh\)

and

(B)
Volume = \(\displaystyle V=\)\(\displaystyle pi\)\(\displaystyle r^2\)\(\displaystyle h\)

Not really sure where to go from here... I assume solve for one variable (like r for h or h for r then plug in and do the derive... but I'm not 100% and before I tell 10 people the wrong way I want to check!!! Any help would be greatly apperciated!!!

P.S. The final is Wednesday, and we have another study group meeting tonight.
Differentiate Surface Area and Volume.

\(\displaystyle \frac{dS}{dr}=....\) multiple both sides by dr.

\(\displaystyle dr=\pm\)error
Plug in numbers and solve
 
May 2010
5
0
Can anyone tell me if I'm headed down the right path???

from what I have gather I need to solve for a variable, then substitute and get the derivative, once I have that solve using differentials (a.k. ds=<something>dr). Then plug in the values. The problem is I keep getting 0 for the derivative…

Ok, first I need to get the surface area.

\(\displaystyle S=2\pi*r^2+2\pi*r*h\)
\(\displaystyle r=6, h=20\)
\(\displaystyle S=2\pi*6^2+2\pi*6*20\)
\(\displaystyle S=72\pi+240\pi\)
\(\displaystyle S=312\pi\)


Now that I have the surface area, I need to set this back equal to the equation and pick which variable I will solve for… Let’s say h

\(\displaystyle 312\pi=2\pi*r^2+2\pi*r*h\)

\(\displaystyle \frac{312\pi-2\pi*r^2}{2\pi*r}=h\)

or

\(\displaystyle \frac{156-r^2}{r}=h\)

So now that I have h I use that in the function instead of h to have a function of r.

\(\displaystyle S=2\pi*r^2+2\pi*r*(\frac{156-r^2}{r})\)

\(\displaystyle S=2\pi*r^2+\frac{312\pi*r-2\pi*r^3}{r}\)

\(\displaystyle S=2\pi*r^2+312\pi-2\pi*r^2\)

So if this is the new function when I do the derivative I get:

\(\displaystyle \frac{dS}{dr}=4\pi+0-4\pi\)

or....

\(\displaystyle \frac{dS}{dr}=0\)

As you see if I then multiply out the \(\displaystyle dr\) to the other side no matter what my error would be 0… this doesn’t seem right?

Could anyone tell me what I’m doing wrong???
 

dwsmith

MHF Hall of Honor
Mar 2010
3,093
582
Florida
Take the derivative with respect to r before you plug in the values.
 
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May 2010
5
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Take the derivative with respect to r before you plug in the values.
Thank you for you're help, do you mean do the derivative of the original surface formula? If so, I'm not sure how to do a derivative with two variables? h and r. Would I just treat h as a constant? Thanks
 

dwsmith

MHF Hall of Honor
Mar 2010
3,093
582
Florida
Thank you for you're help, do you mean do the derivative of the original surface formula? If so, I'm not sure how to do a derivative with two variables? h and r. Would I just treat h as a constant? Thanks
h is just a constant.

Also, for one answer, the variable will be h since the error is in height and the other it is in r.
 
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May 2010
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Ok, wow that's a lot easier than I thought.... Thanks!!!!!
 
May 2010
5
0
Thought I had it...

OK, I lost it again....

The two derivative w/ respect to r and h

r
\(\displaystyle dS=4\pi*r+2\pi*h(dr)\)

h
\(\displaystyle dS=2\pi*r(dh)\)

So then I plug in my values?

r=6
h=20
dr=.01
dh=.02

I get,
r
\(\displaystyle 64\pi*.01=\frac{16\pi}{25}\)

h
\(\displaystyle 12\pi*.02=\frac{6\pi}{25}\)


...now what... lol
 

dwsmith

MHF Hall of Honor
Mar 2010
3,093
582
Florida
Here is an example to aid you.

Radius .07 of a ball bearing

If the measurement is correct to withing .01 in, estimate the propagated error in the volume of the ball bearing.

\(\displaystyle V=\frac{4\pi r^3}{3}\)

\(\displaystyle r=.07\) and \(\displaystyle -.01\leq\Delta r\leq .01\)

\(\displaystyle \frac{dV}{dr}=4\pi r^2\)

\(\displaystyle dV=4\pi r^2dr=4\pi (.07)^2(\pm .01)=\pm .06158\) \(\displaystyle in^3\)

The propagated error is .06 cubic inches.

Is this large or small?

\(\displaystyle \frac{dV}{V}=\frac{4\pi r^2dr}{\frac{4\pi r^3}{3}}=\frac{3dr}{r}=\frac{3(\pm .01)}{.07}=\pm .0429\)

or 4.29%