# Thread: Using differentials to solve propagated error problem?

1. ## Using differentials to solve propagated error problem?

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 = $S=2$ $pi$ $r^2$ $+2$ $pi$ $rh$

and

(B)
Volume = $V=$ $pi$ $r^2$ $h$

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.

2. Originally Posted by TcAllen247
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 = $S=2$ $pi$ $r^2$ $+2$ $pi$ $rh$

and

(B)
Volume = $V=$ $pi$ $r^2$ $h$

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.

$\frac{dS}{dr}=....$ multiple both sides by dr.

$dr=\pm$error
Plug in numbers and solve

3. ## 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.

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

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

$312\pi=2\pi*r^2+2\pi*r*h$

$\frac{312\pi-2\pi*r^2}{2\pi*r}=h$

or

$\frac{156-r^2}{r}=h$

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

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

$S=2\pi*r^2+\frac{312\pi*r-2\pi*r^3}{r}$

$S=2\pi*r^2+312\pi-2\pi*r^2$

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

$\frac{dS}{dr}=4\pi+0-4\pi$

or....

$\frac{dS}{dr}=0$

As you see if I then multiply out the $dr$ 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???

4. Take the derivative with respect to r before you plug in the values.

5. Originally Posted by dwsmith
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

6. Originally Posted by TcAllen247
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.

7. Ok, wow that's a lot easier than I thought.... Thanks!!!!!

8. ## Thought I had it...

OK, I lost it again....

The two derivative w/ respect to r and h

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

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

So then I plug in my values?

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

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

h
$12\pi*.02=\frac{6\pi}{25}$

...now what... lol

9. 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.

$V=\frac{4\pi r^3}{3}$

$r=.07$ and $-.01\leq\Delta r\leq .01$

$\frac{dV}{dr}=4\pi r^2$

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

The propagated error is .06 cubic inches.

Is this large or small?

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

or 4.29%

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