Results 1 to 9 of 9

Math Help - Using differentials to solve propagated error problem?

  1. #1
    Newbie
    Joined
    May 2010
    Posts
    5

    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.
    Last edited by TcAllen247; May 10th 2010 at 08:52 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5
    Quote Originally Posted by TcAllen247 View Post
    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=\pmerror
    Plug in numbers and solve
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    May 2010
    Posts
    5

    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???
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5
    Take the derivative with respect to r before you plug in the values.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    May 2010
    Posts
    5
    Quote Originally Posted by dwsmith View Post
    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
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5
    Quote Originally Posted by TcAllen247 View Post
    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.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    May 2010
    Posts
    5
    Ok, wow that's a lot easier than I thought.... Thanks!!!!!
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Newbie
    Joined
    May 2010
    Posts
    5

    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
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5
    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%
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Estimate relative error using differentials
    Posted in the Calculus Forum
    Replies: 1
    Last Post: August 8th 2010, 07:02 PM
  2. Replies: 2
    Last Post: July 11th 2009, 09:09 PM
  3. Relative Error through Differentials
    Posted in the Calculus Forum
    Replies: 2
    Last Post: July 6th 2009, 08:02 PM
  4. Differentials to estimate the maximum error
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 27th 2009, 07:07 PM
  5. Estimating error using differentials
    Posted in the Calculus Forum
    Replies: 3
    Last Post: March 20th 2007, 08:08 PM

Search Tags


/mathhelpforum @mathhelpforum