Results 1 to 12 of 12

Math Help - wow linear approximation

  1. #1
    Newbie
    Joined
    Nov 2007
    Posts
    7

    wow linear approximation

    - its killing me. ive been fine in calculus but for some reason linear approximation is just not clicking for me. here's the problem:



    Use differentials (linear approximation) to estimate . Base the approximation at .



    ive tried solving this many different ways, always different answers, always incorrect. if anyone can help, with a step by step procedure so when i see another similar problem ill be able to solve it?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,668
    Thanks
    298
    Awards
    1
    Quote Originally Posted by aslan View Post
    - its killing me. ive been fine in calculus but for some reason linear approximation is just not clicking for me. here's the problem:



    Use differentials (linear approximation) to estimate . Base the approximation at .



    ive tried solving this many different ways, always different answers, always incorrect. if anyone can help, with a step by step procedure so when i see another similar problem ill be able to solve it?
    First, \frac{d}{dx}sin(x) = cos(x), so
    sin(x) \approx sin(x_0) + cos(x_0) \cdot (x - x_0)
    to first order, where x_0 is the value of the argument that you are using as an estimate.

    So
    sin(-2) \approx sin \left ( -\frac{2\pi}{3} \right ) + cos \left ( - \frac{2\pi}{3} \right ) \cdot \left ( -\frac{2\pi}{3} - - 2 \right )

    sin(-2) \approx -\frac{\sqrt{3}}{2} - \frac{1}{2} \cdot \left ( - 2 -- \frac{2\pi}{3} \right )

    In terms of decimals:
    sin(-2) \approx -0.913223

    The actual value of sin(-2) = -0.909297 so we are talking about a difference of 0.004317%.

    -Dan
    Last edited by topsquark; November 3rd 2007 at 07:58 AM. Reason: Goofed.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Nov 2007
    Posts
    7
    Quote Originally Posted by topsquark View Post
    First, \frac{d}{dx}sin(x) = cos(x), so
    sin(x) \approx sin(x_0) + cos(x_0) \cdot (x - x_0)
    to first order, where x_0 is the value of the argument that you are using as an estimate.

    So
    sin(-2) \approx sin \left ( -\frac{2\pi}{3} \right ) + cos \left ( - \frac{2\pi}{3} \right ) \cdot \left ( -\frac{2\pi}{3} - - 2 \right )

    sin(-2) \approx -\frac{\sqrt{3}}{2} - \frac{1}{2} \cdot \left ( - \frac{2\pi}{3} + 2 \right )

    In terms of decimals:
    sin(-2) \approx -0.818828

    The actual value of sin(-2) = -0.909297 so we are talking about a difference of 0.99%.

    -Dan
    im starting to understand this. but here:
    sin(-2) \approx -\frac{\sqrt{3}}{2} - \frac{1}{2} \cdot \left ( - \frac{2\pi}{3} + 2 \right )
    wouldnt the ride side be (-2 + 2pi/3) instead of (-2pi/3 + 2) since it is (x-a)?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,668
    Thanks
    298
    Awards
    1
    Quote Originally Posted by aslan View Post
    im starting to understand this. but here:
    sin(-2) \approx -\frac{\sqrt{3}}{2} - \frac{1}{2} \cdot \left ( - \frac{2\pi}{3} + 2 \right )
    wouldnt the ride side be (-2 + 2pi/3) instead of (-2pi/3 + 2) since it is (x-a)?
    You are correct, I switched the order, which improves the error tremendously. (I thought it looked a little high.) I have fixed this in my original post.

    See, you are understanding this!

    -Dan
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Nov 2007
    Posts
    7
    Quote Originally Posted by topsquark View Post
    You are correct, I switched the order, which improves the error tremendously. (I thought it looked a little high.) I have fixed this in my original post.

    See, you are understanding this!

    -Dan
    Thanks a lot man, yeah i think somehow linear approximation just became a lot more understandable.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by aslan View Post
    - its killing me. ive been fine in calculus but for some reason linear approximation is just not clicking for me. here's the problem:



    Use differentials (linear approximation) to estimate . Base the approximation at .



    ive tried solving this many different ways, always different answers, always incorrect. if anyone can help, with a step by step procedure so when i see another similar problem ill be able to solve it?
    Please use a different image hosting service, the one you are using
    does not have acceptable security certificates for many of our computers.

    RonL
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Nov 2007
    Posts
    7
    oh, sorry about the imagine hosting then.

    any help with this problem on a different subject:


    The circumference of a sphere was measured to be 73 cm with a possible error of 0.5 cm. Use differentials to estimate the maximum error in the calculated surface area. ________

    Now estimate the relative error in the calculated surface area. ________


    im supposing i'd use the formula:
    E(h) = f(a+h) - [f(a) + h*derivative of f(a)]
    where h is the difference between x and a. but using this formula doesnt tell me anything about the maximum or relative error..

    btw instead of solving this if anyone would think it was easier to just direct me to a page that explains all applications of linear approximation, id be glad. i dont have a textbook for my calc class because of money problems and for some reason its been hard for me to find sample problems like these online.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Newbie
    Joined
    Nov 2007
    Posts
    7
    Quote Originally Posted by aslan View Post
    oh, sorry about the imagine hosting then.

    any help with this problem on a different subject:


    The circumference of a sphere was measured to be 73 cm with a possible error of 0.5 cm. Use differentials to estimate the maximum error in the calculated surface area. ________

    Now estimate the relative error in the calculated surface area. ________


    im supposing i'd use the formula:
    E(h) = f(a+h) - [f(a) + h*derivative of f(a)]
    where h is the difference between x and a. but using this formula doesnt tell me anything about the maximum or relative error..

    btw instead of solving this if anyone would think it was easier to just direct me to a page that explains all applications of linear approximation, id be glad. i dont have a textbook for my calc class because of money problems and for some reason its been hard for me to find sample problems like these online.
    bump
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Newbie
    Joined
    Nov 2007
    Posts
    7
    Quote Originally Posted by aslan View Post
    oh, sorry about the imagine hosting then.

    any help with this problem on a different subject:


    The circumference of a sphere was measured to be 73 cm with a possible error of 0.5 cm. Use differentials to estimate the maximum error in the calculated surface area. ________

    Now estimate the relative error in the calculated surface area. ________


    im supposing i'd use the formula:
    E(h) = f(a+h) - [f(a) + h*derivative of f(a)]
    where h is the difference between x and a. but using this formula doesnt tell me anything about the maximum or relative error..

    btw instead of solving this if anyone would think it was easier to just direct me to a page that explains all applications of linear approximation, id be glad. i dont have a textbook for my calc class because of money problems and for some reason its been hard for me to find sample problems like these online.
    BUMP for this

    equations:
    C=2r*pi
    S=4pi*r^2


    my process was using the error for circumference to solve for "dr", and then using circumference to solve for the "r." then with those values, differentiate the surface area equation and solve for maximum error. but i am continuously getting the wrong answer.

    anyone with any help?
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,668
    Thanks
    298
    Awards
    1
    Quote Originally Posted by aslan View Post
    BUMP for this

    equations:
    C=2r*pi
    S=4pi*r^2


    my process was using the error for circumference to solve for "dr", and then using circumference to solve for the "r." then with those values, differentiate the surface area equation and solve for maximum error. but i am continuously getting the wrong answer.

    anyone with any help?
    I'm not answering this one because I don't know what to do with the errors.

    But I wanted to say this:
    "Bumping" is rude! If someone is going to answer your question they will answer it their own time. Bumping to get more attention paid to your question is irritating (the whole point of it, really) and makes no sense at all. I mean, if someone was going to answer your question, they'd likely do so the first time they read it.

    -Dan
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Newbie
    Joined
    Nov 2007
    Posts
    7
    Quote Originally Posted by topsquark View Post
    I'm not answering this one because I don't know what to do with the errors.

    But I wanted to say this:
    "Bumping" is rude! If someone is going to answer your question they will answer it their own time. Bumping to get more attention paid to your question is irritating (the whole point of it, really) and makes no sense at all. I mean, if someone was going to answer your question, they'd likely do so the first time they read it.

    -Dan
    what? the whole point of bumping is to call attention to a question. this is the urgent section, i see, and if there hasnt been a response in say, 12 hours, wouldnt a person think the OP no longer needed the help? bumping lets people know the subject at hand is still a concern.

    im not new to forums lol, though i am new to seeing moderators refer to bumping as rude. ridiculous.
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,668
    Thanks
    298
    Awards
    1
    Quote Originally Posted by aslan View Post
    what? the whole point of bumping is to call attention to a question. this is the urgent section, i see, and if there hasnt been a response in say, 12 hours, wouldnt a person think the OP no longer needed the help? bumping lets people know the subject at hand is still a concern.

    im not new to forums lol, though i am new to seeing moderators refer to bumping as rude. ridiculous.
    I'm not a mod, I simply help here.

    It could simply be that no one knows how to deal with your question. In that case, why bump?

    It could be that a person that can answer your question hasn't logged on yet. In that case they will see your question when they do log on. So again, why bump? This person isn't going to log in any sooner because you bumped!

    As much as I'd like to not say this, there are many questions that posters have asked that never get a response. In most cases it's because no one knows how to answer the question. (In some cases, it's because the poster is such a jerk that no one wants to, but that isn't the case with you, I assure you!)

    Bumping is a means to an end: the idea is to call attention to your message, therefore prioritizing your message over everyone else's. I find this kind of behavior rude. If you don't agree, so be it.

    -Dan
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Linear Approximation
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 2nd 2010, 09:56 PM
  2. Linear Approximation
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 11th 2009, 05:40 PM
  3. linear approximation
    Posted in the Calculus Forum
    Replies: 3
    Last Post: September 23rd 2008, 10:23 PM
  4. linear approximation y=
    Posted in the Calculus Forum
    Replies: 2
    Last Post: September 10th 2008, 05:01 PM
  5. Linear approximation
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 27th 2006, 10:20 AM

Search Tags


/mathhelpforum @mathhelpforum