Results 1 to 4 of 4

Math Help - Linear Aprox / differentiatials.

  1. #1
    Junior Member
    Joined
    Dec 2006
    Posts
    40

    Linear Aprox / differentiatials.

    Looking back on this problem, i have one question..
    Its wanting to use a linear approximation to estimate the number..
    (2.001)^5

    Here's my work: x=2..
    f(x)= x^5 f(2) = 32 = (which is y),
    f'(x)= 5x^4. f'(2) = 80 = (which is m)

    Y-32=80(x-2)
    y=80x-128

    so i know (2.001)^5 actually is 32.08008004...
    and y=80(2)-128 = 32...
    and y=80(.001) - 128 = .08 - 127.92 = -127.92...

    then the answers were that the ▲y = .08008 and dy=.08
    what are ▲y=.08008 and dy from?
    If the .08 is from y=80(.001), where did the -128 go?
    thanks...
    Follow Math Help Forum on Facebook and Google+

  2. #2
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by 3deltat View Post
    Looking back on this problem, i have one question..
    Its wanting to use a linear approximation to estimate the number..
    (2.001)^5

    Here's my work: x=2..
    f(x)= x^5 f(2) = 32 = (which is y),
    f'(x)= 5x^4. f'(2) = 80 = (which is m)

    Y-32=80(x-2)
    y=80x-128

    so i know (2.001)^5 actually is 32.08008004...
    and y=80(2)-128 = 32...
    and y=80(.001) - 128 = .08 - 127.92 = -127.92...

    then the answers were that the ▲y = .08008 and dy=.08
    what are ▲y=.08008 and dy from?
    If the .08 is from y=80(.001), where did the -128 go?
    thanks...
    there are two basic ways to approach a problem like this:

    the one i learned is to use the linear approximation equation:

    f(x) \approx f(a) + f'(a)(x - a)

    here, x is the value that you want to evaluate the function at, while a is a value close to x that you already know the value of the function for


    the other way is one i picked up while tutoring.

    f(x + dx) = f(x) + dy ....er, or something like that... i know how to use it, that's what's important .

    anyway, here x is the value you can evaluate the function at, dx is a small change to this value, f(x) is the value of f at x and dy = f'(x) \cdot dx (we get this by solving \frac {dy}{dx} = f'(x) for dy)

    which method do you want to use?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Dec 2006
    Posts
    40
    The f(x)=f(a) + f'(a)(x-a) is one i've learned before aye. For some reason the teacher didnt like it and went to the slope formula directly.

    still, it'd come up with y=80x -128 no?
    Last edited by 3deltat; January 7th 2008 at 06:01 PM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by 3deltat View Post
    The f(x)=f'(a)(x-a) is one i've learned before aye. For some reason the teacher didnt like it and went to the slope formula directly.

    still, it'd come up with y=80x -128 no?
    apparently you have a different method from mine.

    my method:

    define f(x) = x^5

    then (2.001)^5 = f(2.001) \approx 2^5 + f'(2)(2.001 - 2)

    now just work out the right hand side to get the answer. which would be 32.08
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 7
    Last Post: October 10th 2011, 03:06 PM
  2. Replies: 1
    Last Post: August 1st 2011, 10:00 PM
  3. Replies: 5
    Last Post: September 6th 2010, 03:55 PM
  4. Regression fundamentals (linear, non linear, simple, multiple)
    Posted in the Advanced Statistics Forum
    Replies: 13
    Last Post: November 12th 2009, 06:47 AM
  5. Replies: 7
    Last Post: August 30th 2009, 10:03 AM

Search Tags


/mathhelpforum @mathhelpforum