Results 1 to 3 of 3

Math Help - Using differentials to approximate

  1. #1
    Newbie
    Joined
    Sep 2008
    Posts
    13

    Using differentials to approximate

    I'm going to write this problem word for word because I'm not sure how else to describe it.

    Q1. For y=f(x), differentials are used to approximate the change in y for a given change in x. Suppose you have misplaced your calculator and need a decent approximation of the square root of 27. The yknow 25^1/2 = 5 so you assume that 27^1/2 must be slightly larger than 5.

    A. Use differentials to approximate the square root of 27.

    Step 1. f(x) = x^1/2, x=25, dx=2.

    Calculate dy = f '(x) dx and evaluate it.

    __________________________________________________ ___________

    In addition to doing it for 27, I have to do it for 21, 39, and 110. The main problem is I don't know which values to put where. I think the x is always 25 and only dx changes? So for 27, dy = .5(25^.50)*2 which is .2 I believe. So my approximation would be 5.2? Any help or insight is appreciated. Thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Banned
    Joined
    Aug 2008
    Posts
    530
    Quote Originally Posted by Zabulius View Post
    I'm going to write this problem word for word because I'm not sure how else to describe it.

    Q1. For y=f(x), differentials are used to approximate the change in y for a given change in x. Suppose you have misplaced your calculator and need a decent approximation of the square root of 27. The yknow 25^1/2 = 5 so you assume that 27^1/2 must be slightly larger than 5.

    A. Use differentials to approximate the square root of 27.

    Step 1. f(x) = x^1/2, x=25, dx=2.

    Calculate dy = f '(x) dx and evaluate it.

    __________________________________________________ ___________

    In addition to doing it for 27, I have to do it for 21, 39, and 110. The main problem is I don't know which values to put where. I think the x is always 25 and only dx changes? So for 27, dy = .5(25^.50)*2 which is .2 I believe. So my approximation would be 5.2? Any help or insight is appreciated. Thanks
    yes, sqrt(27) = 5.2

    Please see

    Mathwords: Approximation by Differentials
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Grandad's Avatar
    Joined
    Dec 2008
    From
    South Coast of England
    Posts
    2,570
    Thanks
    1

    Approximate square roots

    Hello Zabulius
    Quote Originally Posted by Zabulius View Post
    dy = .5(25^.50)*2
    I don't think your formula is quite right. You need to divide by 25^0.5

    y = \sqrt{x} \Rightarrow \frac{dy}{dx} = \frac{1}{2\sqrt{x}}

    \Rightarrow \delta y \approx \frac{1}{2\sqrt{x}}\delta x

    So to find \sqrt{27}, put x = 25 and \delta x = 2:

    \delta y \approx \frac{1}{2 \cdot 5}\cdot 2 = 0.2

    So y+ \delta y = 5.2 \approx \sqrt {27}

    To find \sqrt{21}, use the fact that 21 = 25 - 4. So you can again use x = 25, but this time \delta x = -4.

    The nearest perfect square to 39 is 36. So to find \sqrt{39}, use x = 36. And since 39 = 36 + 3, use \delta x = 3.

    Can you see what values to use for x and \delta x to find \sqrt{110}?

    Grandad
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: December 1st 2010, 03:37 PM
  2. Replies: 3
    Last Post: March 1st 2010, 09:24 PM
  3. Use of differentials to approximate
    Posted in the Calculus Forum
    Replies: 5
    Last Post: February 10th 2010, 01:00 PM
  4. Differentials
    Posted in the Calculus Forum
    Replies: 1
    Last Post: May 28th 2009, 01:29 PM
  5. Use differentials to approximate?
    Posted in the Calculus Forum
    Replies: 5
    Last Post: November 30th 2006, 10:29 AM

Search Tags


/mathhelpforum @mathhelpforum