theorangedog.net

Differentials and Linear Approximation

by theorangedog on Jan.11, 2008, under Skills

Q. Find the linear change in volume dV if the sides of a cube change from 10 to 10.1.

A. We can use the definition of small dy = (dy/dx)dx to determine the estimated linear change. In this problem, we start with the cubic volume equation of small V = x^3 where small x is the length of one side of the cube. The derivative is then small f^{prime}(V) = 3x^2. Filling in for linear approximation we get small dV = 3x^2dx. This results in small dV = 3(10^2)(.1) = 30. We can see that the linear approximation is not entirely accurate, hence the name approximation, as small Delta V = 10.1^3 - 10^3 = 30.301.

Related equations and relationships:
If small df/dx > 0 then small f(x) is increasing. If small df/dx < 0 then small f(x) is decreasing.

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3 comments for this entry:
  1. just-passing-by
    February 16th, 2008 on 1:26 am

    the latter statement is not true. consider f(x)=x^2sin(\frac{1}{x})+\frac{x}{2} if x\neq 0 and f(x) = 0 if x=0. If x=0 f’(x) = 1/2, but it is obviously is not increasing in any neighborhood of 0.

  2. just-passing-by
    February 16th, 2008 on 1:28 am

    something wrong with tex stuff in my comment
    well, just to be on a safe side

    f(x) = x^2sin(1/x) + x/2 if x !=0
    f(0) = 0

    function is cont.
    f’(0) = 1/2
    f(x) is not increasing in any neigh. of x=0

  3. foq
    February 16th, 2008 on 8:35 pm

    Thanks for the comment - seems there is a mimeTex issue, again… guessing the link was changed or the requirements for it changed.

    It gives me something else to add to my list for this weekend - I’ll clear up the issue and consolidate your comments.

    Again, thanks for stopping by.

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