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Differential question

  1. Dec 7, 2007 #1
    1. The problem statement, all variables and given/known data

    Find the value of dy and deltay for y=2x^3-4x, x=2 and dx = deltax = 0.1.

    3. The attempt at a solution

    dy/dt= 6(4)(.1)-4(.1) = 2

    This is correct. However, I do not understand how to get deltay ? I tried 2/2*2^3-4(2) = .25 -- however, the answer is 2.12. how do they get this value?
  2. jcsd
  3. Dec 7, 2007 #2


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    Homework Helper

    Well first thing note that delta-y is simply the change in the value of y due to delta-x.

    Hence delta-y= y2 - y1

    where y2 is the new value of y due to delta-x (ie. y2 = 2(x+delta-x)^3 - 4(x+delta-x))

    and y1 is already given as y in the equation in the problem.
    Last edited: Dec 7, 2007
  4. Dec 7, 2007 #3


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    Staff Emeritus
    Science Advisor

    Actually that is far from correct, but only because it is written incorrectly! There is NO "dy/dt" because y is not a function of t. What you meant was simply
    dy= (dy/dx)dx= (6x^2- 4)= (6(4)- 4)(0.1)

    y(x+ deltax)= y(2.1= 2(2.1)^3- 4(2.1)
    y(x)= y(2)= 2(2^3)- 4(2)

    delta y is the difference of those two: deltay= f(x+deltax)- f(x).
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