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Linear Approximation

  1. Oct 31, 2007 #1
    1. The problem statement, all variables and given/known data
    To the right is the graph of 5x^3y-3xy^2+y^3=6. Verify that (1,2) is a point on the curve. There's a nearby point on the curve whose point is (1.07,u). What is the approx. value for u? There's a nearby point on the curve whose coordinates are (.98,v). What is the approx. value for v? Theres a nearby point on the curve whose point is (w,2.04). What is an approximate value for w?

    2. Relevant equations

    the derivative of the equation is (15x^2-3y^2)/(-5x^3+3x^2y-3y^2)

    I know how to do the first part, you plug in the values for x and y and they should = 6, which they do. However, for the remainder of the problem i am completely lost. I'm totally
    lost, so if you can offer any input i would appreciate it. Thanks
    Last edited: Oct 31, 2007
  2. jcsd
  3. Oct 31, 2007 #2


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    The tangent line to a curve, y= f(x), through point [itex](x_0, f(x_0))[/itex] on the line is [itex]y= f'(x_0)(x- x_0)+ f(x_0)[/itex] and will give (approximately) the y value for an x value close to [itex]x_0[/itex]. You are told that [itex]x_0[/itex]= 1 and that [itex]f(x_0)[/itex]= 2. What is [itex]f'(x_0)[/itex]? That gives you all the information you need to construct the equation of the tangent line. Then use it with the two new points given to find the missing values.
    Last edited: Oct 31, 2007
  4. Oct 31, 2007 #3
    Thanks. That helped me, I'm pretty sure I have it now.
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