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Finding linear approximation

  1. Nov 6, 2015 #1
    1. The problem statement, all variables and given/known data
    Find linear approximation of the surface z2 = xy + y + 3 at the point (0,6, -3) and use it to approximate f(-0.01, 6.01, -2.98)




    2. Relevant equations


    3. The attempt at a solution


    upload_2015-11-6_2-6-24.png

    so this means the total surface has decrease by -0.17?
     

    Attached Files:

  2. jcsd
  3. Nov 6, 2015 #2
    its not about the surface thats decreased, its about how accurate your approximation is. in this case, you used the newly found equation of the tangent plane to approximate f(-0.01, 6.01, -2.98). you want to compare using the original function to see how accurate it is.
     
  4. Nov 6, 2015 #3
    so you comparing it to the actual change?
    which should be final - initial :
    F(-0.01, 6.01, -2.98) - F(0,6, -3)
     
  5. Nov 6, 2015 #4
    this is not what the question is asking though, you have already answered the question.
    but, if you want to see how close your approximation is, use z^2 = xy + y + 3 as the initial function to find the exact value of the point that you are trying to approximate with newly found tangent plane.
     
  6. Nov 6, 2015 #5
    hm..if you look at this way: the tangent plane and the original function are like twin brothers at a specific time[ point in our case] . we are testing to see how close the brothers are at that time[point] , but as we move away from that time[point], they become not so similar anymore, and if we go far enough, they are completely different from each other.
     
  7. Nov 6, 2015 #6

    HallsofIvy

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    You are asked to do two different problems but seem to have mixed them together! You have, as your final answer "L(x, y, z)= -0.17". That is certainly not true since the left side, L(x, y, z), the linear approximation to the function, is NOT a constant.

    You need two different answers to the two different problems. First you need to write the equation for the tangent approzimation, then evaluate that at (-0.01, 6.01, -2.98).

    Yes, given [itex]F(x, y, z)= xy+ y- z^2+ 3[/itex], then [itex]F_x= -y[/itex], [itex]F_y= -x- 1[/itex], and [itex]F_z= 2z[/itex].
    At (0, 6, -3) we have [itex]F_x=[/itex]

    ..... etc, see how you go with this now. [mod edit]
     
    Last edited by a moderator: Nov 6, 2015
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