1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    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


    User Avatar
    Science Advisor

    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
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted

Similar Discussions: Finding linear approximation
  1. Linear approximation (Replies: 1)

  2. Linear Approximation (Replies: 4)

  3. Linear Approximations (Replies: 3)