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anonymous

  • one year ago

A tough curve question; (paraboloid - tangent) z = x^2 + y^2/9 - 9 How do I find the equation for the tangent plane @ point (0, 3) ? The part that confuses me with this, is that the given equation (it's an elliptic paraboloid) have three variables; not just two. On which level is the plane? Does it matter? Can I just set z = 0 and work with x and y? Here is a piece of my math scribbles; http://i.imgur.com/nCctKKT.png Am I heading the right direction? Would it be wrong to set Z to 0? Help is very much appreciated! <3

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  1. anonymous
    • one year ago
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    The tangent can be found by the following: f(a,b) + f_x(a,b)*(x-a)+f_y(a,b)*(y-b)

  2. anonymous
    • one year ago
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    You have learned about implicit differentiation right?

  3. anonymous
    • one year ago
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    A little - but only with two variables; not three. How do I solve it with three? I need to find the tangent plane (z=?)

  4. amistre64
    • one year ago
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    can you clean up the notation any? to avoid confusion

  5. amistre64
    • one year ago
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    you have values for x,y,and z for a point in space ... find Fx, Fy, and Fz

  6. anonymous
    • one year ago
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    I dont know if I have any specifically for Z - All I know I got a point (x=0,y=3) and there is no mention of z; because the tangent plane should be z... as far as I understand , however - I do not know how to use the original equation (z=(...)) and make it usable for x=0 and y=3

  7. amistre64
    • one year ago
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    if z = (yada) let f = (yada) - z, if that makes sense

  8. amistre64
    • one year ago
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    z is defined by x and y values, so the z value at x=0, y=3 can be determined

  9. anonymous
    • one year ago
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    Big Fx is derivation? Integration? Differential? etc.

  10. anonymous
    • one year ago
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    z = -8 if i put 0 and 3 in it, but I still do not know how to proceed :X

  11. amistre64
    • one year ago
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    Fx is just a partial with respect to x

  12. anonymous
    • one year ago
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    Because I study in another language; I am not familiar with the term "partial" (mathematically)

  13. amistre64
    • one year ago
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    the notation is amgibuous, can you properly notate: z= ..... partial means that you treat all other variables as constants. \[z=\frac{x^2+y^2}{9}-9\] let \[f=\frac{x^2+y^2}{9}-9-z\] \[f_x = \frac{2x+0}{9}-0-0\] \[f_y=\frac{0+2y}{9}-0-0\] \[f_z=\frac{0+0}{9}-0-1\]

  14. amistre64
    • one year ago
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    at x=0, y=3 fx = 0 fy = 6/9 = 2/3 fz = -1 or since the direction is what we need and the magnitude is irrelevant (0,2,-3) is in the same direction is (fx,fy,fz) and is simpler to play with i think

  15. amistre64
    • one year ago
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    partial derivative simply means that if we are working with respect to some variable (like y), then all other variables are considered to act like a constant. say f=2xy^2 fx = 2y^2 fy = 4xy

  16. anonymous
    • one year ago
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    Ah - but what did you mean with Fx, Fy, Fz? (Capital instead of fx, fy, etc.) I understand what you meant with partial now - I have done something similar in other tasks

  17. amistre64
    • one year ago
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    I was simply trying to make x,y,and z stand out is all. It looked fine in my head :)

  18. amistre64
    • one year ago
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    another way to approach this, is to let x=0, and determine the slope of the tangent to z, at y=3 find dz/dy when x=0 find dz/dx when y=3

  19. anonymous
    • one year ago
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    I will do some more work, and try it out - hopefully a puzzle piece will pop into my head and allow me to learn this :)

  20. amistre64
    • one year ago
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    |dw:1442531343203:dw|

  21. amistre64
    • one year ago
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    |dw:1442531676879:dw|

  22. amistre64
    • one year ago
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    cross the vectors to find the normal -(0z+0x+3y) x y z x y 1 0 0 1 0 0 2 3 0 2 +(0x+0y+2z) <0,-3, 2> if im working this right for a normal

  23. amistre64
    • one year ago
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    i think my dz/dy is a little backwards ... translating slope to vector might be off

  24. amistre64
    • one year ago
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    a slope of 2/3 tells us z rises by 2 when y moves by 3 <0,3,2> which then gets us the cross product to conform to fx, fy,fz

  25. anonymous
    • one year ago
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    x^2 is not divided by 9 in the original equation though - so I assume it is a typo. However, I am going to work on this - hopefully I will be able to get it

  26. phi
    • one year ago
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    Here is some background (first part of the lecture) http://ocw.mit.edu/courses/mathematics/18-02-multivariable-calculus-fall-2007/video-lectures/lecture-9-max-min-and-least-squares/

  27. phi
    • one year ago
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    I assume you have answered this How do I find the equation for the tangent plane @ point (0, 3) ? you are given z = x^2 + y^2/9 - 9 so at x=0, y=3, z = -8 and the point is (0,3,-8)

  28. phi
    • one year ago
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    Lecture 12 introduces the gradient, and that is a better way to approach the problem.

  29. phi
    • one year ago
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    but using the approximation idea \[ \Delta z = \frac{\partial z }{\partial x} \Delta x + \frac{\partial z }{\partial y} \Delta y \] or, \[ z - z_0 = \frac{\partial z }{\partial x} (x-x_0) + \frac{\partial z }{\partial y} (y-y_0) \] as you know, the partial derivative with respect to x, evaluating at (0,3,-8) \[ \frac{\partial z }{\partial x}= 2x = 0 \] and \[ \frac{\partial z }{\partial y}= \frac{2y}{9}= \frac{2}{3} \] and the equation of the tangent plane is \[ z +8 = \frac{2}{3} (y - 3) \] which can be written in standard form as \[ 2y-3z=30\]

  30. anonymous
    • one year ago
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    Thanks - I am going to work some more on this

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