anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
The tangent can be found by the following: f(a,b) + f_x(a,b)*(x-a)+f_y(a,b)*(y-b)
anonymous
  • anonymous
You have learned about implicit differentiation right?
anonymous
  • anonymous
A little - but only with two variables; not three. How do I solve it with three? I need to find the tangent plane (z=?)

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amistre64
  • amistre64
can you clean up the notation any? to avoid confusion
amistre64
  • amistre64
you have values for x,y,and z for a point in space ... find Fx, Fy, and Fz
anonymous
  • anonymous
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
amistre64
  • amistre64
if z = (yada) let f = (yada) - z, if that makes sense
amistre64
  • amistre64
z is defined by x and y values, so the z value at x=0, y=3 can be determined
anonymous
  • anonymous
Big Fx is derivation? Integration? Differential? etc.
anonymous
  • anonymous
z = -8 if i put 0 and 3 in it, but I still do not know how to proceed :X
amistre64
  • amistre64
Fx is just a partial with respect to x
anonymous
  • anonymous
Because I study in another language; I am not familiar with the term "partial" (mathematically)
amistre64
  • amistre64
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 \[f=\frac{x^2+y^2}{9}-9-z\] \[f_x = \frac{2x+0}{9}-0-0\] \[f_y=\frac{0+2y}{9}-0-0\] \[f_z=\frac{0+0}{9}-0-1\]
amistre64
  • amistre64
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
amistre64
  • amistre64
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
anonymous
  • anonymous
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
amistre64
  • amistre64
I was simply trying to make x,y,and z stand out is all. It looked fine in my head :)
amistre64
  • amistre64
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
anonymous
  • anonymous
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 :)
amistre64
  • amistre64
|dw:1442531343203:dw|
amistre64
  • amistre64
|dw:1442531676879:dw|
amistre64
  • amistre64
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
amistre64
  • amistre64
i think my dz/dy is a little backwards ... translating slope to vector might be off
amistre64
  • amistre64
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
anonymous
  • anonymous
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
phi
  • phi
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/
phi
  • phi
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)
phi
  • phi
Lecture 12 introduces the gradient, and that is a better way to approach the problem.
phi
  • phi
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 \[ 2y-3z=30\]
anonymous
  • anonymous
Thanks - I am going to work some more on this

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