- 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

- schrodinger

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- 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

You have learned about implicit differentiation right?

- 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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## More answers

- amistre64

can you clean up the notation any? to avoid confusion

- amistre64

you have values for x,y,and z for a point in space ... find Fx, Fy, and Fz

- 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

if z = (yada)
let f = (yada) - z, if that makes sense

- amistre64

z is defined by x and y values, so the z value at x=0, y=3 can be determined

- anonymous

Big Fx is derivation? Integration? Differential? etc.

- anonymous

z = -8 if i put 0 and 3 in it, but I still do not know how to proceed :X

- amistre64

Fx is just a partial with respect to x

- anonymous

Because I study in another language; I am not familiar with the term "partial" (mathematically)

- 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

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

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

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

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

- 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

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

|dw:1442531343203:dw|

- amistre64

|dw:1442531676879:dw|

- 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

i think my dz/dy is a little backwards ... translating slope to vector might be off

- 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

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

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

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

Lecture 12 introduces the gradient, and that is a better way to approach the problem.

- 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

Thanks - I am going to work some more on this

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