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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
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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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The tangent can be found by the following: f(a,b) + f_x(a,b)*(xa)+f_y(a,b)*(yb)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You have learned about implicit differentiation right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0A little  but only with two variables; not three. How do I solve it with three? I need to find the tangent plane (z=?)

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1can you clean up the notation any? to avoid confusion

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1you have values for x,y,and z for a point in space ... find Fx, Fy, and Fz

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I 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
 one year ago
Best ResponseYou've already chosen the best response.1if z = (yada) let f = (yada)  z, if that makes sense

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1z is defined by x and y values, so the z value at x=0, y=3 can be determined

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Big Fx is derivation? Integration? Differential? etc.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0z = 8 if i put 0 and 3 in it, but I still do not know how to proceed :X

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1Fx is just a partial with respect to x

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Because I study in another language; I am not familiar with the term "partial" (mathematically)

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1the 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 \[f=\frac{x^2+y^2}{9}9z\] \[f_x = \frac{2x+0}{9}00\] \[f_y=\frac{0+2y}{9}00\] \[f_z=\frac{0+0}{9}01\]

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1at 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
 one year ago
Best ResponseYou've already chosen the best response.1partial 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
 one year ago
Best ResponseYou've already chosen the best response.0Ah  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
 one year ago
Best ResponseYou've already chosen the best response.1I was simply trying to make x,y,and z stand out is all. It looked fine in my head :)

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1another 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
 one year ago
Best ResponseYou've already chosen the best response.0I 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
 one year ago
Best ResponseYou've already chosen the best response.1dw:1442531343203:dw

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1dw:1442531676879:dw

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1cross 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
 one year ago
Best ResponseYou've already chosen the best response.1i think my dz/dy is a little backwards ... translating slope to vector might be off

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1a 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
 one year ago
Best ResponseYou've already chosen the best response.0x^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
 one year ago
Best ResponseYou've already chosen the best response.0Here is some background (first part of the lecture) http://ocw.mit.edu/courses/mathematics/1802multivariablecalculusfall2007/videolectures/lecture9maxminandleastsquares/

phi
 one year ago
Best ResponseYou've already chosen the best response.0I 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
 one year ago
Best ResponseYou've already chosen the best response.0Lecture 12 introduces the gradient, and that is a better way to approach the problem.

phi
 one year ago
Best ResponseYou've already chosen the best response.0but using the approximation idea \[ \Delta z = \frac{\partial z }{\partial x} \Delta x + \frac{\partial z }{\partial y} \Delta y \] or, \[ z  z_0 = \frac{\partial z }{\partial x} (xx_0) + \frac{\partial z }{\partial y} (yy_0) \] 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 \[ 2y3z=30\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thanks  I am going to work some more on this
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