Determine gradient of a function f(x,y)

  • #1
-EquinoX-
564
1

Homework Statement


View the curve below as a contour of f(x,y).
(y-x)^2 + 2 = xy - 3

Use gradf (2,3) to find a vector normal to the curve at (2,3).


Homework Equations





The Attempt at a Solution


I am not sure how do I get the vector normal to the curve, is it using a cross product>?
 
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  • #2


No. Express your curve as f(x,y)=C and use the gradient. That's what they told you to do.
 
  • #3


Dick said:
No. Express your curve as f(x,y)=C and use the gradient. That's what they told you to do.

what's c here?
 
  • #4


The constant you get by putting x= 2, y= 3 into the function. Since it will drop out of the derivative, its value is not important and you can just leave it as "c". Since the gradient always points in the direction of fastest increase it is always normal to level curves.
 
  • #5


so the gradient will be the normal to the level curves??
 
  • #6


-EquinoX- said:
so the gradient will be the normal to the level curves??

That's exactly what Halls said, isn't it? C doesn't matter. The gradient will depend only on x and y.
 
  • #7


well the gradient I have is:

(-3y + 2x) i + (2y - 3x) j

is this true?
 
  • #8


-EquinoX- said:
well the gradient I have is:

(-3y + 2x) i + (2y - 3x) j

is this true?

Sure. Now just evaluate it at (2,3).
 
  • #9


okay I get -5i , if then I am asked to find an equation for the tangent line to the curve at (2,3) and determine whether it's vertical , diagonal, or horizontal. How do I do this?
 
  • #10


The tangent is perpendicular to the normal, isn't it?
 
  • #11


so then I am assuming it's vertical?
 
  • #12


I wouldn't say your "assuming" anything. It is.
 

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