# Finding a tangent line to a level curve - Don't understand solution

1. Apr 17, 2013

### mrcleanhands

1. The problem statement, all variables and given/known data
If f(x,y) = xy, find the gradient vector $\nabla f(3,2)$ and use it to find the tangent line to the level curve f(x,y) = 6 at the point (3,2)

2. Relevant equations

3. The attempt at a solution
$f(x,y)=xy \Rightarrow\nabla f(x,y)=<y,x>,\nabla f(3,2)=<2,3>$
$\nabla f(3,2)$ is perpendicular to the tanget line, so the tangent line has equation
$\nabla f(3,2)\cdot<x-3,y-2>=0$... and so on

I understand that the dot product must be 0 if the two vectors are perpendicular.
What I don't get is how they pick the vector <x-3, y-2> given that the point were concered with is (3,2)

2. Apr 17, 2013

### Simon Bridge

As opposed to doing: <2,3>.<x,y>=0 you mean?
The set of vectors that are perpendicular to <2,3> lie in a plane don't they?

3. Apr 17, 2013

### mrcleanhands

yeah they're in a plane. But <x-3,y-2> is not a plane?

4. Apr 17, 2013

### Simon Bridge

If it were [the plane perpendicular to <2,3>] then the dot product would be zero for every value of x and y so the relation would be useless.

It follows that just knowing that <2,3> is perpendicular to the tangent you want is not good enough - you need a way to select the one you want out of the whole plane. How would you go about that? What is special about this tangent?

5. Apr 17, 2013

### tiny-tim

hi mrcleanhands!
if (x,y) lies on the tangent plane, then the vector from (3,2) to (x,y) is a tangent at (3,2), and is parallel to <x-3, y-2>

6. Apr 17, 2013

### mrcleanhands

I see, it has to lie on the point (2,3) which you get by taking a line from (3,2) to (x,y)

thanks to both of you!