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

[mrcleanhands]

Homework Statement

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)

Homework Equations

The Attempt at a Solution

f(x,y)=xy<br /> \Rightarrow\nabla f(x,y)=&lt;y,x&gt;,\nabla f(3,2)=&lt;2,3&gt;
\nabla f(3,2) is perpendicular to the tanget line, so the tangent line has equation
\nabla f(3,2)\cdot&lt;x-3,y-2&gt;=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)

 

[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?

 

[mrcleanhands]

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

 

[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?

 

[tiny-tim]

hi mrcleanhands!

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)

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>

 

[mrcleanhands]

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?

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!