Multivariable Question.

Main Question or Discussion Point

Hi, I had problem solving one question.
1. IF f(x,y) = x^2 +4y^2, find the gradient vector f(2,1) and use it to find the tanenet line to the level curve f(x,y) = 8 in the xy-plane at the point (2,1).

I solved this way:

f(x,y) = x^2 + 4y^2
fsubx = 2x
fsuby = 8y
f(x,y) = 8

HOW DO I FIND THE EQUATION OF TANGENT LINE?????
- I know how to find the equation of tangent plane, but i dont know how to find line. If you can tell me how to find it, i would greatly appreciate your help and intellect.

AND, please make sure that What i did until now is right.
Thank you so much.

You want to find ANY vector that is parallel to the gradient at the point specified, and then simplify it.

Do this using the fact that the dot-product of two vectors is zero when they are parallel.

That is, if X, Y are vectors, then they are parallel if and only if
X dot Y = 0 .

So, let X = <x,y> and we have grad(F(2, 1)) = <4,8>.

The equation of the tangent is thus <x,y> dot <4, 8> = 0.

Or, 4x + 8y = 0.
Or, y = -x/2.

HallsofIvy
Homework Helper
pnaj- you used "parallel" when you meant "perpendicular"!

Also, 4x+ 8y= 0 is NOT tangent to the level curve f(x,y) = 8 in the xy-plane at the point (2,1). For one thing, it doesn't go through (2, 1)!

Yes, the gradient is perpendicular to the level curve and the tangent to ax+ by= c is perependicular to <a, b> so we can take
a= 4, b= 8. We then calculate c so that the line goes throught (2, 1). At (2, 1), 4x+ 8y= c become 8+ 8= 16. The tangent line to
f(x,y)= 8 at (2, 1) is 4x+ 8y= 16.

Doh!

Thanks for the correction!