Finding the Tangent Line to a Level Curve at a Given Point: How Do I Do It?

[pezzang]

Hi, I had problem solving one question.
Please help me. I included my answers. Please check to make sure that my way is right.
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
gradient f(x,y) = <fsubx, fsuby>
fsubx = 2x
fsuby = 8y
therefore, gradient f(x,y) = <2x,8y>
gradient f(2,1) = <4,8>
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 don't 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.

 

[pnaj]

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]

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.

 

[pnaj]

Doh!

Thanks for the correction!