Accidental double post.
The curve r(t) passing through the point (1,1,1) is defined implicitly by the equations:
x^2y + y^2x + xyz + z^2 = 4
x + y + z = 3
Find the equation of the tangent line to the curve at the given point.
Attempt at a solution
GRAD f(x,y,z) = (2xy + y^2)i + (x^2 + 2xy)j + (xy + 2z)k
At (1,1,1):
GRAD f(1,1,1) = 3i + 3j + 3k
Tangent line has normal of (3,3,3).
Therefore:
3(x - 1) + 3(y - 1) + 3(z - 1) = 0
3x + 3y + 3z = 9
x + y + z = 3
I'm not sure what I'm doing wrong. Am I supposed to use both equations given in the question?
Since the curve is formed by both surfaces, don't you think that would be a good idea? A single curve does not define a surface, much less its tangent line.
But you started correctly: (3,3,3) is normal to the first surface and so is normal to any line in the surface. But your answer x+ y+ z= 3 is the equation of the plane tangent to the surface there.
The difficulty with using just one surface is that there exist an infinite number of vectors perpendicular to a give vector. There exist and infinite number of tangent lines to that surface.
Instead, find the normal to the second surface (which is actually a plane) at (1, 1, 1). The vector you want must be perpendicular to both of those- in other words it is their cross product.
Now that I look at it more closely, I see a problem with that: the two given surfaces are tangent at (1, 1, 1). The don't intersect in a curve to begin with! There is NO curve "defined implicitly" by those equations.
I think you do. If you eliminate using the plane, then you have
or
which is the curve projected into the plane.
@ little pony. As for the tangent line. It you let , (so ) then from (1) you can find For the last so now you have the three conponents of the tangent vector. You use the point to get the tangent line.