Implicitly Defined Curve: Tangent Line at (1,1,1) and Parameterized Derivatives

[mathman44]

Homework Statement

The curve t(t) passing thru the point (1,1,1) is defined implicitly by:

x²y + y²x + xyz + z² = 4
and
x + y + z = 3

a) find the equation for the tangent line to this curve at the given point
b) assume you may choose x=t as the parameter. Find the formula for:

dy/dt and d²y/dt²

The Attempt at a Solution

I honestly have no idea where to start! Can anyone give me hints to get me going? Ty.

 

[mathman44]

Any hints?

 

[Dick]

Ok, take x to be the parameter. So consider y(x) and z(x) to be functions of x. Take d/dx of both equations and differentiate implicitly. Then put x=1, y=1 and z=1. Doesn't that give you two equations in two unknowns for dy/dx and dz/dx? If you solve for them, then the tangent vector is (dx/dx,dy/dx,dz/dx)=(1,dy/dx,dz/dx), correct?

 

[lanedance]

i'm assuming you don't just want to substitute in, I haven't worked it through totally, but here's some geomteric interpretation I hope helps

the curve is contained in a plane, so its tangent vector will be parallel to the plane, so perpindicular to the normal of the plane

the curve is also contained in the other surface, so its tangent vector will be parallel to its tangent surface, consider a level surface of the function
F(x,y,z) = x^2y + y2x + xyz + z^2 - 4

taking the gradient of will give will give the rate of maximum change, perpindicular to the surface's tangent plane.

you know have two vectors perpindicular to the tangent vector, so you can find the direction of the tangent vector - this will give the relative lengths of dx/dt, dy/dt, dz/dt

now y" = d^2y/dt^2 is a bit trickier, but you know:
- as the curve is contained in a plane you know its acceleration vector must also be parallel to the plane..
- x" is zero,
these 2 should be enough to get the direction, though not too sure on the magnitude... any ideas..? maybe implicit differentiation of the first function?