Question

Hey guys, can you help me with Exam 2 problem 6? I fail to see why y is a function of x along the intersection, and how the idea of "eliminate dz" is derived in order to get dy/dx? Thanks in advance.

Answer 1

Looking at the solutions, I think "the two surfaces intersect in a curve along which y is a function of x" simply refers to the fact that dy/dx can be found without dz. That is merely speculation however. when you take the total differential, you get -4z^3dz + 2xdx + 3y^2dy = 0 for the first equation and (3z^2 + x)dz + (z+y)dx + xdy = 0 for the second. Since you eventually want to end up with dy/dx (without dz), you would want to eliminate dz which can be done by solving for dz in each equation and setting them equal to each other.

Answer 2

As Amaterasu5 said, the statement that "the two curves intersect on some plane where y is a function of x" means, geometrically, that the two level surfaces, call them f(x,y,z) = c and g(x,y,z) = d, have an intersection that lies on some plane z = c, so that y is a function of x only and does not depend on z.

A little less abstractly, this means that if you drew the two surfaces in 3-d space, the line where they intersect would define a plane parallel to z = 0 (in other words, it would be flat, and if you looked at it from above, you'd be able to represent it as a one variable function in 2-d). This is the function y(x) to which the problem refers.

The question is asking you to find dy/dx in order to demonstrate the power of differentials, that you can solve for dy/dx without ever having to solve for y, or for some messy functions of z for that matter.

The method used to find dy/dx is explained perfectly by Amaterasu5, using the differentials obtained in part (a) of the problem and solving for dy/dx.

Answer 3

Those curves do not intersect with z constant. If you put z constant in each equation you will get different types of curve.

Just because y can be found as a function of x does not mean that z is constant. Take for example <cost ,sin t , t>.

I don't think that it's at all obvious how the surfaces in the question intersect but that's not part of the question, that's given.

I'm no authority on this however. I'm only up to lecture 15.