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## Main Question or Discussion Point

Hello,

I often encounter some confusion.

For example, we have the equation for a circle of radius one given by:

x

Now we can't express this as a function in R2 because it's a relation. however, we can still find derivatives like dy/dx or dx/dy using implicit differentiation.

Something that gets me though is, how does this implicit function relate to a function of two varaibles? For example, we could create the function

z=x

Now for example if we let z=1, won't this be the circle of radius one, lying in the plane z=1?

How do the derivatives of these two things relate to each other? For example, in multivariable equations, we wouldn't just say "what's the derivative of z" because that doesn't make sense, we'd think, "well, in terms of what variable?". If we wanted dz/dx we'd take the partial derivative of z in terms of x. But for example, if we go back to the 2D relation x

d/dx(x

2x+2y(dy/dx)=0 --> (dy/dx)=(-x/y)

But if we now go back to our 3D equation z=x

I often encounter some confusion.

For example, we have the equation for a circle of radius one given by:

x

^{2}+y^{2}=1Now we can't express this as a function in R2 because it's a relation. however, we can still find derivatives like dy/dx or dx/dy using implicit differentiation.

Something that gets me though is, how does this implicit function relate to a function of two varaibles? For example, we could create the function

z=x

^{2}+y^{2}Now for example if we let z=1, won't this be the circle of radius one, lying in the plane z=1?

How do the derivatives of these two things relate to each other? For example, in multivariable equations, we wouldn't just say "what's the derivative of z" because that doesn't make sense, we'd think, "well, in terms of what variable?". If we wanted dz/dx we'd take the partial derivative of z in terms of x. But for example, if we go back to the 2D relation x

^{2}+y^{2}=1, if we want to find the derivative of y in terms of x, we use implicit differentiation and go:d/dx(x

^{2}+y^{2})=d/dx(1)2x+2y(dy/dx)=0 --> (dy/dx)=(-x/y)

But if we now go back to our 3D equation z=x

^{2}+y^{2}, say we want the derivative of y in terms of x when z = 1. Would this end up being the same thing?