Derivative of x^2+y^2=2y at (1,1) & Tangent Line

[Eshi]

Homework Statement

What is the derivative of x^2 + y^2 = 2y, and find the tangent line to this equation at (1,1)

Homework Equations

The Attempt at a Solution

I get y' = x / (1-y). However, how do I find the tangent line to this? When I plug in the values it divides by zero! (1 / (1-1))

Is this a trick question?

 

[phsopher]

So the tangent line has an infinite slope. What does that mean? Alternatively you could take the derivative with respect to y instead of x if that makes it clearer.

 

[Eshi]

so as the derivative goes to that point, it goes to infinity, telling us that the function must also be growing infinitly.

But bottom line, the tangent line does not exist at (1,1), right? b/c there is no derivative at the point (1,1), there is just a limit

 

[Dick]

The function isn't growing to infinity, the slope is. You are missing phsopher's point. There are perfectly ordinary lines that have an undefined slope, aren't there? Like y=2?

 

[Eshi]

I'm still a little confused, if the slope was infinity wouldn't the function be increasing infinitly as well? I'm a little confused

 

[Dick]

Not at all. Take a simpler example. y^2=x. Sketch the graph and draw the tangent line at x=0, y=0. The 'function' isn't going to infinity.

 

[Eshi]

ok, yea that makes more sense.

So the answer to the question is that the tangent is undefined

 

[Dick]

No! The tangent is defined. It's a vertical line. Just the slope is undefined.

 

[Eshi]

aahhhh, lol. Ok, so how do i know that the tangent line is simply x = 1?

edit: oh wait, i think i got it. So i realize the slope is divided by 0, which means that the limit approaches infinity, and the slope is undefined, what i don't get is how u make the next connection and say that the tangent line is vertical because of the previous information.

edit 2: ok wait, i think i got it, so the if the tangent line were horizontal the derivative would be zero, if the tangent line were vertical the derivative would be a undefined, which was this case. So the tangent line becomes x = 1

 

[phsopher]

If it's more clear this way solve for the tangent in the yx-plane instead of xy-plane. That involves taking derivative in respect to y which is zero at (1,1). That means that in yx-plane the tangent is a horizontal line.

 

[Mark44]

The function isn't growing to infinity, the slope is. You are missing phsopher's point. There are perfectly ordinary lines that have an undefined slope, aren't there? Like y=2?

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