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

In summary, the derivative of x^2 + y^2 = 2y is undefined, and the tangent line to this equation at (1,1) does not exist.
  • #1
Eshi
27
0

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?
 
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  • #2
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.
 
  • #3
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
 
  • #4
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?
 
  • #5
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
 
  • #6
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.
 
  • #7
ok, yea that makes more sense.

So the answer to the question is that the tangent is undefined
 
  • #8
No! The tangent is defined. It's a vertical line. Just the slope is undefined.
 
  • #9
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
 
Last edited:
  • #10
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.
 
  • #11
Dick said:
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?

More like x = 2 :wink:
 

1. What is the formula for finding the derivative of a curve at a given point?

The formula for finding the derivative of a curve at a given point is dy/dx = (dy/dx)/(dx/dx). This can also be written as dy/dx = (dy/dx)/(1) since dx/dx is always equal to 1.

2. How do you find the derivative of a curve that is defined implicitly?

To find the derivative of a curve defined implicitly, you can use the implicit differentiation method. This involves differentiating both sides of the equation with respect to x, and then solving for dy/dx.

3. What is the equation for the tangent line of a curve at a given point?

The equation for the tangent line of a curve at a given point is y-y1 = m(x-x1), where (x1, y1) is the given point and m is the slope of the tangent line.

4. How do you find the slope of the tangent line of a curve at a given point?

The slope of the tangent line of a curve at a given point can be found by taking the derivative of the curve at that point. This will give you the slope of the curve at that point, which is also the slope of the tangent line.

5. Can the derivative of a curve at a given point be negative?

Yes, the derivative of a curve at a given point can be negative. This indicates that the curve is decreasing at that point, and the tangent line would have a negative slope. The sign of the derivative can also change depending on the direction from which the point is approached.

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