Very very short question on second derivative

[CookieSalesman]

What does it mean when I have to find the second derivative of a circle at a given point? (Implicit diffing)

In specifics, the equation is 9x² +y² =9
At the point (0,3)
You don't really need the rest at all, but it was just my process.

This seems to make no sense.
first D'v 18x+2yy^'=0
Second derivative gives 18+y^''=0
Doesn't it?
How am I supposed to substitute coordinates?

 

[Mark44]

What does it mean when I have to find the second derivative of a circle at a given point? (Implicit diffing)

In specifics, the equation is 9x² +y² =9
At the point (0,3)

This seems to make no sense.
first D'v 18x+2yy^'=0

That equation is not the first derivative. You need to solve for y' algebraically.

Second derivative gives 18+y^''=0
Doesn't it?

No. When you have isolated y', differentiate it to get y''. Then, to find the value of y'' at (0, 3), substitute these coordinates in your formula.

How am I supposed to substitute coordinates?

 

[HallsofIvy]

What does it mean when I have to find the second derivative of a circle at a given point? (Implicit diffing)

Strictly speaking, it doesn't mean anything! You do not take the derivative of geometric objects, you take the derivative of functions. What you are asking about the relation describing the circle.

In specifics, the equation is 9x² +y² =9
At the point (0,3)

That's not a circle, it's an ellipse.

This seems to make no sense.
first D'v 18x+2yy^'=0
Second derivative gives 18+y^''=0
Doesn't it?
How am I supposed to substitute coordinates?

Using "implicit differentiation" as you say, you get 18x+ 2yy'= 0 where y' is the derivative with respect to x.
Doing that again you do NOT get 18+ y''= 0. You forgot to use the product rule on 2yy'. Implicit differentiation gives, rather,

 

[CookieSalesman]

Okay, so if I differentiate once implicitly, I get 18+2yy^'
Then, algebra, so that it's -9/y = y^'
But if I differentiate that, then I have y prime and y double prime...
Oh, so I can find y prime from the earlier equation, then resubsitute?

Thanks.

 

[Mark44]

Okay, so if I differentiate once implicitly, I get 18+2yy^'
Then, algebra, so that it's -9/y = y^'
But if I differentiate that, then I have y prime and y double prime...
Oh, so I can find y prime from the earlier equation, then resubsitute?

Thanks.

That's the idea.