# Thread: Slope of a Circle

1. ## Slope of a Circle

...is $\displaystyle y'=\frac{-x}{y}$

My book says the second derivative of a circle is also $\displaystyle y''=\frac{-x}{y}$
Every time I do it, I get $\displaystyle y''=\frac{-y}{x}$

So what is the 2nd derivative of a circle?
Por favor show some steps or tell me how you did it?

Ty
Editops! wrong secition. This isn't homework.

2. Originally Posted by Truthbetold
My book says the second derivative of a circle is also $\displaystyle y''=\frac{-x}{y}$
Every time I do it, I get $\displaystyle y''=\frac{-y}{x}$

3. Originally Posted by Krizalid
Got it solved by help from somewhere else.

The book doesn't say y''= -x/y
It doesn't say anything about the derivative of -x/y, even thouhg they ask you to do it and its an odd problem. Very strange this book.

y'' solved by complicated methods i've never heard before = $\displaystyle \frac{36}{y^3}$

4. Originally Posted by Truthbetold
...is $\displaystyle y'=\frac{-x}{y}$

...
Hi, obviously you are refering to a circle with the equation

$\displaystyle x^2 + y^2 = r^2$

Then the first derivative (using implicite derivation) is

$\displaystyle y' = \frac{-x}{y}$

To get the second derivative use quotient rule and chain rule:

$\displaystyle y''=\frac{y \cdot (-1) - (-x) \cdot y \cdot y'}{y^2}$

Substitute y':

$\displaystyle y''=\dfrac{y \cdot (-1) - (-x) \cdot y \cdot \frac{-x}{y}}{y^2} = \frac{-x^2 - y^2}{y^3}$

Use the first equation and you'll get:

$\displaystyle y''=\frac{-r^2}{y^3}$

Actually you have to consider 2 cases:

1. The conditions of the upper hemisphere
2. The conditions of the lower hemisphere