# Slope of a Circle

• December 12th 2007, 04:04 PM
Truthbetold
Slope of a Circle
...is $y'=\frac{-x}{y}$

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

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

Ty
Edit:oops! wrong secition. This isn't homework.
• December 12th 2007, 05:16 PM
Krizalid
Quote:

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

• December 12th 2007, 05:30 PM
Truthbetold
Quote:

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 = $\frac{36}{y^3}$
• December 12th 2007, 08:32 PM
earboth
Quote:

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

...

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

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

Then the first derivative (using implicite derivation) is

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

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

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

Substitute y':

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

Use the first equation and you'll get:

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

Actually you have to consider 2 cases:

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