# how to rotate a curve

• Jan 26th 2010, 11:11 AM
dynamo
how to rotate a curve
how do you rotate a curve ,whether it is a quadratic ,parabolic any type of curve and get the equation of the new/transformed curve,after rotating by x degrees.
• Jan 27th 2010, 01:55 AM
Corum
Are you familier with polar coordinates?
Imagine your equation for the curve. Y=...? in polar coordinates (with x=r cos $\displaystyle \theta$ and y=r sin $\displaystyle \theta$) it is expressed in r(distance from origin) and $\displaystyle \theta$, angle created with oX+. with this it's easy to change $\displaystyle \theta$ to the appropriate angle.
I hope this helps
• Jan 27th 2010, 02:29 AM
HallsofIvy
Dynamo, to rotate y= f(x) through angle $\displaystyle \theta$, around the origin, replace x with $\displaystyle x'= x cos(\theta)+ y sin(\theta)$ and y with $\displaystyle y= -x sin(\theta)+ y cos(\theta)$.

To rotate y= f(x) through angle $\displaystyle \theta$, about point $\displaystyle (x_0, y_0)$, first translate [tex](x_0, y_0) to the origin by subtracting $\displaystyle x_0$ from x and $\displaystyle y_0$ from y, then do the rotation, then translate back.

That is, replace x with $\displaystyle x'= (x- x_0)cos(\theta)+ (y- y_0)sin(\theta)+ x_0$ and replace y with $\displaystyle y'= -(x- x_0)sin(\theta)+ (y- y_0)cos(\theta)+ y_0$
• Jan 27th 2010, 04:15 AM
dynamo
Quote:

Originally Posted by Corum
Are you familier with polar coordinates?
Imagine your equation for the curve. Y=...? in polar coordinates (with x=r cos $\displaystyle \theta$ and y=r sin $\displaystyle \theta$) it is expressed in r(distance from origin) and $\displaystyle \theta$, angle created with oX+. with this it's easy to change $\displaystyle \theta$ to the appropriate angle.
I hope this helps

thanks
• Jan 27th 2010, 04:32 AM
dynamo
thanks
Quote:

Originally Posted by HallsofIvy
Corum, on this forum use "\", not "#". Also, the final "/MATH must be completely inclosed in [ ].

"[ math ] \theta [ /math ] (without the spaces) gives $\displaystyle \theta$.

To see the LaTex code for any expression in a post, click on that expression.

Dynamo, to rotate y= f(x) through angle $\displaystyle \theta$, around the origin, replace x with $\displaystyle x'= x cos(\theta)+ y sin(\theta)$ and y with $\displaystyle y= -x sin(\theta)+ y cos(\theta)$.

To rotate y= f(x) through angle $\displaystyle \theta$, about point [tex](x_0, y_0), first translate [tex](x_0, y_0) to the origin by subtracting $\displaystyle x_0$ from x and $\displaystyle y_0$ from y, then do the rotation, then translate back.

That is, replace x with $\displaystyle x'= (x- x_0)cos(\theta)+ (y- y_0)sin(\theta)+ x_0$ and replace y with $\displaystyle y'= -(x- x_0)sin(\theta)+ (y- y_0)cos(\theta)+ y_0$

thanks