# how to rotate a curve

• Jan 26th 2010, 12:11 PM
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, 02:55 AM
Corum
Are you familier with polar coordinates?
Imagine your equation for the curve. Y=...? in polar coordinates (with x=r cos $\theta$ and y=r sin $\theta$) it is expressed in r(distance from origin) and $\theta$, angle created with oX+. with this it's easy to change $\theta$ to the appropriate angle.
I hope this helps
• Jan 27th 2010, 03:29 AM
HallsofIvy
Dynamo, to rotate y= f(x) through angle $\theta$, around the origin, replace x with $x'= x cos(\theta)+ y sin(\theta)$ and y with $y= -x sin(\theta)+ y cos(\theta)$.

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

That is, replace x with $x'= (x- x_0)cos(\theta)+ (y- y_0)sin(\theta)+ x_0$ and replace y with $y'= -(x- x_0)sin(\theta)+ (y- y_0)cos(\theta)+ y_0$
• Jan 27th 2010, 05:15 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 $\theta$ and y=r sin $\theta$) it is expressed in r(distance from origin) and $\theta$, angle created with oX+. with this it's easy to change $\theta$ to the appropriate angle.
I hope this helps

thanks
• Jan 27th 2010, 05:32 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 $\theta$.

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

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

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

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

thanks