Cartesian coordinates

**karatekid_81** · September 20th 2007, 04:46 PM

I am trying to find the equation of the tangent line in cartestian coordinates of the curve given in polar coordinates by

r=3-2 costheta, at theta=pi/3

i have that

from there I am stuck on the rest I know that

now can someone plz help me solve it

**Jhevon** · September 20th 2007, 08:45 PM

Originally Posted by karatekid_81

I am trying to find the equation of the tangent line in cartestian coordinates of the curve given in polar coordinates by

r=3-2 costheta, at theta=pi/3

i have that

from there I am stuck on the rest I know that

now can someone plz help me solve it

here's what you want to do. first, find $\frac {dy}{dx}$ . how? just plug in all the values and solve.

$\frac {dy}{dx} = \frac {r \cos \theta + \frac {dr}{d \theta} \sin \theta}{-r \sin \theta + \frac {dr}{d \theta} \cos \theta}$

we know $r$ and you found $\frac {dr}{d \theta}$

$\Rightarrow \frac {dy}{dx} = \frac {(3 - 2 \cos \theta) \cos \theta + (2 \sin \theta) \sin \theta}{-(3 - 2 \cos \theta) \sin \theta + (2 \sin \theta) \cos \theta}$

simplify that, and plug in $\theta = \frac {\pi}3$ when done. or you can just plug it in now and calculate the result, it's up to you

now, when $\theta = \frac {\pi}3$ , $r = 2$

so the point we are concerned with is $(r, \theta) = \left( 2, \frac {\pi}3 \right)$

now we must change this coordinate to Cartesian coordinates:

recall, to go from Polar coordinates $(r , \theta)$ to Cartesian coordinates $(x,y)$ , we use the fact that:

$x = r \cos \theta$ and $y = r \sin \theta$

use those equations to find the point $(x,y)$

now, finally, use the point slope form to get the equation of the tangent line.

using $m = \frac {dy}{dx}$ and $(x_1,y_1)$ as the point you found just above, we have by the point slope form:

$y - y_1 = m(x - x_1)$

plug in the values and solve for $y$ and that's your line

**karatekid_81** · September 21st 2007, 02:58 AM

these whole trig functions mess with me, what would it look like simplified, i understand from there on but i dont get the whole simplify thing,

**Jhevon** · September 21st 2007, 09:15 AM

Originally Posted by karatekid_81

these whole trig functions mess with me, what would it look like simplified, i understand from there on but i dont get the whole simplify thing,

as i said. you don't HAVE TO simplify, it would be nice, but it is not necessary. you could just plug in $\theta = \frac {\pi}3$ into $\frac {dy}{dx} = \frac {(3 - 2 \cos \theta) \cos \theta + (2 \sin \theta) \sin \theta}{-(3 - 2 \cos \theta) \sin \theta + (2 \sin \theta) \cos \theta}$ and work it out.

however, if you were simplifying, you would do something like this:

$\frac {dy}{dx} = \frac {(3 - 2 \cos \theta) \cos \theta + (2 \sin \theta) \sin \theta}{-(3 - 2 \cos \theta) \sin \theta + (2 \sin \theta) \cos \theta}$

$= \frac {3 \cos \theta - 2 \cos^2 \theta + 2 \sin^2 \theta}{-3 \sin \theta + 2 \sin \theta \cos \theta + 2 \sin \theta \cos \theta}$

$= \frac {3 \cos \theta - 2 \left( {\color {red} cos^2 \theta - \sin^2 \theta } \right)}{-3 \sin \theta + 2 ({\color {blue} 2 \sin \theta \cos \theta})}$

what is in red is: $\cos 2 \theta$
what is in blue is: $\sin 2 \theta$

so plug those in, we get:

$\frac {dy}{dx} = \frac {3 \cos \theta - 2 \cos 2 \theta}{2 \sin 2 \theta - 3 \sin \theta}$

now plug in $\theta = \frac {\pi}3$

**karatekid_81** · September 21st 2007, 01:49 PM

thanks I believe I got it.

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