# Point on a CSC graph

• Oct 2nd 2007, 05:56 PM
Mr_Green
Point on a CSC graph
Find the points on the graph of y = -csc(x), where the tangent is parallel to 3y-2x=4. Use exact values.

0 is less than or equal to x which is less than or equal to 2*pi

Thanks
• Oct 2nd 2007, 06:00 PM
ThePerfectHacker
Quote:

Originally Posted by Mr_Green
Find the points on the graph of y = -csc(x), where the tangent is parallel to 3y-2x=4. Use exact values.

0 is less than or equal to x which is less than or equal to 2*pi

Thanks

$\displaystyle 3y - 2x = 4\implies y = (2/3)x+(4/3)$.

To be parallel it means the derivative must be equal to 2/3. Now find all point with derivative 2/3.
• Oct 2nd 2007, 06:07 PM
Mr_Green
so thh derivate of -csc is =

So I set this equal to 2/3?
• Oct 2nd 2007, 06:17 PM
ThePerfectHacker
Quote:

Originally Posted by Mr_Green
so thh derivate of -csc is =

So I set this equal to 2/3?

Yes.
• Oct 2nd 2007, 06:38 PM
Mr_Green
so i get

pi/3

Is this correct? Is this the only point?
• Oct 4th 2007, 02:40 AM
Mr_Green
i found that out by plugging into my calculator and using the intersect ability on the graph. Is there anything mathematical that I could do to solve it?
• Oct 4th 2007, 03:47 AM
topsquark
Quote:

Originally Posted by Mr_Green
Find the points on the graph of y = -csc(x), where the tangent is parallel to 3y-2x=4. Use exact values.

0 is less than or equal to x which is less than or equal to 2*pi

Thanks

Quote:

Originally Posted by Mr_Green
so i get

pi/3

Is this correct? Is this the only point?

$\displaystyle y = -csc(x) = -\frac{1}{sin(x)}$

$\displaystyle \frac{dy}{dx} = - \left ( - \frac{cos(x)}{sin^2(x)} \right ) = \frac{cos(x)}{sin^2(x)}$

So we need to solve
$\displaystyle \frac{cos(x)}{sin^2(x)} = \frac{2}{3}$

$\displaystyle 3cos(x) = 2sin^2(x) = 2(1 - cos^2(x))$

$\displaystyle 2cos^2(x) + 3cos(x) - 2 = 0$

Let $\displaystyle z = cos(x)$. Then this equation becomes:
$\displaystyle 2z^2 + 3z - 2 = 0$

This has solutions:
$\displaystyle z = -2$ or $\displaystyle z = \frac{1}{2}$

Thus
$\displaystyle cos(x) = -2$ <-- Impossible for real x.
or
$\displaystyle cos(x) = \frac{1}{2} \implies x = \frac{\pi}{3}, \frac{5 \pi}{3}$