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

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- Oct 2nd 2007, 05:56 PMMr_GreenPoint 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 PMThePerfectHacker
- Oct 2nd 2007, 06:07 PMMr_Green
so thh derivate of -csc is =

So I set this equal to 2/3? - Oct 2nd 2007, 06:17 PMThePerfectHacker
- Oct 2nd 2007, 06:38 PMMr_Green
so i get

pi/3

Is this correct? Is this the only point? - Oct 4th 2007, 02:40 AMMr_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 AMtopsquark
$\displaystyle y = -csc(x) = -\frac{1}{sin(x)}$

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

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

$\displaystyle cos(x) = \frac{1}{2} \implies x = \frac{\pi}{3}, \frac{5 \pi}{3}$

So there are two answers.

-Dan - Oct 5th 2007, 08:03 AMMr_Green
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