X-Coordinates on Tangent Line

**FalconPUNCH!** · October 23rd 2007, 03:58 PM

Find the x-coordinates of all points on the curve $y = sin2x - 2sinx$ at which the tangent line is horizontal.

I don't know what to do and where to begin.

**Jhevon** · October 23rd 2007, 04:08 PM

Originally Posted by FalconPUNCH!

Find the x-coordinates of all points on the curve $y = sin2x - 2sinx$ at which the tangent line is horizontal.

I don't know what to do and where to begin.

the tangent line is horizontal when the derivative is zero. find the derivative, set it equal to zero and solve for x

**FalconPUNCH!** · October 23rd 2007, 04:10 PM

Originally Posted by Jhevon

the tangent line is horizontal when the derivative is zero. find the derivative, set it equal to zero and solve for x

Thanks I'll try that.

**FalconPUNCH!** · October 23rd 2007, 04:16 PM

I got $y' = (cos(2)) (2x) - 2sin(x)cos(x) = 0$ as the derivative and I pretty much don't know how to solve it for x.

**Jhevon** · October 23rd 2007, 04:19 PM

Originally Posted by FalconPUNCH!

I got $y' = (cos(2)) (2x) - 2sin(x)cos(x) = 0$ as the derivative and I pretty much don't know how to solve it for x.

that's because your derivative is incorrect

$y' = 2 \cos 2x - 2 \cos x$

**FalconPUNCH!** · October 23rd 2007, 04:21 PM

Originally Posted by Jhevon

that's because your derivative is incorrect

$y' = 2 \cos 2x - 2 \cos x$

could you please explain how you got that because every time I try I get a different answer.

**Jhevon** · October 23rd 2007, 04:32 PM

Originally Posted by FalconPUNCH!

could you please explain how you got that because every time I try I get a different answer.

for the first part we use the chain rule:

$\frac d{dx}f(g(x)) = f'(g(x)) \cdot g'(x)$

for $\sin 2x$ : $f(x) = \sin x$ and $g(x) = 2x$

the other piece is trivial. $\frac d{dx} \sin x = \cos x$ so $\frac d{dx} -2 \sin x = -2 \frac d{dx} \sin x = -2 \cos x$

**FalconPUNCH!** · October 23rd 2007, 04:37 PM

Originally Posted by Jhevon

for the first part we use the chain rule:

$\frac d{dx}f(g(x)) = f'(g(x)) \cdot g'(x)$

for $\sin 2x$ : $f(x) = \sin x$ and $g(x) = 2x$

the other piece is trivial. $\frac d{dx} \sin x = \cos x$ so $\frac d{dx} -2 \sin x = -2 \frac d{dx} \sin x = -2 \cos x$

Oh I know what I did wrong. I forgot to take the -2 like this $-2 \frac d{dx} sinx$

**Jhevon** · October 23rd 2007, 04:39 PM

Originally Posted by FalconPUNCH!

Oh I know what I did wrong. I forgot to take the -2 like this $-2 \frac d{dx} sinx$

ok. now set the derivative to zero and continue

**FalconPUNCH!** · October 23rd 2007, 04:46 PM

Originally Posted by Jhevon

ok. now set the derivative to zero and continue

I got cos2x = cosx

**Jhevon** · October 23rd 2007, 05:04 PM

Originally Posted by FalconPUNCH!

I got cos2x = cosx

are you forgetting your identities?

$\cos 2x - \cos x = 0$

$\Rightarrow 2 \cos^2 x - 1 - \cos x = 0$

this is quadratic in cos(x). you can take it from here right?

**FalconPUNCH!** · October 23rd 2007, 05:05 PM

Originally Posted by Jhevon

ok. now set the derivative to zero and continue

Originally Posted by Jhevon

are you forgetting your identities?

$\cos 2x - \cos x = 0$

$\Rightarrow 2 \cos^2 x - 1 - \cos x = 0$

this is quadratic in cos(x). you can take it from here right?

To tell you the truth I can't remember any of the identities since I haven't used them in a year. I'll try and take it from there.

**Jhevon** · October 23rd 2007, 05:07 PM

Originally Posted by FalconPUNCH!

To tell you the truth I can't remember any of the identities since I haven't used them in a year. I'll try and take it from there.

you should look them up. for double angles you must know that:

$\cos 2x = \cos^2 x - \sin^2 x = 2 \cos^2 x - 1 = 1 - 2 \sin^2 x$

$\sin 2x = 2 \sin x \cos x$

ok, continue. tell me if you have any problems

**FalconPUNCH!** · October 23rd 2007, 05:14 PM

Originally Posted by Jhevon

you should look them up. for double angles you must know that:

$\cos 2x = \cos^2 x - \sin^2 x = 2 \cos^2 x - 1 = 1 - 2 \sin^2 x$

$\sin 2x = 2 \sin x \cos x$

ok, continue. tell me if you have any problems

Yeah I don't think I can do it. I'm just going in circles. I was never good with functions

**Jhevon** · October 23rd 2007, 05:22 PM

Originally Posted by FalconPUNCH!

Yeah I don't think I can do it. I'm just going in circles. I was never good with functions

are you taking me for a ride? or are you really trying? *suspicious smiley (we still need one of these)*

$2 \cos^2 x - \cos x - 1 = 0$ ....this is a quadratic, see if you can factorize

$\Rightarrow (2 \cos x + 1)(\cos x - 1) = 0$

$\Rightarrow 2 \cos x + 1 = 0 \mbox{ or } \cos x - 1 = 0$

$\Rightarrow \cos x = - \frac 12 \mbox{ or } \cos x = 1$

what angles does this happen for?

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