# X-Coordinates on Tangent Line

Show 40 post(s) from this thread on one page
Page 1 of 2 12 Last
• Oct 23rd 2007, 03:58 PM
FalconPUNCH!
X-Coordinates on Tangent Line
Find the x-coordinates of all points on the curve $\displaystyle y = sin2x - 2sinx$ at which the tangent line is horizontal.

I don't know what to do and where to begin. :confused:
• Oct 23rd 2007, 04:08 PM
Jhevon
Quote:

Originally Posted by FalconPUNCH!
Find the x-coordinates of all points on the curve $\displaystyle y = sin2x - 2sinx$ at which the tangent line is horizontal.

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

the tangent line is horizontal when the derivative is zero. find the derivative, set it equal to zero and solve for x
• Oct 23rd 2007, 04:10 PM
FalconPUNCH!
Quote:

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.
• Oct 23rd 2007, 04:16 PM
FalconPUNCH!
I got $\displaystyle y' = (cos(2)) (2x) - 2sin(x)cos(x) = 0$ as the derivative and I pretty much don't know how to solve it for x.
• Oct 23rd 2007, 04:19 PM
Jhevon
Quote:

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

that's because your derivative is incorrect

$\displaystyle y' = 2 \cos 2x - 2 \cos x$
• Oct 23rd 2007, 04:21 PM
FalconPUNCH!
Quote:

Originally Posted by Jhevon
that's because your derivative is incorrect

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

could you please explain how you got that because every time I try I get a different answer.
• Oct 23rd 2007, 04:32 PM
Jhevon
Quote:

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:

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

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

the other piece is trivial. $\displaystyle \frac d{dx} \sin x = \cos x$ so $\displaystyle \frac d{dx} -2 \sin x = -2 \frac d{dx} \sin x = -2 \cos x$
• Oct 23rd 2007, 04:37 PM
FalconPUNCH!
Quote:

Originally Posted by Jhevon
for the first part we use the chain rule:

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

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

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

Oh I know what I did wrong. I forgot to take the -2 like this $\displaystyle -2 \frac d{dx} sinx$
• Oct 23rd 2007, 04:39 PM
Jhevon
Quote:

Originally Posted by FalconPUNCH!
Oh I know what I did wrong. I forgot to take the -2 like this $\displaystyle -2 \frac d{dx} sinx$

ok. now set the derivative to zero and continue
• Oct 23rd 2007, 04:46 PM
FalconPUNCH!
Quote:

Originally Posted by Jhevon
ok. now set the derivative to zero and continue

I got cos2x = cosx :confused:
• Oct 23rd 2007, 05:04 PM
Jhevon
Quote:

Originally Posted by FalconPUNCH!
I got cos2x = cosx :confused:

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

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

this is quadratic in cos(x). you can take it from here right?
• Oct 23rd 2007, 05:05 PM
FalconPUNCH!
Quote:

Originally Posted by Jhevon
ok. now set the derivative to zero and continue

Quote:

Originally Posted by Jhevon

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

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

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.
• Oct 23rd 2007, 05:07 PM
Jhevon
Quote:

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:

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

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

ok, continue. tell me if you have any problems
• Oct 23rd 2007, 05:14 PM
FalconPUNCH!
Quote:

Originally Posted by Jhevon
you should look them up. for double angles you must know that:

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

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

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 :(
• Oct 23rd 2007, 05:22 PM
Jhevon
Quote:

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)*

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

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

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

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

what angles does this happen for?
Show 40 post(s) from this thread on one page
Page 1 of 2 12 Last