the question is

find the equation of the normal line and tangent line to the graph of y = cos^2 x at x = Pi/3

how do i do this question without knowing the equation (something like 2x^2 + 2y^2 = 9xy that i can diffrentiate or derivate) ?????

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- Feb 17th 2010, 05:07 AMhaebinparkthe equation of normal lline/tangent line
the question is

find the equation of the normal line and tangent line to the graph of y = cos^2 x at x = Pi/3

how do i do this question without knowing the equation (something like 2x^2 + 2y^2 = 9xy that i can diffrentiate or derivate) ????? - Feb 17th 2010, 05:44 AMe^(i*pi)
You can differentiate $\displaystyle \cos^2(x)$ directly using the chain rule:

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

If you prefer to use an identity: $\displaystyle \cos^2(x) = \frac{\cos(2x)}{2}+\frac{1}{2}$

$\displaystyle \frac{d}{dx} \left[\frac{1}{2}\cos(2x)\right] = -\sin(2x)$

Note that $\displaystyle \frac{d}{dx}\, 0.5 = 0$

These answers are equal as $\displaystyle \sin(2x) = 2\sin(x) \cos(x)$

To find your gradient find $\displaystyle f' \left(\frac{\pi}{3}\right)$ - Feb 17th 2010, 09:43 AMhaebinpark
i found slope of tangent

f'(pi/3) = -2(cos pi/3)(sin pi/3)

= -2(1/2)(sqrt 3 /2)

= - sqrt(3) / 2

i also found the point (pi/3 , 1/4)

then what do i do next? - Feb 17th 2010, 11:27 AMhaebinpark
here's what i tried

if i did something wrong, could you please point me out?

because i still am not sure if i did it right or wrong... the answer doesn't seem to right...

f(x)=y=cos^2 x

f(pi/3)= (cos pi/3)^2 = (1/2)^2 = 1/4

point: (pi/3 , 1/4) (x0,y0)

f'(x)= -2(cosx)(sinx)

f'(pi/3) = -2(1/2)(sqrt3/2) = -sqrt3 / 2 <-- slope of tangent

y=m(x-x0)+y0

y= -sqrt3/2 (x-pi/3)+1/4

= -(sqrt3/2 x)+[(2sqrt3 pi + 3)/12] <-- eq. of tangent

slope of normal line = -1/-sqrt3 / 2 = 2/sqrt3 = 2sqrt3/3

y=m(x-x0)+y0

y= 2sqrt3/3 (x-pi/3) + 1/4

= 2sqrt3/3x - [(8sqrt3 + 9)/36] <--- eq. of normal line - Feb 17th 2010, 11:34 AMicemanfan
- Feb 17th 2010, 11:37 AMhaebinpark
= 2sqrt3/3x - [(8sqrt3 pi + 9)/36]

is this what you're taking about? - Feb 17th 2010, 11:47 AMicemanfan