the equation of normal lline/tangent line

**haebinpark** · February 17th 2010, 05:07 AM

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) ?????

**e^(i*pi)** · February 17th 2010, 05:44 AM

Originally Posted by haebinpark

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) ?????

You can differentiate $\cos^2(x)$ directly using the chain rule:

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

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

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

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

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

To find your gradient find $f' \left(\frac{\pi}{3}\right)$

**haebinpark** · February 17th 2010, 09:43 AM

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?

**haebinpark** · February 17th 2010, 11:27 AM

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

**icemanfan** · February 17th 2010, 11:34 AM

Originally Posted by haebinpark

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...

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

You made an error in the last step of this series of equations (what happened to the factor of pi?)

**haebinpark** · February 17th 2010, 11:37 AM

= 2sqrt3/3x - [(8sqrt3 pi + 9)/36]

is this what you're taking about?

**icemanfan** · February 17th 2010, 11:47 AM

Originally Posted by haebinpark

= 2sqrt3/3x - [(8sqrt3 pi + 9)/36]

is this what you're taking about?

Yes. Although the correct equation is actually

$y = \frac{2\sqrt{3}}{3}x + \frac{9 - 8\pi\sqrt{3}}{36}$

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