Question

Find k(x) if f(x) = sin(x). Now find where k(x) is maximum on the interval [0, pi]...I have absolutely no idea how to do this, if someone could do each step and post it, I would be eternally grateful. AWARD AT END

Answer 1

This is what they said we would use : y = f(x) we choose x as the parameter...Then they show the formula... k(x) = |f''(x)| / [1 + (y')2]^(3/2)

Answer 2

I will actually be signing off, gotta get some sleep before class tomorrow(geez 2 hours of sleep)...If this could be answered in the morning, whoever, does that..Seriously you would make a college students day. Thanks so much! Have a wonderful day/night! :) "We all need help, so lend a hand, make that change." ~Anonymous

Answer 3

use the information of graph

Answer 4

The derivative of the function will equal zero at critical points,
and the second derivative of the function will be negative for a maximum critical point

Answer 5

first make the rough graph of function

Answer 6

@zepdrix, \(|f''(x)|=|-\sin x|=|\sin x|\). There's still a negative part to the function

Answer 7

Actually it doesn't matter, since \(\sin x>0\) for \(0\le x\le\pi\).
\[k(x)=\frac{\sin x}{\left(1+\cos^2x\right)^{3/2}}\]
So yeah, we take the derivative:
\[k'(x)=\frac{\cos x(1+\cos^2x)^{3/2}-\dfrac{3}{2}\sin x(1+\cos^2x)^{1/2}(-2\cos x \sin x)}{(1+\cos^2x)^3}\]
\[k'(x)=\frac{\cos x(1+\cos^2x)^{3/2}+3\sin^2 x(1+\cos^2x)^{1/2}\cos x}{(1+\cos^2x)^3}\]
\[k'(x)=\cos x(1+\cos^2x)^{1/2}\frac{1+\cos^2x+3\sin^2 x}{(1+\cos^2x)^3}\]
\[k'(x)=\cos x\frac{2+2\sin^2 x}{(1+\cos^2x)^{5/2}}\]
Set equal to 0 and solve for the critical points.

Answer 8

The fraction part is never zero, so you just find when \(\cos x=0\) in \([0,\pi]\).

Answer 9

@Loser66 at the end after Sith made his last comment, is it cosx=0 which is where k(x) is at the maximum?

Answer 10

I understand it as: You have to find out the max/ min from k'(x) by let it =0, except cos =0, k' =0, the second part is never =0 ( because the max value of sin^2 , cos^2 =1, so that 2+1 is not 0 for numerator)
That's why prof Sith asked you to find the value of x in the interval which makes cos =0 to have k'=0

Answer 11

Well..cos pi/2 = 0. Is that it then?

Answer 12

The maximum is pi/2 which = 1.5707963267948966

Answer 13

so, when x =pi/2 , k(x) =1
that is max of k(x)

Answer 14

Yeah, according to the plot it looks like there's a max at \(\dfrac{\pi}{2}\):
http://www.wolframalpha.com/input/?i=Sin%5Bx%5D%2F%281%2BCos%5E2%5Bx%5D%29%5E%283%2F2%29