- 28 Jan '06 20:54so a math problem that i'm having trouble with. somebody please help. this isn't the entire problem...just the part a can't figure out.

so along the curve y=sin(x) the velocity of an object is described as V=1+5cos^2(x^2). How does one determine the time it takes for the object to move between two points. Specifically between the two points in which the curve y=sin(x) intersects y=x^2 (which is 0 and .8767)

Thanking whoever can help with this bastard in advance. Mahalo nui. NN - 29 Jan '06 10:00

A clue: If the curve had been a straight line (instead of a sinus curve), would it be easier to solve the problem?*Originally posted by Hawaiianhomegrown***so along the curve y=sin(x) the velocity of an object is described as V=1+5cos^2(x^2). How does one determine the time it takes for the object to move between two points. Specifically between the two points in which the curve y=sin(x) intersects y=x^2 (which is 0 and .8767)**

I never give the answers, I always give clues to progress. - 29 Jan '06 20:19

You need to find the length of the curve y = sin(x) between the two points in question. The formula for curve length is:*Originally posted by Hawaiianhomegrown***so a math problem that i'm having trouble with. somebody please help. this isn't the entire problem...just the part a can't figure out.**

so along the curve y=sin(x) the velocity of an object is described as V=1+5cos^2(x^2). How does one determine the time it takes for the object to move between two points. Specifically between the two points ...[text shortened]... h is 0 and .8767)

Thanking whoever can help with this bastard in advance. Mahalo nui. NN

integral from (a to b) of (sqrt(1 + (d/dx(f(x)))^2)) dx. This will be your distance.

So, we have: integral of(sqrt(1+cos^2(x))dx).

You'll have to do some fancy footwork with trig identities to integrate that function, but I hope that this helps you. - 01 Feb '06 07:04

How 'bout this:*Originally posted by Hawaiianhomegrown***oh...that was sarcasm...it didn't help at all. The problem I'm having is that the velocity is a function of where you are along the x-axis. I got the distance of the curve but how long does it take to go through the distance as a function of x?**

arc length = L(x) = int[x0,x] of SQRT(1+cos^2(x))

speed = v = 1+5cos^2(x^2)

Also, v = dL/dt = dL/dx * dx/dt by the chain rule.

Now, dL/dx = SQRT(1+cos^2(x)), and subbing this into the expression derived from the chain rule and solving for dx/dt, we get:

dx/dt = [1+5cos^2(x^2)]/[SQRT(1+cos^2(x))]

This is a separable differential equation, so we can separate the like terms to get:

int(x0,x) of [SQRT(1+cos^2(x))]/[1+5cos^2(x^2)] dx = int(t0,t) of dt

So there's your expression for the time as a function of x. Now you just have to solve it. If it's wrong, just PM royalchicken. He eats this kind of crap for breakfast. Then he picks his teeth with the Navier-Stokes equations. Then he takes a nap, but that's just for show. - 03 Feb '06 16:42

Yeah, but we invoked your name a few times. Better get used to being the RHP math messiah.*Originally posted by royalchicken***But I wasn't involved at all .**

Oh, BTW, I'm going to kidnap a few random pedestrians later this week and shave their heads until they promise to be your acolytes, so if you could swing by, umm, maybe next Wednesday to anoint their shorn skulls, that would be great!

Prosperity and long life, royal one! Baacock!

- 03 Feb '06 17:23

Funnily enough, the original and probably greatest RHP maths messiah was Acolyte. I met him in person about a year ago, but he seems to have vanished from RHP, unfivetunately.*Originally posted by PBE6***Yeah, but we invoked your name a few times. Better get used to being the RHP math messiah.**

Oh, BTW, I'm going to kidnap a few random pedestrians later this week and shave their heads until they promise to be your*acolytes*, so if you could swing by, umm, maybe next Wednesday to anoint their shorn skulls, that would be great!

Prosperity and long life, royal one! Baacock!

Do I have to shave my head into some sort of rooster mohawk? I'm totally up for that.

BAAAacock back atcha. - 03 Feb '06 19:18

I remember Acolyte...his avatar was a giant digestive cookie, wasnt' it? Whatever happened to him? I remember he had some pretty interesting puzzles. Too bad he's gone, I love digestive cookies. They're so close to poop, yet sweet like candy. Truly nature's perfect food.*Originally posted by royalchicken***Funnily enough, the original and probably greatest RHP maths messiah was Acolyte. I met him in person about a year ago, but he seems to have vanished from RHP, unfivetunately.**

Do I have to shave my head into some sort of rooster mohawk? I'm totally up for that.

BAAAacock back atcha.

The Royal One's wishes are law, Royal One. I will send several peons to gather a team of capon stylists so that your ex-salted-ness may pick and choose in a manner befitting Reality TV or grade school sports teams.

O! Let stringed instruments of every kind ring out, and the corn in the fields be shucked in joy, and the wheat in the fields sheaved in mirth!