Question

calc 2

Answer 1

Given,\[\Large\bf\sf x(t)=\cos4t, \qquad\qquad y(t)=\sin7t\]The formula for arc length in parametric form is uhhhh,
\[\Large\bf\sf L\quad=\quad \int\limits_0^{2\pi}\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}~ dt\]Something like that right?
So where you getting stuck?

Answer 2

yeah not sure where to go from there

Answer 3

You need a couple derivatives that you can plug into the integral.

\[\Large\bf\sf x(t)=\cos4t\]\[\Large\bf\sf \frac{dx}{dt}=?\]

Answer 4

Hmm this is a weird problem...
Feels like it's going to work out similarly to the one we had yesterday, where it can't be integrated.

Answer 5

yeah thats what i though too. what do i do in that situation

Answer 6

Never mind, I was wrong.

Answer 7

let me try once:
we have arcsin y = \(\dfrac{7}{4} arccos x\)
so, y = 7/4 sin(arccos x)
dy/dx = 7/4 cos (arccos x)* (arccos x)' = 7/4 *x * (-1/ sqrt (1-x^2))
from this, put all into the formula
\[L = \int_0^{2\pi} \sqrt{1-(dy/dx)^2}dx\]
Not sure, need check

Answer 8

Well the thing is: \[
\frac{dy}{dt}=\frac{dy}{dx} \frac{dx}{dt}
\]Even when you solve for \(dy/dx\) here, you still get a nasty integrand.

Answer 9

\[\frac{dy}{dx}= -\frac{7}{4}\frac{x}{\sqrt{1-x^2}}\] and (dy/dx)^2 is quite "beautiful"

Answer 10

Well... \(t=(1/4)\arccos(x) \), so \(y=\sin((7/4)\arccos(x) )\)

Answer 11

oh, you see, to x = cos 4t, from 0 to 2pi . According to the formula, t =0, x = 1, t =2pi, x =1 too. therefore, when applying the formula, we get integral from 1 to 1 (respect to x) =0

Answer 12

You can't just pull out the \(7/4\) though.

Answer 13

\[L = \int_a^b \sqrt{1+(\frac{dy}{dx})^2}dx\] where \(a = f(\alpha\) ) and b = \(f(\beta\))
so, we convert a, b from x = cos 4t, (0, 2pi) to (1,1)
therefore int =0

Answer 14

Am I right? from my note, x = f(t) , y = g(t) , the integral can be calculate respect to x on that way.

Answer 15

Well, \([a,b] = [x(0), x(2\pi)]\), but I'm not sure if your \(y(x)\) is correct.

Answer 16

if we go on that way, no need to calculate integrand, right? since the limit shows the value of integral =0 for all y(x)

Answer 17

Well, if \(y(x)\) isn't even a function to begin with, then it is possible that \(y(1) \neq y(1)\),

Answer 18

Which is why not all parametrizations work.

Answer 19

I don't get what you mean by y(1) != y(1) if it is an odd function.

Answer 20

If you actually draw out the curve, you will realize that its length is not 0.

Answer 21

oh, got you, parametric curve. !!

Answer 22

The relationship between \(y\) and \(x\) is not that of a function, since \(x(t_1)=x(t_2)\not\to y(t_1)=y(t_2)\).

Answer 23

You could split it up inter certain sections, and then \(y(x)\) would be a function on those sections.

Answer 24

wow, it's so complicated.