Find the exact arc length of the curve over the stated interval. `x=cos(3t)`   ,   `y=sin(3t)`   ,   `0lt=tlt=pi`

1 Answer | Add Yours

embizze's profile pic

embizze | High School Teacher | (Level 1) Educator Emeritus

Posted on

The formula for the length of a curve given parametrically is `s=int_a^b sqrt(((dx)/(dt))^2+((dy)/(dt))^2) ` where the function is defined parametrically as x=g(t),y=h(t) on `a<=t<=b ` . However, this formula can only be applied if the curve does not intersect itself more than a finite number of times. In particular, it cannot intersect itself on a subinterval of the interval for t.

Consider the function defined by x=cos(3t), y=sin(3t). On the interval `0<=t<=(2pi)/3 ` this is a circle centered at the origin with radius 1.

The interval we are given is `0<=t<=pi ` ; on this interval the curve intersects itself on `(2pi)/3<t<=pi ` which is not allowed by the theorem. (The circle is traversed, and then one half of the circle is traced over again.)

If you applied the theorem you get:

`s=int_0^(pi) sqrt(d/(dt)(cos(3t))^2+d/(dt)(sin(3t))^2)dt `

`=int_0^(pi)sqrt((-3sin(3t))^2+(3cos(3t))^2)dt `

`=int_0^(pi) sqrt(9sin^2(3t)+9cos^2(3t))dt `

`=int_0^(pi) sqrt(9(sin^2(3t)+cos^2(3t)))dt `

`=int_0^(pi)sqrt(9)dt `

`=3t|_0^(pi)=3pi `

But this is clearly incorrect as the arclength is the circumference of a circle with radius 1 which is `2pi ` .

---------------------------------------------------------------------------------------------------------------

The arclength is `2pi `

---------------------------------------------------------------------------------------------------------------

We’ve answered 318,911 questions. We can answer yours, too.

Ask a question