Circumference of a circle in parametrics

[hangainlover]

Homework Statement

x^2+y^2=1

Homework Equations

length of curve square root (1+(dx/dy)^2)dy or square root ((dx/dt)^2 + (dy/dt)^2) dt




3. The Attempt at a Solution [/b

I isolated y and got y= square root (1-x^2)

finding dy/dx = -x/square root (1-x^2)
i plugged that into the formula but i did not get the correct answer.

im aware that if i do integral of square root (1+ (dy/dx)^2 ) i have to do imporper integral as i get undefined for denominator at x=-1,1 so what should i do
we all know that answer should be 2pi

 

[Dick]

Ignore the 'undefined' issue. Just do a trig substitution to solve the integral. Once you've done the trig substitution you won't see the singularity. BTW you'll get pi, not 2*pi. Your parametrization only covers half the circle.

 

[hangainlover]

I know, i can get the answer by defining the circle in trig.

But, shouldn't i be able to do it in conventional way and by paramatrization as well?
i just need to take care of the x values and t values where the function gets undefined.
My teacher wants me to do it in these two ways.

 

[Dick]

Even to do it in the x-y way you need to integrate 1/sqrt(1-x^2) from -1 to 1, right? Even if you haven't parametrized it as a trig function you still need to do a trig substitution to solve that integral. That's what I'm talking about.