# Reparameterization of ellipse c(t)=(2sint, cost), and parabola y=x^2/2

#### nisde

##### New member
Hello,

I'm working on a project regarding plane curves and I need to reparameterizate these two curves using arc lenght

Ellipse: c(t)=(2sint, cost)

Parabola: y=x^2/2

I solved the ellipse, but I don't know if it's the right answer, my solution is c1(s)=(2sin*s/2, cos*s/2)

Also the professor said there is a problem in one of these 2, but I don't know which one, one is maybe unsolvable?

English is not my native language, so sorry if I missed out on some math terminology

Thank you #### Subhotosh Khan

##### Super Moderator
Staff member
Hello,

I'm working on a project regarding plane curves and I need to reparameterizate these two curves using arc lenght

Ellipse: c(t)=(2sint, cost)

Parabola: y=x^2/2

I solved the ellipse, but I don't know if it's the right answer, my solution is c1(s)=(2sin*s/2, cos*s/2)

Also the professor said there is a problem in one of these 2, but I don't know which one, one is maybe unsolvable?

English is not my native language, so sorry if I missed out on some math terminology

Thank you #### HallsofIvy

##### Elite Member
Hello,

I'm working on a project regarding plane curves and I need to reparameterizate these two curves using arc lenght

Ellipse: c(t)=(2sint, cost)
x= 2sin(t), y= cos(t), dx/dt= 2cos(t), dy/dt= -sin(t)
so the arc length is given by $$\displaystyle s= \int\sqrt{4cos^2(t)+ sin^2(t)} dt= \int \sqrt{3cos^2(t)+ 1} dt$$. The "problem" your professor refers to is that this cannot be integrated in terms of an elementary function. There are "elliptic functions" that are defined to be the integrals of the arc length of various ellipses.

Parabola: y=x^2/2
dy/dx= x so the arc length is given by $$\displaystyle s= \int\sqrt{1+ x^2}dx$$. Rembering that $$\displaystyle sin^2(t)+ cos^2(t)= 1$$, dividing both sides by cos(t) gives $$\displaystyle tan^2(t)+ 1= sec^2(t)$$. So if we make the substitution x= tan(t), $$\displaystyle \sqrt{1+ x^2}= \sqrt{1+ tan^2(t)}= sec(t)$$ and $$\displaystyle dx= sec^2(t) dt$$. The integral becomes $$\displaystyle s= \int sec^3(t)dt$$. To integrate that, write it as $$\displaystyle s= \int \frac{1}{cos^3(t)}dt$$. That has cosine to an odd power. Multiply both numerator and denominator by cos(t) to get $$\displaystyle \int\frac{cos(t)}{cos^4(t)}dt= \int\frac{cos(t)}{(1- sin^2(t))^2}dt$$. Let u= sin(t) so that $$\displaystyle du= cos(t)dt$$ and the integral becomes $$\displaystyle s= \int \frac{1}{(1- u^2)^2}du$$. That can be integrated by "partial fractions".

I solved the ellipse, but I don't know if it's the right answer, my solution is c1(s)=(2sin*s/2, cos*s/2)

Also the professor said there is a problem in one of these 2, but I don't know which one, one is maybe unsolvable?

English is not my native language, so sorry if I missed out on some math terminology

Thank you Last edited:

#### nisde

##### New member
Thank you so much HallsofIvy!

I will study your solution, you explained it in great detail

I really can't thank you enaugh!