1, x = 2+t

^{2}, y = t^{2}+ t^{3 }1, x = 2+t^{2} , y = t^{2} + t^{3 }

Tutors, please sign in to answer this question.

Seattle, WA

When finding total length of a curve, you must first find an expression for the differential vector along your path of interest, ds.

You know from the pythagorean theorem that the length of any vector in your space is going to be the sqrt(x^2+y^2+z^2....+...)

In your case, you have differential elements dx and dy in your measure space.

So the differential arc length ds = sqrt(dx^2 + dy^2)

= sqrt(1 + (dy/dx)^2)dx or sqrt(1+(dx/dy)^2)dy whichever you prefer

By the chain rule you know that dx = f'(t)dt and dy = g'(t)dt.

Algebraically, you know that dy/dx = f'(t)dt/g'(t)dt

leaving us with ds = sqrt(1+f'(t)/g'(t))f'(t)dt or sqrt(1+g'(t)/f'(t))g'(t)dt

the length then is simply the integral of ds from a to b.

Or, in a more clear form,

Integrate sqrt((dx/dt)^2 + (dy/dt)^2) with respect to t from 0 to 2p.

I will leave the actual arithmetic to you.

Cheers,

--Tim

John E.

Physics and Math tutor - Masters from Cornell and Columbia University

New York, NY

4.8
(124 ratings)

Edward B.

STAT/MATH/Actuarial Science/MBA/Econ/Fin. - Ivy League Exp & Prof

New York, NY

4.9
(294 ratings)

Hilton T.

Master teacher/tutor of Math, physics, and Chemistry

South Richmond Hill, NY

5.0
(67 ratings)

- Math 6723
- Integration 89
- Calculus 1 302
- Precalculus 1182
- Derivatives 162
- Differentiation 95
- Calculus 2 250
- Multivariable Calculus 12
- Vector Calculus 12
- Algebra 2 2685

Find a tutor fast. Get the app.

Are you a tutor? Get the app for tutors

© 2005 - 2016 WyzAnt, Inc. - All Rights Reserved