- #1

- 78

- 1

## Homework Statement

*Find the length of the path traced out by a particle moving on a curve according to the given equation during the time interval specified in each case.*

**= (c**

*r(t)*^{2}/a)cos

^{3}t

**+ (c**

*i*^{2}/b)sin

^{3}t

*j*where

**and**

*i***are the usual unit vectors, 0 [itex]\leq[/itex] t [itex]\leq[/itex] 2[itex]\pi[/itex], c**

*j*^{2}= a

^{2}- b

^{2}, and 0 < b < a

Also,

**is the position vector-valued function of t. Scalar functions will not be bolded.**

*r(t)*## Homework Equations

I know that the arc length in this case is the integral of ||

*r**'*|| = v(t) from 0 to 2[itex]\pi[/itex].

**(t)**## The Attempt at a Solution

Differentiating gives:

*r**'*= (-3c

**(t)**^{2}/a)cos

^{2}t sin t

**+ (3c**

*i*^{2}/b)sin

^{2}t cos t

**.**

*j*Therefore, v(t) = √{(9c

^{4}/a

^{2})cos

^{4}t sin

^{2}t + (9c

^{4}/b

^{2})sin

^{4}t cos

^{2}t)} = √{(9c

^{4}/a

^{2})cos

^{2}t(cos

^{2}t sin

^{2}t) + (9c

^{4}/b

^{2})sin

^{2}t(sin

^{2}t cos

^{2}t)} =

√{(9c

^{4}/a

^{2})cos

^{2}t(0.25sin

^{2}2t) + (9c

^{4}/b

^{2})sin

^{2}t(0.25sin

^{2}2t)} = √{(9c

^{4}/a

^{2})cos

^{2}t(0.25sin

^{2}2t) + (9c

^{4}/b

^{2})sin

^{2}t(0.25sin

^{2}2t)} =

(3c

^{2}/2)(sin2t)*√{(1/a

^{2})cos

^{2}t + (1/b

^{2})sin

^{2}t} = (3c

^{2}/(2ab))(sin2t)*√{(b

^{2})cos

^{2}t + (a

^{2})sin

^{2}t}

And that is where I get stuck. I'm not sure what the integral of that expression is, nor do I know what I can do to simplify it further. I would say that it has no simple solution, since the argument under the square root looks an awfully lot like an integral for an ellipse, and I have heard one needs something called an "elliptic integral" to solve. However, the answer for this problem is quite simple: (4a

^{3}-4b

^{3})/(ab). Will someone please help me with this?

Last edited: