Question

Set up, but do not evaluate, the integral that represents the length of the curve given by x = 1 + 3t^2, y = 4 + 2t^3 over the interval 0 ≤ t ≤ 1

Answer 1

L

=

∫

t

f

t

i

√

(

d

x

d

t

)

2

+

(

d

y

d

t

)

2

d

t

. Since  

x

and  

y

are perpendicular, it's not difficult to see why this computes the arclength.

It isn't very different from the arclength of a regular function:  

L

=

∫

b

a

√

1

+

(

d

y

d

x

)

2

d

x

. If you need the derivation of the parametric formula, please ask it as a separate question.

We find the 2 derivatives:

d

x

d

t

=

3

−

3

t

2

d

y

d

t

=

6

t

And we substitute these into the integral:

L

=

∫

√

3

0

√

(

3

−

3

t

2

)

2

+

(

6

t

)

2

d

t

And solve:

=

∫

√

3

0

√

9

−

18

t

2

+

9

t

4

+

36

t

2

d

t

=

∫

√

3

0

√

9

+

18

t

2

+

9

t

4

d

t

=

∫

√

3

0

√

(

3

+

3

t

2

)

2

d

t

=

∫

√

3

0

(

3

+

3

t

2

)

d

t

=

3

t

+

t

3

∣

∣

√

3

0

=

3

√

3

+

3

√

3

=6The arclength of a parametric curve can be found using the formula:  

L

=

∫

t

f

t

i

√

(

d

x

d

t

)

2

+

(

d

y

d

t

)

2

d

t

. Since  

x

and  

y

are perpendicular, it's not difficult to see why this computes the arclength.

It isn't very different from the arclength of a regular function:  

L

=

∫

b

a

√

1

+

(

d

y

d

x

)

2

d

x

. If you need the derivation of the parametric formula, please ask it as a separate question.

We find the 2 derivatives:

d

x

d

t

=

3

−

3

t

2

d

y

d

t

=

6

t

And we substitute these into the integral:

L

=

∫

√

3

0

√

(

3

−

3

t

2

)

2

+

(

6

t

)

2

d

t

And solve:

=

∫

√

3

0

√

9

−

18

t

2

+

9

t

4

+

36

t

2

d

t

=

∫

√

3

0

√

9

+

18

t

2

+

9

t

4

d

t

=

∫

√

3

0

√

(

3

+

3

t

2

)

2

d

t

=

∫

√

3

0

(

3

+

3

t

2

)

d

t

=

3

t

+

t

3

∣

∣

√

3

0

=

3

√

3

+

3

√

3

=

6

√

3

Be aware that arclength usually has a difficult function to integrate. Most integrable functions look like the above where a binomial is squared and adding the two terms will flip the sign of the binomial.    

Be aware that arclength usually has a difficult function to integrate. Most integrable functions look like the above where a binomial is squared and adding the two terms will flip the sign of the binomial.