Simplifying Square Roots of a Parametrized Path

[mill]

Homework Statement

Find the arclength of the parametrized path x(t) = (t^2)/2 , y(t) = (t^3)/3 for 1<t<3.

Homework Equations

Arc Length Formula

The Attempt at a Solution

x'=t and y'=t^2.

Putting them into the arc length formula, I get sqrt(t^2 + t^4) inside.

I'm confused about how to simplify this part. The answer (10sqrt(10)-2sqrt(2))/3 suggests the quadratic formula somewhere along the way. I could probably pull out a variable into sqrt(t^2(t^2 + 1)) or tsqrt(t^2+1) but how do I use the quadratic equation in this case?

 

[Dick]

Homework Statement

Find the arclength of the parametrized path x(t) = (t^2)/2 , y(t) = (t^3)/3 for 1<t<3.

Homework Equations

Arc Length Formula

The Attempt at a Solution

x'=t and y'=t^2.

Putting them into the arc length formula, I get sqrt(t^2 + t^4) inside.

I'm confused about how to simplify this part. The answer (10sqrt(10)-2sqrt(2))/3 suggests the quadratic formula somewhere along the way. I could probably pull out a variable into sqrt(t^2(t^2 + 1)) or tsqrt(t^2+1) but how do I use the quadratic equation in this case?

You don't use the quadratic equation. You just integrate $t \sqrt{t^2+1}$. It's not hard, just do it by a substitution.

 

[mill]

Thanks.