# Integrating a square root function

Problem statement:
Find the arc length of the curve defined by x = √t and y = 3t -1 on the interval 0 < t < 1

attempted solution:
dx/dt = 1/2t-1/2 , dy/dt = 3 and dx = dt/ 2√t
dy/dx = 6√t
length = ∫01 √(1 + (6√t)2) .dt/ 2√t
= ∫01 √1 + 36t) dt/2√t
now I'm stuck with a product that is very difficult to solve using integration by parts or by trig substitution. Please shed some light.

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Ray Vickson
Homework Helper
Dearly Missed
Find the arc length of the curve defined by x = √t and y = 3t -1 on the interval 0 < t < 1

attempted solution:
dx/dt = 1/2t-1/2 , dy/dt = 3 and dx = dt/ 2√t
dy/dx = 6√t
length = ∫01 √(1 + (6√t)2) .dt/ 2√t
= ∫01 √1 + 36t) dt/2√t
now I'm stuck with a product that is very difficult to solve using integration by parts or by trig substitution. Please shed some light.

If I was doing the problem I would first perform a simple change-of-variables to make the problem easier. However, PF rules forbid me from saying more, except to suggest you look in your textbook and/or course notes (or on-line) to find similar integrals.

And: please stop using bold fonts in your posts. It is annoying and a bit intimidating.

Svein
Find the arc length of the curve defined by x = √t and y = 3t -1 on the interval 0 < t < 1
Looking at the equation. I can see that y = 3x2-1. Furthermore t=0⇒x=0 and t=1⇒x=1.

Mark44
Mentor
And: please stop using bold fonts in your posts. It is annoying and a bit intimidating.
Fixed in original post...
The excess bolding comes from deleting the template boilerplate, but missing the bold tags.

ande