Finding the length of an arc of a parabola

[HannahBridge]

Homework Statement

y^2 = x from (0,0) to (1,1)

Homework Equations

L = ∫√(1+[g'(y)]^2) dy

The Attempt at a Solution

So this is a problem in my textbook that has been bothering me because I can't seem to come up with the same answer.

1. [bounds 0 to 1] 1/2∫ sec^3θ was obtained using trig substitution with y = 1/2tanθ and dy=1/2sec^2θ which, according to the steps in the textbook, is correct.

2. I use integration by parts which gives me 1/2∫ sec^3θ = 1/2secθtanθ - 1/2∫sec^3θ + 1/2∫secθ
adding - 1/2∫sec^3θ to the other side of the equation it becomes
1/2secθtanθ + 1/2∫secθ=
1/2secθtanθ + 1/2 ln(secθ + tanθ)

then I use y = 1/2tanθ to change my bounds from 0 to θ
and then evaluating for tanθ = 2 I get the answer

L = √5 + ln(√5 +2) / 2

where the book comes up with

L = √5/2 + ln(√5 +2) / 4

and one of it's steps after integration by parts shows

1/4secθtanθ + 1/4 ln(secθ + tanθ)

and I seem to be having trouble how they came up with 1/4 instead of 1/2. Most likely a stupid mistake I made and am overlooking? Thanks in advance for the help! xo <3

 

[Dick]

Homework Statement

y^2 = x from (0,0) to (1,1)

Homework Equations

L = ∫√(1+[g'(y)]^2) dy

The Attempt at a Solution

So this is a problem in my textbook that has been bothering me because I can't seem to come up with the same answer.

1. [bounds 0 to 1] 1/2∫ sec^3θ was obtained using trig substitution with y = 1/2tanθ and dy=1/2sec^2θ which, according to the steps in the textbook, is correct.

2. I use integration by parts which gives me 1/2∫ sec^3θ = 1/2secθtanθ - 1/2∫sec^3θ + 1/2∫secθ
adding - 1/2∫sec^3θ to the other side of the equation it becomes
1/2secθtanθ + 1/2∫secθ=
1/2secθtanθ + 1/2 ln(secθ + tanθ)

then I use y = 1/2tanθ to change my bounds from 0 to θ
and then evaluating for tanθ = 2 I get the answer

L = √5 + ln(√5 +2) / 2

where the book comes up with

L = √5/2 + ln(√5 +2) / 4

and one of it's steps after integration by parts shows

1/4secθtanθ + 1/4 ln(secθ + tanθ)

and I seem to be having trouble how they came up with 1/4 instead of 1/2. Most likely a stupid mistake I made and am overlooking? Thanks in advance for the help! xo <3

Yes. It's simple. You got 1/2∫ sec^3θ = 1/2secθtanθ - 1/2∫sec^3θ + 1/2∫secθ. That tells you ∫ sec^3θ = 1/2secθtanθ + 1/2∫secθ. Which is fine. But you wanted to integrate (1/2)∫ sec^3θ. It's a little confusing with all of the (1/2)'s running around. Easy to miss one.