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I know I'm doing the work right, but I can't get my answers to match and they are pretty close. I think it's just some arithmetic errors.

Help? I've been trying to solve my mistakes forever. I can't find them!

1. The problem statement, all variables and given/known data

Use the parametric equation of an ellipse x = acos(t) and y=bsin(t) from 0<t<2PI (those are or equal to signs)

2. Relevant equations

A=INTEGRAL from 0 - 2PI ydx

3. The attempt at a solution

INTEGRAL from 0-2PI bsin(t)(-asin(t) dt

= -ab INTEGRAL 0 -2PI sin^2(t) dt

= -ab EVALUATED 1/2 (t) - 1/4*sin(2t) from 0-2PI

= -ab ((1/2*(2PI) - 1/4 (sin(2*2PI) - 0)

= -abPI

But the answer is abPI. I don't know where the negative sign goes... or if I am supposed to have another one that cancels mine out

1. The problem statement, all variables and given/known data

Find the exact length of the curve

2. Relevant equations

x = 1+3t^2

y= 4+2t^3

0<t<1 (or equal)

)^2)

L= INTEGRAL from a - b SQUARE ROOT (dx/dt)^2 +(dy/dt)^2 END SQUARE ROOT dt

3. The attempt at a solution

dx/dt = 6t

dy/dt = 6t^2

L= INTEGRAL from 0-1 SQARE ROOT (6t)^2 + (6t^2)^2 END SQUARE ROOT dt

= 6 INTEGRAL from 0-1 t* SR 1+t^2 dt

using u substitution

u = 1+t^2

du = 2t dt

boundaries change to 1 - 2

= 3 * integral from 1-2 square root u du

= 3 * 2*u^3/2 evaluated from 1-2

= 3 * ( 2(2)^3/2 - 2(1)^3/2 )

= 3 * (4 square root (2) - 2)

= 12 square root (2) - 6

answer is supposed to be

4 square root (2) - 1

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# Homework Help: Calc 3 (parametric equations) My answers won't match/can't find arithmetic error

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