Arc length of a curve C described by the parametric equations x=f(t) and y=g(t),`a<=t<=b` , where f'(t) and g'(t) are continuous on [a,b] and C is traversed exactly once as t increases from a to b , then the length of the curve C is given by:

`L=int_a^bsqrt((dx/dt)^2+(dy/dt)^2)dt`

Given parametric equations are :`x=sqrt(t),y=(3t-1) , 0<=t<=1`

`x=sqrt(t)`

`=>dx/dt=1/2(t)^(1/2-1)`

`dx/dt=1/2t^(-1/2)`

`dx/dt=1/(2sqrt(t))`

`y=3t-1`

`dy/dt=3`

Now let's evaluate the length of the arc by using the stated formula,

`L=int_0^1sqrt((1/(2sqrt(t)))^2+3^2)dt`

`L=int_0^1sqrt(1/(4t)+9)dt`

`L=int_0^1sqrt((1+36t)/(4t))dt`

`L=int_0^1sqrt(1+36t)/sqrt(4t)dt`

`L=int_0^1 1/2sqrt(1+36t)/sqrt(t)dt`

Take the constant out,

`L=1/2int_0^1sqrt(1+36t)/sqrt(t)dt`

Let's first evaluate the indefinite integral:`intsqrt(1+36t)/sqrt(t)dt`

Use integral substitution:`u=6sqrt(t)`

`du=6(1/2)t^(1/2-1)dt`

`du=3/sqrt(t)dt`

`intsqrt(1+36t)/sqrt(t)dt=int1/3sqrt(1+u^2)du`

`=1/3intsqrt(1+u^2)du`

Use the following standard integral from the integration tables:

`intsqrt(u^2+a^2)du=1/2(usqrt(u^2+a^2)+a^2ln|u+sqrt(a^2+u^2)|)+C`

`=1/3[1/2(usqrt(u^2+1)+ln|u+sqrt(1+u^2)|)]+C`

`=1/6(usqrt(1+u^2)+ln|u+sqrt(1+u^2)|)+C`

Substitute back:`u=6sqrt(t)`

`=1/6(6sqrt(t)sqrt(1+36t)+ln|6sqrt(t)+sqrt(1+36t)|)+C`

`L=1/2[1/6(6sqrt(t)sqrt(1+36t)+ln|6sqrt(t)+sqrt(1+36t)|)]_0^1`

`L=1/12[6sqrt(t)sqrt(1+36t)+ln|6sqrt(t)+sqrt(1+36t)|]_0^1`

`L=1/12{[6sqrt(37)+ln|6+sqrt(37)|]-[ln|1|]}`

`L=1/12[6sqrt(37)+ln|6+sqrt(37)|]`

Using calculator,

`L=1/12[36.4965751818+ln12.0827625303]`

`L=1/12[38.9883550344]`

`L=3.2490295862`

`L~~3.249`

Arc length of the curve on the given interval `~~3.249`

The equation for arc length in parametric coordinates is:

`L=int_a^b sqrt((dx/dt)^2+(dy/dt)^2) dt`

Where in this case:

`dx/dt=d/dt (t^(1/2))=1/2t^(-1/2)`

`dy/dt=d/dt(3t-1)=3`

`a=0, b=1`

Therefore

`L=int_0^1 sqrt((1/2t^(-1/2))^2+(3)^2) dt `

`L=int_0^1 sqrt((1/4t^(-1))+9) dt`

Solve this integral numerically.

`L~~3.249`

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