`x=sqrt(t) , y=3t-1 , 0<=t<=1` Find the arc length of the curve on the given interval.

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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:


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







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





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

Take the constant out,


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

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





Use the following standard integral from the integration tables:




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






Using calculator,





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


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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)`


`a=0, b=1`



`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.


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