`x=6t^2 , y=2t^3 , 1

The formula of arc length of a parametric equation on the interval `alt=tlt=b` is:

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

The given parametric equation is:

 `x=6t^2`

`y=2t^3`

The derivative of x and y are:

`dx/dt= 12t`

`dy/dt = 6t^2`

 So the integral needed to compute the arc length of the given parametric equation on the interval `1lt=tlt=4` is:

`L= int_1^4 sqrt ((12t)^2+(6t^2)^2) dt`

The simplified form of the integral is:

`L= int_1^4 sqrt (144t^2+36t^4)dt`

`L=int_1^4 sqrt...

The formula of arc length of a parametric equation on the interval `alt=tlt=b` is:

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

The given parametric equation is:

 `x=6t^2`

`y=2t^3`

The derivative of x and y are:

`dx/dt= 12t`

`dy/dt = 6t^2`

 So the integral needed to compute the arc length of the given parametric equation on the interval `1lt=tlt=4` is:

`L= int_1^4 sqrt ((12t)^2+(6t^2)^2) dt`

The simplified form of the integral is:

`L= int_1^4 sqrt (144t^2+36t^4)dt`

`L=int_1^4 sqrt (36t^2(4+t^2))dt`

`L= int_1^4 6tsqrt(4+t^2)dt`

To take the integral, apply u-substitution method.

`u= 4+t^2`

`du=2t dt`

`1/2du=tdt`

`t=1` ,  `u =4+1^2=5`

`t=4` ,  `u = 4+4^2=20`

Expressing the integral in terms of u, it becomes:

`L=6int_1^4 sqrt(4+t^2)* tdt`

`L=6 int _5^20 sqrtu *1/2du`

`L=3int_5^20 sqrtu du`

`L=3int_5^20 u^(1/2)du`

`L=3*u^(3/2)/(3/2)`  `|_5^20`

`L=2u^(3/2)`  `|_5^20`

`L = 2usqrtu`  `|_5^20`

`L=2(20)sqrt20 - 2(5)sqrt5`

`L=40sqrt20-10sqrt5`

`L=40*2sqrt5 - 10sqrt5`

`L=80sqrt5-10sqrt5`

`L=70sqrt5`

Therefore, the arc length of the curve is  `70sqrt5`  units.