# x=6t^2 , y=2t^3 , 1<=t<=4 Find the arc length of the curve on the given interval. 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...

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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.

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