Calculate the length of the path over the given interval: (sin6t, cos6t), 0<t<pi

To calculate the length of the path over the given interval (sin(6t), cos(6t)), where 0 < t < pi, we need to use the concept of arc length in parametric form.

The formula to find the arc length of a parametric curve is given by:

L = ∫[a, b] sqrt[(dx/dt)^2 + (dy/dt)^2] dt

In this case, we have:

x = sin(6t)
y = cos(6t)

To find dx/dt and dy/dt, we need to take the derivatives of x and y with respect to t:

dx/dt = d/dt (sin(6t)) = 6cos(6t)
dy/dt = d/dt (cos(6t)) = -6sin(6t)

Now, we can substitute these values into the arc length formula:

L = ∫[0, pi] sqrt[(6cos(6t))^2 + (-6sin(6t))^2] dt

Simplifying the equation:

L = ∫[0, pi] sqrt[36cos^2(6t) + 36sin^2(6t)] dt

L = ∫[0, pi] sqrt[36(cos^2(6t) + sin^2(6t))] dt

Using the trigonometric identity cos^2(x) + sin^2(x) = 1, we can simplify further:

L = ∫[0, pi] sqrt[36(1)] dt

L = ∫[0, pi] 6 dt

Finally, integrating with respect to t:

L = [6t] from 0 to pi

L = 6pi - 0

L = 6pi

Therefore, the length of the path over the given interval is 6pi units.