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.