Let C be parametrized by x = 3 sin2 t and y = 3 sin t cos t for 0 β€ t β€ π/2. Find the length L of C.
L =
β«0^π/2β(9sin^4t + 9sin^2tcos^2t)dt
=
β«0^π/2β(9sin^2t(sin^2t + cos^2t))dt
=
β«0^π/2β(9sin^2t)dt
=
3β«0^π/2β(sin^2t)dt
=
3β«0^π/2|sint|dt
=
3[t]0^π/2
=
3(π/2)
=
3π/2
How ever did the robot get the right answer, starting out with such a wrong integral? Maybe it just recognized the curve as just 1/4 of a circle of radius 3/2
If you meant
x = 3 sin^2 t and y = 3 sin t cos t
or x = 3/2 (1 - cos 2t) y = 3/2 sin 2t
then
L = β«[0,Ο/2] β((3 sin 2t)^2 + (3 cos 2t)^2) dt
To find the length of the curve C, we can use the arc length formula:
L = β«β(dx/dt)^2 + (dy/dt)^2 dt
Given that x = 3sin^2(t) and y = 3sin(t)cos(t), we can calculate dx/dt and dy/dt as follows:
dx/dt = d/dt (3sin^2(t))
= 6sin(t)cos(t)
dy/dt = d/dt (3sin(t)cos(t))
= 3cos^2(t) - 3sin^2(t)
Now we can substitute these values back into the arc length formula:
L = β«β(6sin(t)cos(t))^2 + (3cos^2(t) - 3sin^2(t))^2 dt
Simplifying the expression by expanding and combining like terms:
L = β«β(36sin^2(t)cos^2(t) + 9cos^4(t) - 18sin^2(t)cos^2(t) + 9sin^4(t)) dt
= β«β(9sin^4(t) - 18sin^2(t)cos^2(t) + 9cos^4(t)) dt
Factoring out a 9:
L = β«β(9(sin^4(t) - 2sin^2(t)cos^2(t) + cos^4(t))) dt
Using the identity sin^2(t) + cos^2(t) = 1, we can rewrite the expression inside the square root as (sin^2(t) + cos^2(t))^2:
L = β«β(9(sin^2(t) + cos^2(t))^2) dt
= β«β(9) dt
= 3t + C
Finally, to find the length L of C, we need to evaluate the integral from 0 to π/2:
L = [3t]_0^(π/2)
= 3(π/2) - 3(0)
= 3π/2
Therefore, the length of the curve C is 3π/2.
To find the length L of the curve C, we will use the arc length formula. The arc length formula for a curve parametrized by x = f(t) and y = g(t) is given by:
L = β« sqrt((dx/dt)^2 + (dy/dt)^2) dt,
where dx/dt and dy/dt are the derivatives of x and y with respect to t.
Let's first compute the derivatives of x and y:
dx/dt = d(3 sin^2 t)/dt
= 6 sin t cos t
dy/dt = d(3 sin t cos t)/dt
= 3(cos^2 t - sin^2 t)
= 3 cos 2t
Now, substitute these derivatives back into the arc length formula:
L = β« sqrt((6 sin t cos t)^2 + (3 cos 2t)^2) dt
= β« sqrt(36 sin^2 t cos^2 t + 9 cos^2 2t) dt
= β« sqrt(9(4 sin^2 t cos^2 t + cos^2 2t)) dt
= β« 3 sqrt(4 sin^2 t cos^2 t + cos^2 2t) dt
Let's now simplify the integrand. We can use the identity cos^2 t = (1 + cos 2t)/2 to rewrite the integrand:
L = β« 3 sqrt(4 sin^2 t ((1 + cos 2t)/2) + cos^2 2t) dt
= β« 3 sqrt(2 sin^2 t + cos^2 2t + 2 sin^2 t cos^2 2t) dt
The integral of this expression can only be evaluated numerically, so we cannot find a closed-form solution for L. However, you can approximate the value of the integral using numerical methods such as numerical integration or using a computer software that can handle symbolic integrals.
Therefore, the length L of curve C cannot be determined without estimating it numerically.