∫0^𝜋/2√(9sin^4t + 9sin^2tcos^2t)dt
=
∫0^𝜋/2√(9sin^2t(sin^2t + cos^2t))dt
=
∫0^𝜋/2√(9sin^2t)dt
=
3∫0^𝜋/2√(sin^2t)dt
=
3∫0^𝜋/2|sint|dt
=
3[t]0^𝜋/2
=
3(𝜋/2)
=
3𝜋/2
Let C be parametrized by x = 3 sin2 t and y = 3 sin t cos t for 0 ≤ t ≤ 𝜋/2. Find the length L of C.
L =
2 answers
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 = ∫[0,π/2] √((3 sin 2t)^2 + (3 cos 2t)^2) dt
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 = ∫[0,π/2] √((3 sin 2t)^2 + (3 cos 2t)^2) dt