Find the exact arc length of the parametric curve without eliminating the parameter.
11. x = cos (2t), y = sin (2t) (0 less than or equal to t less than or equal to (pi/2))
L = definite integral from 0 to pi/2; square root of [-2 sin (2t))^2 + (2 cos (2t))^2] with respect to y.
= square root of [- 4 sin^2 (2t) + 4 cos^2 (2t)]
= square root of [4 (cos^2 (2t) - sin^2 (2t)]
= square root of [4 (cos (4t)]
= 2 square root of [cos (4t)]
I'm stuck. Maybe I did it incorrectly from the start. Learning process FTW!!
5. y = x^(2/3) from x = 1 to x = 8.
L = definite integral from 1 to 8 with respect to x;
square root of [1 + ((2/3x^(1/3))^2]
= square root of [1 + ((4/9x^(2/3)]
= square root of [1 + ((4/9x^(-2/3)]
If I use u-substitution, I don't see how I can get it into the u du form.
13. x = e^t cos t, y = e^t sin t (0 less than or equal to t less than or equal to pi/2)
L = definite integral from 0 to pi/ with respect to t;
square root of [(e^t(cos t - sin t))^2 + ((e^t(cos t + sin t))^2]
= square root of [2((e^t)^2)]
?????????????????
11. x = cos (2t), y = sin (2t) (0 less than or equal to t less than or equal to (pi/2))
L = definite integral from 0 to pi/2; square root of [-2 sin (2t))^2 + (2 cos (2t))^2] with respect to y.
= square root of [- 4 sin^2 (2t) + 4 cos^2 (2t)]
= square root of [4 (cos^2 (2t) - sin^2 (2t)]
= square root of [4 (cos (4t)]
= 2 square root of [cos (4t)]
I'm stuck. Maybe I did it incorrectly from the start. Learning process FTW!!
5. y = x^(2/3) from x = 1 to x = 8.
L = definite integral from 1 to 8 with respect to x;
square root of [1 + ((2/3x^(1/3))^2]
= square root of [1 + ((4/9x^(2/3)]
= square root of [1 + ((4/9x^(-2/3)]
If I use u-substitution, I don't see how I can get it into the u du form.
13. x = e^t cos t, y = e^t sin t (0 less than or equal to t less than or equal to pi/2)
L = definite integral from 0 to pi/ with respect to t;
square root of [(e^t(cos t - sin t))^2 + ((e^t(cos t + sin t))^2]
= square root of [2((e^t)^2)]
?????????????????