Calc question I am stumped on it is: Find the length of the curve x = 5 sin^3t, y = 5 cos^3t, 0 <=t<=pi?

Answer 1

The parametric equations involve CUBE of sine and cosine. Thus, it does not represent circle, but an ASTEROID

The equation of this curve is x=5 sin^3 (t) & y=5 cos^3 (t)
sin (t) = (x/5)^1/3 cos (t) = (y/5)^1/3
sin^2 (t) + cos^2 (t) =1 or (x/5)^2/3 + (y/5)^2/3 = 1 (asteroid)

x = 5 (sin t)^3, y = 5 (cos t)^3, 0 <=t<=pi

dx/dt = 15 sin^2 t *cos t dy/dt = - 15 cos^2 t *sin t

their derivative in Cartesian system
dy/dx = - cos t / sin t = - cot t and dx = 15 sin^2 t *cos t. dt

****** This curve does not allow calculating length [t=0 or (0, 5) to t=pi or (0, -5)] freely as it is symetrical about both x,y.

Therefore length = 2 times {length from t=0 to pi/2)
S = 2s=2 ∫sqrt [1+ (dy/dx)^2] dx or
= 2 ∫sqrt [1+ cot^2 t] 15 sin^2 t *cos t dt (from t=0 to pi/2)
= 2 ∫sqrt [1+ cot^2 t] 15 sin^2 t *cos t dt
= 30 ∫(1/sin t)*sin^2 t *cos t dt (from t=0 to pi/2)
=(30/2) ∫2 sin t *cos t dt (from t=0 to pi/2)
= (15) ∫ sin (2t) dt (from t=0 to pi/2)
=(15/2) [cos (2t)] (from t=0 to pi/2)
=(15/2) [-1- 1] =15

Length = S =15 ((from t=0 to pi)

Answer 2

The graph is just a circle with radius 5 and center (0,0). To find length take the integral of the square root of (dx/dt)^2 + (dy/dt)^2 from 0 to pi/6 and multiply by 6 or just use 3(2*pi*radius) for the circumference of the circle. The 3t within the sin and cos functions alters the amount of circle being drawn per radian so for pi/6 radians half of the circle is defined and therefore from zero to pi the circle is traced three times.

Answer 3

The curve is a circle (radius 5) given with the parameter t.
Since 0 <= t <= pi, length = half the perimeter = pi r = 5pi

Th

Answer 4

dx = 15sin^2tcost dt
dy = -15cos^2tsint dt
(dl) = √[(dx)^2+(dy)^2]
= 15sint cost dt
= 7.5sin2t dt

l
= ∫|7.5sin2t| dt[t:from 0 to pi]
= 2∫7.5sin2t dt[t:from 0 to pi/2]
= 15