A particle moves along a path described by x=cos³t and y=sin³t. The distance that the particle travels along the path from t=0 to t=π/2 is...
Thank you!
if t=0
cos^3 0 = 1 , sin^3 0 = 0
thus position point is (1,0)
if t=π/2
cos^3 π/2 = 0 , sin^3 π/2 = 1
thus position point is (0,1)
Use your distance between 2 points formula to determine the distance between those two points
You should get √2
0.75
To find the distance traveled by the particle along the path, we need to calculate the arc length of the path from t = 0 to t = π/2.
The arc length formula for a curve defined by parametric equations x = f(t) and y = g(t) is given by:
s = ∫(a to b) √[ (dx/dt)² + (dy/dt)² ] dt
Here, we have x = cos³t and y = sin³t. Let's start by finding dx/dt and dy/dt.
dx/dt = d/dt (cos³t) = 3cos²t(-sin t) = -3cos²tsin t
dy/dt = d/dt (sin³t) = 3sin²t(cos t) = 3sin²tcos t
Next, let's calculate (√[ (dx/dt)² + (dy/dt)² ]).
(√[ (dx/dt)² + (dy/dt)² ]) = (√[ (-3cos²tsin t)² + (3sin²tcos t)² ])
= (√[ 9cos⁴t sin²t + 9sin⁴t cos²t ])
= (√[ 9(cos⁴t sin²t + sin⁴t cos²t) ])
= (√[ 9(cos⁴t sin²t + sin⁴t cos²t) ])
= 3√[ cos²t sin²t ]
= 3|cos t sin t|
Now, we need to evaluate this expression from t = 0 to t = π/2.
s = ∫(0 to π/2) 3|cos t sin t| dt
Since cos t sin t is positive in the first and second quadrants (0 ≤ t ≤ π/2), we can simplify the integral by assuming cos t sin t = cos t sin t, which is valid in this interval.
s = 3∫(0 to π/2) cos t sin t dt
Now, we can use the trigonometric identity: cos t sin t = (1/2)sin(2t)
s = 3(1/2)∫(0 to π/2) sin(2t) dt
= (3/2)∫(0 to π/2) sin(2t) dt
Integrating sin(2t) with respect to t, we get:
(3/2)∫(0 to π/2) sin(2t) dt = (3/2)(-1/2)cos(2t) ∣ (0 to π/2)
= (3/2)(-1/2)[cos(π) - cos(0)]
= (3/2)(-1/2)(-1 - 1)
= 3/2(2) = 3
Therefore, the distance that the particle travels along the path from t = 0 to t = π/2 is 3 units.