Final answer:
To find the unit tangent vector at t = 3 of the position vector r(t) = t² - 3t, 1/4t, 1/3t³ - 1/2t², we need to calculate the derivative of r(t) and then normalize it. The derivative of r(t) is 2t - 3, 1/4, t² - t. The magnitude of v(3) is approximately 4.53. Dividing v(t) by its magnitude gives us the unit tangent vector t(t) = (√721(2t - 3)/4, √721/4, √721(t² - t)/4).
Step-by-step explanation:
We are given the position vector r(t) = t² - 3t, 1/4t, 1/3t³ - 1/2t², with t = 3. To find the unit tangent vector at t = 3, we need to calculate the derivative of r(t) and then normalize it to get a unit vector.
The derivative of r(t) will give us the velocity vector v(t), and dividing v(t) by its magnitude will give us the unit tangent vector t(t).
Calculating the derivative of r(t):
r'(t) = 2t - 3, 1/4, t² - t
Calculating the magnitude of v(3):
|v(t)| = √((2t - 3)² + (1/4)² + (t² - t)²)
= √((2(3) - 3)² + (1/4)² + ((3)² - (3))²)
= √((6 - 3)² + (1/4)² + (9 - 3)²)
= √((3)² + (1/4)² + (6)²)
= √(9 + 1/16 + 36)
= √(45 + 1/16)
= √(721/16)
= √(721)/4
≈ 4.53
Dividing v(t) by its magnitude:
t(t) = v(t)/|v(t)| = (2t - 3, 1/4, t² - t) / (√(721)/4) = (√721(2t - 3)/4, √721/4, √721(t² - t)/4)