Final answer:
The total work done in moving a particle in a force field along a curve is calculated by evaluating the line integral of the force vector along the curve. In this case, the force F = 3y i - 5z j + 10x k and the curve is defined by x = t, y = 2t², z = t³ from t = 1 to t = 2. By parameterizing the curve, calculating the dot product of the force and the curve's derivative, and evaluating the resulting integral, the total work done is found to be approximately 48.7 units of work.
Step-by-step explanation:
To find the total work done in moving a particle along a curve in a force field, we need to calculate the line integral of the force vector along the curve. In this case, the force is given as F = (3y)i - (5z)j + (10x)k and the curve is defined by x = t, y = 2t², z = t³ where t ranges from 1 to 2. To calculate the line integral, we need to parameterize the curve and then evaluate the integral. Let's calculate step by step:
- Parameterize the curve: r(t) = ti + (2t²)j + (t³)k
- Calculate the derivative of r with respect to t: dr/dt = i + (4t)j + (3t²)k
- Substitute the values of r(t) and dr/dt into the force vector: F = (3(2t²))i - (5(t³))j + (10(t))k = 6t²i - 5t³j + 10tk
- Calculate the dot product of F and dr/dt: F · dr/dt = (6t²)(1) + (-5t³)(4t) + (10t)(3t²) = 6t² - 20t⁴ + 30t³
- Integrate F · dr/dt with respect to t from 1 to 2: ∫12 (6t² - 20t⁴ + 30t³) dt
- Evaluate the integral: ∫12 (6t² - 20t⁴ + 30t³) dt = (2t³ - (4/5)t⁵ + (15/2)t⁴) evaluated from 1 to 2
- Substitute the values of t into the expression: (2(2)³ - (4/5)(2)⁵ + (15/2)(2)⁴) - (2(1)³ - (4/5)(1)⁵ + (15/2)(1)⁴)
- Simplify the expression: (16 - (64/5) + 60) - (2 - (4/5) + (15/2)) = 48 - (2/5) + (15/2) = 48 - (2/5) + 7.5 = 48.7
Therefore, the total work done in moving the particle along the given curve is approximately 48.7 units of work.