Let a particle move along the curve y(x)=1+x^2 subject to an attractive force of magnitude F=kR direct toward the origin. Determine the work W in moving the particle along the curve from P0(0,1) to P1(1,2).
To determine the work W in moving the particle along the curve from P0(0,1) to P1(1,2), we need to calculate the line integral of the force along the curve.
First, we need to find the parametric equation of the curve. Since we are given the equation y(x) = 1 + x^2, we can parametrize it as follows:
x(t) = t
y(t) = 1 + t^2
Next, we need to find the expression for the force at each point on the curve. We know that the force F is given by F = kR, where R is the position vector pointing from the origin to the particle. In this case, R = xi + yj, where x and y are the respective components of the position vector.
So, the force can be expressed as F = k(xi + yj), where x = x(t) and y = y(t). Substitute the values into the force equation:
F = k(t i + (1 + t^2) j)
Now, we can calculate the line integral of the force F along the curve from P0 to P1:
W = ∫[P0 to P1] F · dr
= ∫[P0 to P1] F · (dx i + dy j)
= ∫[P0 to P1] k(t i + (1 + t^2) j) · (dx i + dy j)
= ∫[P0 to P1] k(t dx + (1 + t^2) dy)
To evaluate this integral, we need to express dx and dy in terms of t. Since x(t) = t, we have dx = dt. Similarly, y(t) = 1 + t^2, so dy = 2t dt.
Substituting the values of dx and dy into the integral expression:
W = ∫[0 to 1] k(t dt + (1 + t^2) (2t dt))
= ∫[0 to 1] (kt dt + 2t^3 + 2t dt)
= ∫[0 to 1] (kt + 2t^3 + 2t) dt
Now we can integrate term by term:
W = ∫[0 to 1] kt dt + ∫[0 to 1] 2t^3 dt + ∫[0 to 1] 2t dt
= (1/2)kt^2 + (1/2)t^4 + t^2 [from 0 to 1]
= (1/2)k(1^2) + (1/2)(1^4) + 1^2 - (1/2)k(0^2) - (1/2)(0^4) - 0^2
= (1/2)k + (1/2) + 1 - 0 - 0 - 0
= (1/2)k + 3/2
Therefore, the work W in moving the particle along the curve from P0(0,1) to P1(1,2) is given by (1/2)k + 3/2.