I am supposed to find the line integral over C of (x+y)ds where C is the line segment from (0,1,1) to (3,2,2)?
So far I have 3ti + (1+t)j + (1+t)k as the parameterisation of C, and when I make an integral from that, I get 3(sqrt(11)). But I am uncertain of this answer, so if anyone could help confirm that (or even just the parameterisation), that would be very helpful. Thank you!
To find the line integral over curve C of the function (x+y)ds, you need to follow these steps:
1. Parameterize the curve C:
Given that C is a line segment from (0, 1, 1) to (3, 2, 2), you correctly used a parameterization that represents points on that line segment. The parameterization you used is:
r(t) = 3ti + (1+t)j + (1+t)k, where t ranges from 0 to 1.
This parameterization correctly traces the line segment from the initial point (0, 1, 1) when t = 0 to the final point (3, 2, 2) when t = 1.
2. Compute the derivative of the parameterization wrt t:
To calculate the derivative of r(t), differentiate each component with respect to t:
r'(t) = 3i + j + k
The derivative represents the tangent vector to the curve C at each point.
3. Calculate the length of the derivative:
The length of the derivative vector is given by the magnitude:
||r'(t)|| = √(32 + 12 + 12) = √14
This magnitude represents the speed of the parameterization, which is the rate at which the curve is traversed.
4. Evaluate the line integral:
The line integral of the function (x+y)ds over curve C is given by:
∫(0 to 1) (r(t) · n(t)) ||r'(t)|| dt
where · denotes the dot product, n(t) is the unit normal vector to the curve, and ||r'(t)|| is the length of the derivative vector.
In this case, n(t) coincides with r'(t) since C is a straight line. Therefore, the line integral simplifies to:
∫(0 to 1) (r(t) · r'(t)) ||r'(t)|| dt
Plugging in the values of r(t) and r'(t), the integral becomes:
∫(0 to 1) (3t + (1+t) + (1+t)) √14 dt
= ∫(0 to 1) (3t + 2 + 3t) √14 dt
= ∫(0 to 1) (6t + 2) √14 dt
= (√14/2) ∫(0 to 1) (6t + 2) dt
= (√14/2) [3t^2 + 2t] evaluated from 0 to 1
= (√14/2) [(3(1)^2 + 2(1)) - (3(0)^2 + 2(0))]
= (√14/2) [3 + 2]
= (√14/2) (5)
= 5√14 / 2
Therefore, the line integral over curve C of (x+y)ds is 5√14 / 2 or approximately 5.92.