Question

The line integral of (2x+9z) ds where the curve is given by the parametric equations x=t, y=t^2, z=t^3 for t between 0 and 1. Please don't respond if you can't explain how to get out of an impossible square root. Thanks

Answer 1

To evaluate the line integral, calculate the arclength element and substitute it into the integral. Expand the expression inside the square root, multiply it by (2x+9z), and integrate term by term.

Step-by-step explanation:

To evaluate the line integral of (2x+9z) ds where the curve is given by the parametric equations x=t, y=t^2, z=t^3 for t between 0 and 1, we need to find the arclength element ds and substitute it into the integral. The arclength element ds can be calculated using the formula ds = sqrt(dx^2 + dy^2 + dz^2). In this case, dx = dt, dy = 2t dt, and dz = 3t^2 dt. Substituting these values into the arclength element formula, we get ds = sqrt(dt^2 + 4t^2 dt^2 + 9t^4 dt^2) = sqrt(1 + 4t^2 + 9t^4) dt.

Substituting ds into the line integral, we get the integral of (2x+9z) * sqrt(1 + 4t^2 + 9t^4) dt. To evaluate this integral, you can expand the expression inside the square root, multiply it by (2x+9z), and integrate term by term.

However, it seems that there might be a typo in the original question regarding the curve parametrization. The curve given by x=t, y=t^2, z=t^3 is actually a parabolic curve, not a line. If you need further clarification or assistance, please let me know.

Answer 2

Let r = (t,t^2,t^3)

Then r' = (1, 2t, 3t^2)

General Line integral is:

`\int_a^b f(r) |r'| dt`

The limits are 0 to 1
f(r) = 2x + 9z = 2t +9t^3
|r'| is magnitude of derivative vector
`√((x')^2 + (y')^2 + (z')^2)`


`\int_0^1 (2t+9t^3) √(1+4t^2 +9t^4) dt`

Fortunately, this simplifies nicely with a 'u' substitution.

Let u = 1+4t^2 +9t^4

du = 8t + 36t^3 dt


`\int_0^1 (2t+9t^3)/(8t+36t^3) √(u) du \\ \\ \int_0^1 (2t+9t^3)/(4(2t+9t^3)) √(u) du \\ \\ (1)/(4) \int_0^1 √(u) du`

After integrating using power rule, replace 'u' with function for 't' and evaluate limits:

`=(1)/(4) |_0^1 ((2)/(3)) (1+4t^2 +9t^4)^(3/2) \\ \\ =(1)/(6) (14^(3/2) - 1)`