Question

Consider the functions below. f(x, y, z) = x i − z j + y k r(t) = 4t i + 6t j − t2 k (a) evaluate the line integral c f · dr, where c is given by r(t), −1 ≤ t ≤ 1.

Answer 1

With

$\vec r(t)=4t\,\vec\imath+6t\,\vec\jmath-t^2\,\vec k$

we have

$\mathrm d\vec r=(4\,\vec\imath+6\,\vec\jmath-2t\,\vec k)\,\mathrm dt$

The vector field evaluated over this parameterization is

$\vec f(x,y,z)=\vec f(x(t),y(t),z(t))=4t\,\vec\imath+t^2\,\vec\jmath+6t\,\vec k$

so the line integral is

$\displaystyle\int_{-1}^1(4t\,\vec\imath+t^2\,\vec\jmath+6t\,\vec k)\cdot(4\,\vec\imath+6\,\vec\jmath-2t\,\vec k)\,\mathrm dt$

$=\displaystyle\int_{-1}^1(16t+6t^2-12t^2)\,\mathrm dt=-4$

Answer 2

Applying the steps, the result of the line integral is -4.

The curve is:

$f(x,y,z) = xi - zj + yk$

The vector field is:

$r(t) = (x(t), y(t), z(t)) = (4t, 6t, -t^2)$

Applying the vector field at the curve, we have that:

$f(t) = (4t, t^2, 6t)$

The derivative of the vector field is:

$r^{\prime}(t) = (4, 6, -2t)$

The dot product of the vector field along the curve with the derivative is:

$f(t)r^{\prime}(t) = (4t, t^2, 6t)(4, 6, -2t) = 16t + 6t - 12t^2 = -6t^2 + 16t$

Hence, the line integral is:

$I = \int_{-1}^{1} (-6t^2 + 16t) dt$

$I = -2t^3 + 8t^2|_{t = -1}^{t = 1}$

$I = -2 + 8 -2 - 8$

$I = -4$

A similar problem is given at https://brainly.com/question/12666512