Consider F and C below.
F(x, y, z) = yz i + xz j + (xy + 14z) k
C is the line segment from (2, 0, −2) to (5, 5, 1)
(a) Find a function f such that F = ∇f.
f(x, y, z) =
(b) Use part (a) to evaluate integral C ∇f · dr along the given curve C.
(a) To find a function f such that F = ∇f, we need to find the potential function for the vector field F.
Let's start by finding the partial derivatives of f with respect to x, y, and z.
∂f/∂x = yz
∂f/∂y = xz
∂f/∂z = xy + 14z
Now, integrate each of these partial derivatives with respect to their respective variables:
f(x, y, z) = ∫(yz) dx + g1(y, z)
= xyz + g1(y, z)
f(x, y, z) = ∫(xz) dy + g2(x, z)
= xyz + g2(x, z)
f(x, y, z) = ∫(xy + 14z) dz + g3(x, y)
= xyz + 14z + g3(x, y)
Since f(x, y, z) should be the same for all three partial derivatives, we can equate the expressions:
xyz + g1(y, z) = xyz + g2(x, z) = xyz + 14z + g3(x, y) = f(x, y, z)
From this, we can see that g1(y, z) = g2(x, z) = 14z + g3(x, y), where g1, g2, and g3 are arbitrary functions of the remaining variables.
Therefore, the function f(x, y, z) = xyz + 14z + g(x, y, z), where g(x, y, z) is an arbitrary function of x, y, and z, is a possible function such that F = ∇f.
(b) The line segment C is defined as the curve from (2, 0, -2) to (5, 5, 1).
To evaluate the line integral C ∇f · dr, we need to parameterize the curve C.
Let's parameterize the curve C as r(t) = (2 + 3t, 3t, -2 + 3t) where t varies from 0 to 1.
Taking the derivative of r(t) with respect to t, dr/dt, we have dr/dt = (3, 3, 3).
Now, let's substitute the parameterized curve r(t) and the derivative dr/dt into ∇f · dr:
∇f · dr = (∂f/∂x, ∂f/∂y, ∂f/∂z) · (dx/dt, dy/dt, dz/dt)
∇f · dr = (yz, xz, xy + 14z) · (3, 3, 3)
∇f · dr = 3yz + 3xz + 3(xy + 14z)
Let's evaluate this expression along the curve C:
∇f · dr = 3(3t)(3t) + 3(2 + 3t)(3t) + 3[(2 + 3t)(3t) + 14(-2 + 3t)]
∇f · dr = 27t^2 + 9t(2 + 3t) + 3(6t^2 + 6t + 42t - 28)
Simplifying further:
∇f · dr = 27t^2 + 18t^2 + 27t^2 + 6t + 126t - 84
∇f · dr = 72t^2 + 132t - 84
Finally, let's integrate this expression with respect to t from 0 to 1:
∫(C) ∇f · dr = ∫[0,1] (72t^2 + 132t - 84) dt
Using the power rule of integration, we get:
∫(C) ∇f · dr = [24t^3 + 66t^2 - 84t] from 0 to 1
Plugging in the values of t, we have:
∫(C) ∇f · dr = (24(1)^3 + 66(1)^2 - 84(1)) - (24(0)^3 + 66(0)^2 - 84(0))
∫(C) ∇f · dr = (24 + 66 - 84) - (0)
∫(C) ∇f · dr = 6
Therefore, the value of the integral C ∇f · dr along the given curve C is 6.
To find a function f such that F = ∇f, we need to find the potential function of the vector field F(x, y, z).
Let's start by determining the partial derivatives of the potential function f(x, y, z):
∂f/∂x = Fx = yz
∂f/∂y = Fy = xz
∂f/∂z = Fz = xy + 14z
Now, we need to integrate each of these partial derivatives with respect to its corresponding variable, adding a constant of integration for each integration.
Integrating ∂f/∂x = yz with respect to x, treating y and z as constants, we get:
f(x, y, z) = ∫ (yz) dx = yzx + C1(y, z)
Next, integrating ∂f/∂y = xz with respect to y, treating x and z as constants, we get:
f(x, y, z) = ∫ (xz) dy = xzy + C2(x, z)
Finally, integrating ∂f/∂z = xy + 14z with respect to z, treating x and y as constants, we get:
f(x, y, z) = ∫ (xy + 14z) dz = xyz + 7z^2 + C3(x, y)
Combining these results, we can write the potential function f(x, y, z) as:
f(x, y, z) = yzx + xzy + xyz + 7z^2 + C
Now, let's move on to part (b) and use the potential function f(x, y, z) to evaluate the line integral of C ∇f · dr.
The line segment C is defined from (2, 0, -2) to (5, 5, 1).
To evaluate the line integral, we need to parametrize the curve C by describing it as a vector function of a single parameter t.
We can define r(t) = (x(t), y(t), z(t)) as the vector function for C.
To find x(t), y(t), and z(t), we need to determine the equations for the line passing through the two given points.
x(t) = 2 + 3t
y(t) = 0 + 5t
z(t) = -2 + 3t
Now, let's calculate ∇f and dr to find ∇f · dr:
∇f = (∂f/∂x, ∂f/∂y, ∂f/∂z) = (yz, xz, xy + 14z)
dr = (dx/dt, dy/dt, dz/dt) = (3, 5, 3)
∇f · dr = (yz, xz, xy + 14z) · (3, 5, 3) = 3yz + 5xz + 3(xy + 14z)
Now, substituting x(t), y(t), and z(t) into the expression for ∇f · dr, we get:
∇f · dr = 3[0 + 5t]z + 5[2 + 3t]z + 3[2 + 3t][5t] + 3[2 + 3t][14(-2 + 3t)]
Next, we need to substitute the values of t where t ranges from 0 to 1 because C goes from (2, 0, -2) to (5, 5, 1).
Finally, we integrate the expression over the given parameter range to evaluate the line integral C ∇f · dr.