Part A: 2-Mark Questions
1. If φ(x, y, z) = x2y - 2y2z3 find ∇φ at the point (1, -1, 2).
Sol:
∇φ = (∂/∂x)\(\vec{i}\) + (∂/∂y)\(\vec{j}\) + (∂/∂z)\(\vec{k}\)
∇φ = (2xy)\(\vec{i}\) + (x2 - 4yz3)\(\vec{j}\) + (-6y2z2)\(\vec{k}\)
At (1, -1, 2):
∇φ = (2(1)(-1))\(\vec{i}\) + ((1)2 - 4(-1)(2)3)\(\vec{j}\) - (6(-1)2(2)2)\(\vec{k}\)
∇φ = (-2)\(\vec{i}\) + (1 - 4(-1)(8))\(\vec{j}\) - (6(1)(4))\(\vec{k}\)
∇φ = -2\(\vec{i}\) + (1 + 32)\(\vec{j}\) - 24\(\vec{k}\)
∇φ = -2\(\vec{i}\) + 33\(\vec{j}\) - 24\(\vec{k}\)
2. If φ(x, y, z) = 3x2y - y3z2 find ∇φ at the point (1, -2, 1).
Sol:
∇φ = (∂/∂x)\(\vec{i}\) + (∂/∂y)\(\vec{j}\) + (∂/∂z)\(\vec{k}\)
∇φ = (6xy)\(\vec{i}\) + (3x2 - 3y2z2)\(\vec{j}\) + (-2y3z)\(\vec{k}\)
At (1, -2, 1):
∇φ = (6(1)(-2))\(\vec{i}\) + (3(1)2 - 3(-2)2(1)2)\(\vec{j}\) - (2(-2)3(1))\(\vec{k}\)
∇φ = (-12)\(\vec{i}\) + (3 - 3(4)(1))\(\vec{j}\) - (2(-8))\(\vec{k}\)
∇φ = -12\(\vec{i}\) + (3 - 12)\(\vec{j}\) + 16\(\vec{k}\)
∇φ = -12\(\vec{i}\) - 9\(\vec{j}\) + 16\(\vec{k}\)
3. Compute curl \(\vec{F}\) at (1, -1, 1) for \(\vec{F}\) = x2z \(\vec{i}\) + y3z2 \(\vec{j}\) + xy2z \(\vec{k}\).
Sol:
curl \(\vec{F}\) = ∇ × \(\vec{F}\) =
| \(\vec{i}\) \(\vec{j}\) \(\vec{k}\) |
| ∂/∂x ∂/∂y ∂/∂z |
| x2z y3z2 xy2z |
= \(\vec{i}\) [(∂/∂y)(xy2z) - (∂/∂z)(y3z2)] - \(\vec{j}\) [(∂/∂x)(xy2z) - (∂/∂z)(x2z)] + \(\vec{k}\) [(∂/∂x)(y3z2) - (∂/∂y)(x2z)]
= \(\vec{i}\) [2xyz - 2y3z] - \(\vec{j}\) [y2z - x2] + \(\vec{k}\) [0 - 0]
At (1, -1, 1):
curl \(\vec{F}\) = \(\vec{i}\) [2(1)(-1)(1) - 2(-1)3(1)] - \(\vec{j}\) [(-1)2(1) - (1)2] + \(\vec{k}\)[0]
= \(\vec{i}\) [-2 - 2(-1)] - \(\vec{j}\) [1 - 1] + \(\vec{k}\)[0]
= \(\vec{i}\) [-2 + 2] - \(\vec{j}\) [0] = 0\(\vec{i}\) + 0\(\vec{j}\) + 0\(\vec{k}\) = \(\vec{0}\)
4. If \(\vec{F}\) = xz3 \(\vec{i}\) - 2xyz \(\vec{j}\) + xz \(\vec{k}\). Find curl \(\vec{F}\) at (1, 2, 0).
Sol:
curl \(\vec{F}\) = ∇ × \(\vec{F}\) =
| \(\vec{i}\) \(\vec{j}\) \(\vec{k}\) |
| ∂/∂x ∂/∂y ∂/∂z |
| xz3 -2xyz xz |
= \(\vec{i}\) [(∂/∂y)(xz) - (∂/∂z)(-2xyz)] - \(\vec{j}\) [(∂/∂x)(xz) - (∂/∂z)(xz3)] + \(\vec{k}\) [(∂/∂x)(-2xyz) - (∂/∂y)(xz3)]
= \(\vec{i}\) [0 - (-2xy)] - \(\vec{j}\) [z - 3xz2] + \(\vec{k}\) [-2yz - 0]
= (2xy) \(\vec{i}\) - (z - 3xz2) \(\vec{j}\) - (2yz) \(\vec{k}\)
At (1, 2, 0):
curl \(\vec{F}\) = (2(1)(2)) \(\vec{i}\) - (0 - 3(1)(0)2) \(\vec{j}\) - (2(2)(0)) \(\vec{k}\)
= 4\(\vec{i}\) - 0\(\vec{j}\) - 0\(\vec{k}\) = 4\(\vec{i}\)
5. Find the directional derivative of φ = 4xz2 + x2yz at (1, -2, -1) in the direction 2\(\vec{i}\) - \(\vec{j}\) - 2\(\vec{k}\).
Sol: Directional Derivative (DD) = ∇φ · \(\hat{a}\)
1. Find ∇φ:
∇φ = (4z2 + 2xyz)\(\vec{i}\) + (x2z)\(\vec{j}\) + (8xz + x2y)\(\vec{k}\)
At (1, -2, -1):
∇φ = (4(-1)2 + 2(1)(-2)(-1))\(\vec{i}\) + ((1)2(-1))\(\vec{j}\) + (8(1)(-1) + (1)2(-2))\(\vec{k}\)
∇φ = (4 + 4)\(\vec{i}\) + (-1)\(\vec{j}\) + (-8 - 2)\(\vec{k}\)
∇φ = 8\(\vec{i}\) - \(\vec{j}\) - 10\(\vec{k}\)
2. Find unit vector \(\hat{a}\):
\(\vec{a}\) = 2\(\vec{i}\) - \(\vec{j}\) - 2\(\vec{k}\)
|\(\vec{a}\)| = √(22 + (-1)2 + (-2)2) = √(4 + 1 + 4) = √9 = 3
\(\hat{a}\) = (1/3) (2\(\vec{i}\) - \(\vec{j}\) - 2\(\vec{k}\))
3. Calculate DD:
DD = (8\(\vec{i}\) - \(\vec{j}\) - 10\(\vec{k}\)) · (1/3) (2\(\vec{i}\) - \(\vec{j}\) - 2\(\vec{k}\))
DD = (1/3) [ (8)(2) + (-1)(-1) + (-10)(-2) ]
DD = (1/3) [ 16 + 1 + 20 ] = 37/3
6. Find the directional derivative of φ = xyz - xy2z3 at (1, 2, -1) in the direction \(\vec{i}\) - \(\vec{j}\) - 3\(\vec{k}\).
Sol: DD = ∇φ · \(\hat{a}\)
1. Find ∇φ:
∇φ = (yz - y2z3)\(\vec{i}\) + (xz - 2xyz3)\(\vec{j}\) + (xy - 3xy2z2)\(\vec{k}\)
At (1, 2, -1):
∇φ = ((2)(-1) - (2)2(-1)3)\(\vec{i}\) + ((1)(-1) - 2(1)(2)(-1)3)\(\vec{j}\) + ((1)(2) - 3(1)(2)2(-1)2)\(\vec{k}\)
∇φ = (-2 - 4(-1))\(\vec{i}\) + (-1 - 4(-1))\(\vec{j}\) + (2 - 3(1)(4)(1))\(\vec{k}\)
∇φ = (-2 + 4)\(\vec{i}\) + (-1 + 4)\(\vec{j}\) + (2 - 12)\(\vec{k}\)
∇φ = 2\(\vec{i}\) + 3\(\vec{j}\) - 10\(\vec{k}\)
2. Find unit vector \(\hat{a}\):
\(\vec{a}\) = \(\vec{i}\) - \(\vec{j}\) - 3\(\vec{k}\)
|\(\vec{a}\)| = √(12 + (-1)2 + (-3)2) = √(1 + 1 + 9) = √11
\(\hat{a}\) = (1/√11) (\(\vec{i}\) - \(\vec{j}\) - 3\(\vec{k}\))
3. Calculate DD:
DD = (2\(\vec{i}\) + 3\(\vec{j}\) - 10\(\vec{k}\)) · (1/√11) (\(\vec{i}\) - \(\vec{j}\) - 3\(\vec{k}\))
DD = (1/√11) [ (2)(1) + (3)(-1) + (-10)(-3) ]
DD = (1/√11) [ 2 - 3 + 30 ] = 29/√11
7. Find the normal derivative of φ = xy + yz + zx at (-1, 1, 1).
Sol: The normal derivative is the magnitude of the gradient, |∇φ|.
∇φ = (∂/∂x)\(\vec{i}\) + (∂/∂y)\(\vec{j}\) + (∂/∂z)\(\vec{k}\)
∇φ = (y + z)\(\vec{i}\) + (x + z)\(\vec{j}\) + (y + x)\(\vec{k}\)
At (-1, 1, 1):
∇φ = (1 + 1)\(\vec{i}\) + (-1 + 1)\(\vec{j}\) + (1 + (-1))\(\vec{k}\)
∇φ = 2\(\vec{i}\) + 0\(\vec{j}\) + 0\(\vec{k}\)
|∇φ| = √(22 + 02 + 02) = √4 = 2
8. Find the normal derivative of φ = x3 - y3 + z at (1, -1, -1).
Sol: Normal derivative = |∇φ|.
∇φ = (∂/∂x)\(\vec{i}\) + (∂/∂y)\(\vec{j}\) + (∂/∂z)\(\vec{k}\)
∇φ = (3x2)\(\vec{i}\) + (-3y2)\(\vec{j}\) + (1)\(\vec{k}\)
At (1, -1, -1):
∇φ = (3(1)2)\(\vec{i}\) + (-3(-1)2)\(\vec{j}\) + \(\vec{k}\)
∇φ = 3\(\vec{i}\) - 3\(\vec{j}\) + \(\vec{k}\)
|∇φ| = √(32 + (-3)2 + 12) = √(9 + 9 + 1) = √19
9. What are the conditions for solenoidal and irrotational.
Sol:
- A vector field \(\vec{F}\) is solenoidal if its divergence is zero.
Condition: ∇ · \(\vec{F}\) = 0
- A vector field \(\vec{F}\) is irrotational if its curl is zero.
Condition: ∇ × \(\vec{F}\) = \(\vec{0}\)
10. Prove that \(\vec{F}\) = z \(\vec{i}\) + x \(\vec{j}\) + y \(\vec{k}\) is solenoidal.
Sol: To prove \(\vec{F}\) is solenoidal, we must show that its divergence is zero (∇ · \(\vec{F}\) = 0).
∇ · \(\vec{F}\) = (∂/∂x)(F1) + (∂/∂y)(F2) + (∂/∂z)(F3)
∇ · \(\vec{F}\) = (∂/∂x)(z) + (∂/∂y)(x) + (∂/∂z)(y)
∇ · \(\vec{F}\) = 0 + 0 + 0 = 0
Since ∇ · \(\vec{F}\) = 0, the vector field \(\vec{F}\) is solenoidal.