Question

5 Given that ϕ=2x^2y−xz^3 find ▽^2ϕ
6 If A=xz^3i−2x^2yzj+2yz^4, find ▽×A at point (1,-1,1).
7 Given that A=A1i+A2j+A3k and r=xi+yj+zk, evaluate ▽⋅(A×r) if ▽×A=0
8 Let A=x^2yi−2xzj+2yzk, find Curl curl A.
9 Given A=2x^2i−3yzj+xz^2k and ϕ=2z−x^3y, find A⋅▽ϕ at point (1,-1,1).
10 Find the directional derivative of ϕ=x^2yz+4xz^2 at (1,-2,-1) in the direction 2i−j−2k

Answer 1

Answer on Question #59208 – Math – Calculus

Question

5. Given that $\varphi = 2x^2 y - xz^3$ find $\nabla^2\varphi$

Solution

$$\varphi = 2 x ^ {2} y - x z ^ {3}$$ $$\nabla^ {2} \varphi = \left(\frac {\partial^ {2}}{\partial x ^ {2}} + \frac {\partial^ {2}}{\partial y ^ {2}} + \frac {\partial^ {2}}{\partial z ^ {2}}\right) \varphi$$ $$\frac {\partial^ {2}}{\partial x ^ {2}} \varphi = 4 y, \frac {\partial^ {2}}{\partial y ^ {2}} \varphi = 0, \frac {\partial^ {2}}{\partial z ^ {2}} \varphi = - 6 z x$$ $$\nabla^ {2} \varphi = 4 y - 6 x z$$

Answer:

$$\nabla^ {2} \varphi = 4 y - 6 x z$$

Question

6. If $A = xz^3 i - 2x^2 yz^2 + 2yz^4 k$, find $\nabla \times A$ at point $(1, -1, 1)$.

Solution

$$\nabla \times A = \left| \begin{array}{ccc} i & j & k \\ \frac {\partial}{\partial x} & \frac {\partial}{\partial y} & \frac {\partial}{\partial z} \\ A _ {x} & A _ {y} & A _ {z} \end{array} \right| = i \left(\frac {\partial A _ {z}}{\partial y} - \frac {\partial A _ {y}}{\partial z}\right) - j \left(\frac {\partial A _ {x}}{\partial x} - \frac {\partial A _ {x}}{\partial z}\right) + k \left(\frac {\partial A _ {y}}{\partial x} - \frac {\partial A _ {x}}{\partial y}\right) = i (2 z ^ {4} + 2 x ^ {2} y) - j (- 3 x z ^ {2}) + k (- 4 x y z) = 2 i (z ^ {4} + x ^ {2} y) + 3 x z ^ {2} j - 4 x y z k$$ $$\nabla \times A (1, - 1, 1) = 2 i (1 - 1) + 3 j + 4 k = 3 j + 4 k = (0, 3, 4)$$

Answer:

$$\nabla \times A (1, - 1, 1) = 3 j + 4 k = (0, 3, 4)$$

Question

7. Given that $A = A1i + A2j + A3k$ and $r = xi + yj + zk$, evaluate $\nabla \cdot (A \times r)$ if $\nabla \times A = 0$

Solution

$$A \times r = \left| \begin{array}{ccc} i & j & k \\ x & y & z \\ A _ {1} & A _ {2} & A _ {3} \end{array} \right| = i (y A _ {3} - z A _ {2}) - j (x A _ {3} - z A _ {1}) + k (x A _ {2} - y A _ {1})$$ $$\nabla \cdot (A \times r) = \frac {\partial (A \times r) _ {x}}{\partial x} + \frac {\partial (A \times r) _ {y}}{\partial y} + \frac {\partial (A \times r) _ {z}}{\partial z} = \frac {\partial (y A _ {3} - z A _ {2})}{\partial x} + \frac {\partial (- x A _ {3} + z A _ {1})}{\partial y} + \frac {\partial (x A _ {2} - y A _ {1})}{\partial z}$$ $$\nabla \times A = \left| \begin{array}{ccc} i & j & k \\ \frac {\partial}{\partial x} & \frac {\partial}{\partial y} & \frac {\partial}{\partial z} \\ A _ {1} & A _ {2} & A _ {3} \end{array} \right| = i \left(\frac {\partial A _ {3}}{\partial y} - \frac {\partial A _ {2}}{\partial z}\right) - j \left(\frac {\partial A _ {3}}{\partial x} - \frac {\partial A _ {1}}{\partial z}\right) + k \left(\frac {\partial A _ {2}}{\partial x} - \frac {\partial A _ {1}}{\partial y}\right) = 0$$ $$\begin{array}{l} \frac {\partial \left(y A _ {3} - z A _ {2}\right)}{\partial x} + \frac {\partial \left(- x A _ {3} + z A _ {1}\right)}{\partial y} + \frac {\partial \left(x A _ {2} - y A _ {1}\right)}{\partial z} = \\ = y \frac {\partial A _ {3}}{\partial x} - y \frac {\partial A _ {1}}{\partial z} - z \frac {\partial A _ {2}}{\partial x} + z \frac {\partial A _ {1}}{\partial y} - x \frac {\partial A _ {3}}{\partial y} + x \frac {\partial A _ {2}}{\partial z} = 0 \end{array}$$

Answer:

$$\nabla \cdot (\mathbf {A} \times \mathbf {r}) = 0$$

Question

8. Let $A = x^2 y_i - 2xz_j + 2yzk$, find $\operatorname{Curl} \, \operatorname{curl} A$

Solution

$$\begin{array}{l} \operatorname{curl} A = \left| \begin{array}{ccc} i & j & k \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ A_x & A_y & A_z \end{array} \right| = i \left( \frac{\partial A_z}{\partial y} - \frac{\partial A_x}{\partial z} \right) - j \left( \frac{\partial A_z}{\partial x} - \frac{\partial A_x}{\partial z} \right) + k \left( \frac{\partial A_y}{\partial x} - \frac{\partial A_x}{\partial y} \right) = \\ = i(2z + 2x) - j(0) + k(-2z - x^2) \\ \operatorname{curl} \operatorname{curl} A = \left| \begin{array}{ccc} i & j & k \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ (\operatorname{curl} A)_x & (\operatorname{curl} A)_y & (\operatorname{curl} A)_z \end{array} \right| = \\ = i \left( \frac{\partial (\operatorname{curl} A)_z}{\partial y} - \frac{\partial A (\operatorname{curl} A)_x}{\partial z} \right) - j \left( \frac{\partial (\operatorname{curl} A)_z}{\partial x} - \frac{\partial (\operatorname{curl} A)_x}{\partial z} \right) + k \left( \frac{\partial (\operatorname{curl} A)_y}{\partial x} - \frac{\partial (\operatorname{curl} A)_x}{\partial y} \right) \end{array}$$

$\operatorname{curl} \, \operatorname{curl} A = i(0 - 0) - j(-2x - 2) + k(0 - 0) = 2j(x - 1)$

Answer:

$$\operatorname{curl} \, \operatorname{curl} A = 2j(x - 1)$$

Question

9. Given $A = 2x^2 i - 3yz_j + xz^2k$ and $\varphi = 2z - x^3y$, find $A \cdot \nabla \varphi$ at point $(1, -1, 1)$.

Solution

$$\nabla \varphi = i \frac{\partial}{\partial x} + j \frac{\partial}{\partial y} + k \frac{\partial}{\partial z} = -3x^2 y_i - x^3 j + 2k$$ $$A \nabla \varphi = -6x^4 y + 3x^3 yz + 2xz^2$$

Answer:

$$A \nabla \varphi = -6x^4 y + 3x^3 yz + 2xz^2$$

Question

10. Find the directional derivative of $\varphi = x^2 yz + 4xz^2$ at $(1, -2, -1)$ in the direction $2i - j - 2k$

Solution

Let $\vec{i} = 2\vec{i} + \vec{j} - 2\vec{k} \rightarrow |\vec{i}| = \sqrt{4 + 1 + 4} = 3$

$$A = (1, -2, -1)$$

So, the direction cosines are

$$\cos \alpha = \frac{2}{3}, \cos \beta = \frac{1}{3}, \cos \gamma = -\frac{2}{3}$$ $$\frac{\partial \varphi}{\partial x}(A) = (2xyz + 4z^2)(A) = 4 + 4 = 8;$$ $$\frac{\partial \varphi}{\partial y}(A) = (x^2 z)(A) = -1$$ $$\begin{array}{l} \frac{\partial \varphi}{\partial z}(A) = (x^{2} y + 8 x z)(A) = -2 - 8 = -10 \\ \frac{\partial \varphi}{\partial l}(A) = \frac{\partial \varphi}{\partial x}(A) \cos \alpha + \frac{\partial \varphi}{\partial y}(A) \cos \beta + \frac{\partial \varphi}{\partial z}(A) \cos \gamma = 8 * \frac{2}{3} - 1 * \frac{1}{3} - 10 * \frac{2}{3} = \frac{16}{3} - \frac{1}{3} - \frac{20}{3} \\ \frac{\partial \varphi}{\partial l}(A) = -\frac{5}{3} \end{array}$$

Answer:

$$-\frac{5}{3}$$

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