Answer on Question #58425 – Math – Vector Calculus
Question
1. φ = 2 x z 4 − x 2 y \varphi = 2xz^4 - x^2y φ = 2 x z 4 − x 2 y ∣ ∇ φ ⃗ ∣ |\nabla \vec{\varphi}| ∣∇ φ ∣
Solution
∇ φ ⃗ = ∂ φ ∂ x i ⃗ + ∂ φ ∂ y j ⃗ + ∂ φ ∂ z k ⃗ = ( 2 z 4 − 2 x y ) i ⃗ − x 2 j ⃗ + 8 x z 3 k ⃗ ∣ ∇ φ ⃗ ∣ = ( 2 z 4 − 2 x y ) 2 + x 4 + 64 x 2 z 6 \begin{array}{l}
\nabla \vec{\varphi} = \frac{\partial \varphi}{\partial x} \vec{i} + \frac{\partial \varphi}{\partial y} \vec{j} + \frac{\partial \varphi}{\partial z} \vec{k} = (2z^4 - 2xy)\vec{i} - x^2 \vec{j} + 8xz^3 \vec{k} \\
|\nabla \vec{\varphi}| = \sqrt{(2z^4 - 2xy)^2 + x^4 + 64x^2z^6}
\end{array} ∇ φ = ∂ x ∂ φ i + ∂ y ∂ φ j + ∂ z ∂ φ k = ( 2 z 4 − 2 x y ) i − x 2 j + 8 x z 3 k ∣∇ φ ∣ = ( 2 z 4 − 2 x y ) 2 + x 4 + 64 x 2 z 6
Point ( x , y , z ) (x,y,z) ( x , y , z )
Question
2. φ = 3 x 2 y − y 3 z 2 \varphi = 3x^{2}y - y^{3}z^{2} φ = 3 x 2 y − y 3 z 2 ∇ φ ⃗ \nabla \vec{\varphi} ∇ φ
Solution
∇ φ ⃗ = ∂ φ ∂ x i ⃗ + ∂ φ ∂ y j ⃗ + ∂ φ ∂ z k ⃗ = 6 x y i ⃗ + ( 3 x 2 − 3 y 2 z 2 ) j ⃗ − 2 z y 3 k ⃗ \nabla \vec{\varphi} = \frac{\partial \varphi}{\partial x} \vec{i} + \frac{\partial \varphi}{\partial y} \vec{j} + \frac{\partial \varphi}{\partial z} \vec{k} = 6xy \vec{i} + (3x^2 - 3y^2z^2)\vec{j} - 2zy^3 \vec{k} ∇ φ = ∂ x ∂ φ i + ∂ y ∂ φ j + ∂ z ∂ φ k = 6 x y i + ( 3 x 2 − 3 y 2 z 2 ) j − 2 z y 3 k
Then we put point's coordinates in the previous expression:
∇ φ ⃗ ( 1 ; − 2 ; − 1 ) = − 12 i ⃗ − 9 j ⃗ − 16 k ⃗ \nabla \vec{\varphi}(1; -2; -1) = -12\vec{i} - 9\vec{j} - 16\vec{k} ∇ φ ( 1 ; − 2 ; − 1 ) = − 12 i − 9 j − 16 k
Answer: ∇ φ ⃗ ( 1 ; − 2 ; − 1 ) = − 12 i ⃗ − 9 j ⃗ − 16 k ⃗ \nabla \vec{\varphi}(1; -2; -1) = -12\vec{i} - 9\vec{j} - 16\vec{k} ∇ φ ( 1 ; − 2 ; − 1 ) = − 12 i − 9 j − 16 k
Question
3. Find unit normal to a surface: x 2 y + 2 x z = 4 x^{2}y + 2xz = 4 x 2 y + 2 x z = 4
Solution
First, we rewrite the surface equation in the form of F ( x , y , z ) = 0 \mathrm{F(x,y,z)} = 0 F ( x , y , z ) = 0
x 2 y + 2 x z − 4 = 0. x^2y + 2xz - 4 = 0. x 2 y + 2 x z − 4 = 0.
A normal to a surface can be found as ( F x ′ ( A ) ; F y ′ ( A ) ; F z ′ ( A ) ) \left(F_x'(A); F_y'(A); F_z'(A)\right) ( F x ′ ( A ) ; F y ′ ( A ) ; F z ′ ( A ) ) A A A
F x ′ ( A ) = 2 x y + 2 z = − 4 F_x'(A) = 2xy + 2z = -4 F x ′ ( A ) = 2 x y + 2 z = − 4 F y ′ ( A ) = x 2 = 4 F_y'(A) = x^2 = 4 F y ′ ( A ) = x 2 = 4 F z ′ ( A ) = 2 x = 4 F_z'(A) = 2x = 4 F z ′ ( A ) = 2 x = 4
Vector is ( − 4 ; 4 ; 4 ) (-4; 4; 4) ( − 4 ; 4 ; 4 ) ( 23 ; − 23 ; − 23 ) (23; -23; -23) ( 23 ; − 23 ; − 23 )
Answer: ( − 4 ; 4 ; 4 ) (-4; 4; 4) ( − 4 ; 4 ; 4 ) ( 23 ; − 23 ; − 23 ) (23; -23; -23) ( 23 ; − 23 ; − 23 )
Question
4. φ = x y 2 z A ⃗ = x z i ⃗ − x y 2 j ⃗ + y z 2 k ⃗ \varphi = xy^2z \vec{A} = xz\vec{i} - xy^2\vec{j} + yz^2\vec{k} φ = x y 2 z A = x z i − x y 2 j + y z 2 k ∂ 3 φ A ⃗ ∂ x 2 ∂ z \frac{\partial^3\varphi\vec{A}}{\partial x^2\partial z} ∂ x 2 ∂ z ∂ 3 φ A
Solution
φ A ⃗ = x 2 y 2 z 2 i ⃗ − x 2 y 4 z j ⃗ + x y 3 z 3 k ⃗ \varphi \vec{A} = x^2y^2z^2\vec{i} - x^2y^4z\vec{j} + xy^3z^3\vec{k} φ A = x 2 y 2 z 2 i − x 2 y 4 z j + x y 3 z 3 k ∂ 3 ( φ A ⃗ ) ∂ x 2 ∂ z = ∂ 2 ∂ ( φ A ⃗ ) ∂ z ∂ x 2 = ∂ 2 ( 2 x 2 y 2 z i ⃗ − x 2 y 4 j ⃗ + 3 x y 3 z 2 k ⃗ ) ∂ x 2 = 4 y 2 z i ⃗ − 2 y 4 j ⃗ \frac{\partial^3(\varphi \vec{A})}{\partial x^2\partial z} = \frac{\partial^2 \frac{\partial(\varphi \vec{A})}{\partial z}}{\partial x^2} = \frac{\partial^2 (2x^2y^2z\vec{i} - x^2y^4\vec{j} + 3xy^3z^2\vec{k})}{\partial x^2} = 4y^2z\vec{i} - 2y^4\vec{j} ∂ x 2 ∂ z ∂ 3 ( φ A ) = ∂ x 2 ∂ 2 ∂ z ∂ ( φ A ) = ∂ x 2 ∂ 2 ( 2 x 2 y 2 z i − x 2 y 4 j + 3 x y 3 z 2 k ) = 4 y 2 z i − 2 y 4 j
Answer should be based on point (not given in task), but most appropriate from answers given:
∂ 3 ( φ A ⃗ ) ∂ x 2 ∂ z = 4 y 2 z i ⃗ − 2 y 4 j ⃗ = 4 i ⃗ − 2 j ⃗ . \frac{\partial^3(\varphi \vec{A})}{\partial x^2\partial z} = 4y^2z\vec{i} - 2y^4\vec{j} = 4\vec{i} - 2\vec{j}. ∂ x 2 ∂ z ∂ 3 ( φ A ) = 4 y 2 z i − 2 y 4 j = 4 i − 2 j . Question
5. φ = 2 x 2 y − x z 3 \varphi = 2x^{2}y - xz^{3} φ = 2 x 2 y − x z 3 ∇ 2 φ ⃗ \nabla^2\vec{\varphi} ∇ 2 φ
Solution
∇ φ ⃗ = ∂ φ ∂ x i ⃗ + ∂ φ ∂ y j ⃗ + ∂ φ ∂ z k ⃗ = ( 4 x y − z 3 ) i ⃗ + 2 x 2 j ⃗ − 3 x z 2 k ⃗ , ∇ 2 φ ⃗ = 16 x 2 y 2 + z 6 − 8 x y z 3 + 4 x 4 + 9 x 2 z 4 \begin{array}{l}
\nabla \vec{\varphi} = \frac{\partial \varphi}{\partial x} \vec{i} + \frac{\partial \varphi}{\partial y} \vec{j} + \frac{\partial \varphi}{\partial z} \vec{k} = (4xy - z^{3})\vec{i} + 2x^{2}\vec{j} - 3xz^{2}\vec{k}, \\
\nabla^{2} \vec{\varphi} = 16x^{2}y^{2} + z^{6} - 8xyz^{3} + 4x^{4} + 9x^{2}z^{4}
\end{array} ∇ φ = ∂ x ∂ φ i + ∂ y ∂ φ j + ∂ z ∂ φ k = ( 4 x y − z 3 ) i + 2 x 2 j − 3 x z 2 k , ∇ 2 φ = 16 x 2 y 2 + z 6 − 8 x y z 3 + 4 x 4 + 9 x 2 z 4
Answer: ∇ 2 φ ⃗ = 16 x 2 y 2 + z 6 − 8 x y z 3 + 4 x 4 + 9 x 2 z 4 \nabla^2\vec{\varphi} = 16x^2y^2 + z^6 - 8xyz^3 + 4x^4 + 9x^2z^4 ∇ 2 φ = 16 x 2 y 2 + z 6 − 8 x y z 3 + 4 x 4 + 9 x 2 z 4
Question
6. A ⃗ = x z 3 i ⃗ − 2 x 2 y z j ⃗ + 2 y z k ⃗ \vec{A} = xz^{3}\vec{i} - 2x^{2}yz\vec{j} + 2yz\vec{k} A = x z 3 i − 2 x 2 yz j + 2 yz k [ ∇ ∗ A ⃗ ] [\nabla * \vec{A}] [ ∇ ∗ A ]
Solution
[ ∇ ∗ A ⃗ ] = ( − ∂ 2 y z ∂ z − ∂ 2 x 2 y z ∂ y ) i ⃗ + ( − ∂ 2 y z ∂ x + ∂ x z 3 ∂ z ) j ⃗ + ( − ∂ x z 3 ∂ y − ∂ 2 x 2 y z ∂ x ) k ⃗ = ( − 2 y − 2 x 2 z ) i ⃗ + 3 x z 2 j ⃗ − 4 x y z k ⃗ \left[\nabla * \vec{A}\right] = \left(-\frac{\partial 2yz}{\partial z} - \frac{\partial 2x^{2}yz}{\partial y}\right)\vec{i} + \left(-\frac{\partial 2yz}{\partial x} + \frac{\partial xz^{3}}{\partial z}\right)\vec{j} + \left(-\frac{\partial xz^{3}}{\partial y} - \frac{\partial 2x^{2}yz}{\partial x}\right)\vec{k} = (-2y - 2x^{2}z)\vec{i} + 3xz^{2}\vec{j} - 4xyz\vec{k} [ ∇ ∗ A ] = ( − ∂ z ∂ 2 yz − ∂ y ∂ 2 x 2 yz ) i + ( − ∂ x ∂ 2 yz + ∂ z ∂ x z 3 ) j + ( − ∂ y ∂ x z 3 − ∂ x ∂ 2 x 2 yz ) k = ( − 2 y − 2 x 2 z ) i + 3 x z 2 j − 4 x yz k
Now we put point’s coordinates x = 1 , y = − 1 , z = 1 x = 1, y = -1, z = 1 x = 1 , y = − 1 , z = 1
[ ∇ ∗ A ⃗ ] = 3 j ⃗ + 4 k ⃗ . \left[\nabla * \vec{A}\right] = 3\vec{j} + 4\vec{k}. [ ∇ ∗ A ] = 3 j + 4 k .
Answer: [ ∇ ∗ A ⃗ ] = 3 j ⃗ + 4 k ⃗ \left[\nabla * \vec{A}\right] = 3\vec{j} + 4\vec{k} [ ∇ ∗ A ] = 3 j + 4 k
Question
7. A ⃗ = A 1 i ⃗ + A 2 j ⃗ + A 3 k ⃗ \vec{A} = A1\vec{i} + A2\vec{j} + A3\vec{k} A = A 1 i + A 2 j + A 3 k r ⃗ = x i ⃗ + y j ⃗ + z k ⃗ \vec{r} = x\vec{i} + y\vec{j} + z\vec{k} r = x i + y j + z k ∇ ⃗ ⋅ ( A ⃗ ∗ r ⃗ ) \vec{\nabla} \cdot (\vec{A} * \vec{r}) ∇ ⋅ ( A ∗ r )
Solution
∇ ⃗ ⋅ ( A ⃗ ∗ r ⃗ ) = ∇ A ⃗ ⋅ ( A ⃗ ∗ r ⃗ ) + ∇ r ⃗ ⋅ ( A ⃗ ∗ r ⃗ ) = r ⃗ ⋅ ( ∇ A ⃗ ∗ A ⃗ ) − A ⃗ ⋅ ( ∇ r ⃗ ∗ r ⃗ ) . \vec{\nabla} \cdot (\vec{A} * \vec{r}) = \vec{\nabla_{\mathrm{A}}} \cdot (\vec{A} * \vec{r}) + \vec{\nabla_{\mathrm{r}}} \cdot (\vec{A} * \vec{r}) = \vec{\mathrm{r}} \cdot (\vec{\nabla_{\mathrm{A}}} * \vec{A}) - \vec{\mathrm{A}} \cdot (\vec{\nabla_{\mathrm{r}}} * \vec{r}). ∇ ⋅ ( A ∗ r ) = ∇ A ⋅ ( A ∗ r ) + ∇ r ⋅ ( A ∗ r ) = r ⋅ ( ∇ A ∗ A ) − A ⋅ ( ∇ r ∗ r ) .
As A ⃗ \vec{A} A ∇ A ⃗ ∗ A ⃗ = 0 \vec{\nabla_{\mathrm{A}}} * \vec{A} = 0 ∇ A ∗ A = 0 ∇ r ⃗ ∗ r ⃗ = 0 \vec{\nabla_{\mathrm{r}}} * \vec{r} = 0 ∇ r ∗ r = 0
Answer: ∇ ⃗ ⋅ ( A ⃗ ∗ r ⃗ ) = 0 \vec{\nabla} \cdot (\vec{A} * \vec{r}) = 0 ∇ ⋅ ( A ∗ r ) = 0
Question
8. A ⃗ = x 2 y i ⃗ − 2 x z j ⃗ + 2 y z k ⃗ \vec{A} = x^{2}y\vec{i} - 2xz\vec{j} + 2yz\vec{k} A = x 2 y i − 2 x z j + 2 yz k C u r l c u r l A ⃗ Curl\ curl\ \vec{A} C u r l c u r l A
Solution
Curl curl A ⃗ \vec{A} A [ ∇ ⃗ ∗ [ ∇ ⃗ ∗ A ⃗ ] ] [\vec{\nabla} * [\vec{\nabla} * \vec{A}]] [ ∇ ∗ [ ∇ ∗ A ]]
[ ∇ ⃗ ∗ [ ∇ ⃗ ∗ A ⃗ ] ] = grad ( d i v A ⃗ ) − Δ A ⃗ = grad ( 2 x y + 2 y ) − ( 2 y i ⃗ ) = 2 y i ⃗ + ( 2 x + 2 ) j ⃗ − 2 y i ⃗ = = ( 2 x + 2 ) j ⃗ . \begin{array}{l}
\left[\vec{\nabla} * \left[\vec{\nabla} * \vec{A}\right]\right] = \operatorname{grad}(div\vec{A}) - \Delta\vec{A} = \operatorname{grad}(2xy + 2y) - (2y\vec{i}) = 2y\vec{i} + (2x + 2)\vec{j} - 2y\vec{i} = \\
= (2x + 2)\vec{j}.
\end{array} [ ∇ ∗ [ ∇ ∗ A ] ] = grad ( d i v A ) − Δ A = grad ( 2 x y + 2 y ) − ( 2 y i ) = 2 y i + ( 2 x + 2 ) j − 2 y i = = ( 2 x + 2 ) j .
Answer: [ ∇ ⃗ ∗ [ ∇ ⃗ ∗ A ⃗ ] ] = ( 2 x + 2 ) j ⃗ \left[\vec{\nabla} * [\vec{\nabla} * \vec{A}]\right] = (2x + 2)\vec{j} [ ∇ ∗ [ ∇ ∗ A ] ] = ( 2 x + 2 ) j
Question
9. A ⃗ = 2 x 2 i ⃗ − 3 y z j ⃗ + x z 2 k ⃗ \vec{A} = 2x^{2}\vec{i} - 3yz\vec{j} + xz^{2}\vec{k} A = 2 x 2 i − 3 yz j + x z 2 k φ = 2 z − x 3 y \varphi = 2z - x^{3}y φ = 2 z − x 3 y A ⃗ ⋅ ∇ φ \vec{A} \cdot \nabla \varphi A ⋅ ∇ φ
Solution
∇ φ = − 3 x 2 y i ⃗ − x 3 j ⃗ + 2 k ⃗ \nabla \varphi = -3x^2 y \vec{i} - x^3 \vec{j} + 2 \vec{k} ∇ φ = − 3 x 2 y i − x 3 j + 2 k A ⃗ ⋅ ∇ φ = − 6 x 4 y + 3 y z x 3 + 2 x z 2 \vec{A} \cdot \nabla \varphi = -6x^4 y + 3yzx^3 + 2xz^2 A ⋅ ∇ φ = − 6 x 4 y + 3 yz x 3 + 2 x z 2
Put point’s coordinates x = 1 , y = − 1 , z = 1 x = 1, y = -1, z = 1 x = 1 , y = − 1 , z = 1
A ⃗ ⋅ ∇ φ ( 1 ; − 1 ; 1 ) = 6 − 3 + 2 = 5 \vec{A} \cdot \nabla \varphi(1; -1; 1) = 6 - 3 + 2 = 5 A ⋅ ∇ φ ( 1 ; − 1 ; 1 ) = 6 − 3 + 2 = 5
Answer: A ⃗ ⋅ ∇ φ ( 1 ; − 1 ; 1 ) = 5 \vec{A} \cdot \nabla \varphi(1; -1; 1) = 5 A ⋅ ∇ φ ( 1 ; − 1 ; 1 ) = 5
Question
10. Find the directional derivative of φ = x 2 y z + 4 x z 2 \varphi = x^2 yz + 4xz^2 φ = x 2 yz + 4 x z 2 l = 2 i ⃗ − j ⃗ − 2 k ⃗ l = 2\vec{i} - \vec{j} - 2\vec{k} l = 2 i − j − 2 k
Solution
∇ φ = ( 2 x y z + 4 z 2 ) i ⃗ + x 2 z j ⃗ + ( x 2 y + 8 x z ) k ⃗ \nabla \varphi = (2xyz + 4z^2) \vec{i} + x^2 z \vec{j} + (x^2 y + 8xz) \vec{k} ∇ φ = ( 2 x yz + 4 z 2 ) i + x 2 z j + ( x 2 y + 8 x z ) k
Directional derivative is l ⋅ ∇ φ = 2 ( 2 x y z + 4 z 2 ) − x 2 z − 2 ( x 2 y + 8 x z ) l \cdot \nabla \varphi = 2(2xyz + 4z^2) - x^2 z - 2(x^2 y + 8xz) l ⋅ ∇ φ = 2 ( 2 x yz + 4 z 2 ) − x 2 z − 2 ( x 2 y + 8 x z )
Putting values x = 1 , y = − 2 , z = − 1 x = 1, y = -2, z = -1 x = 1 , y = − 2 , z = − 1
l ⋅ ∇ φ = 2 ( 4 + 4 ) + 1 − 2 ( − 2 − 8 ) = 16 + 1 + 20 = 37 l \cdot \nabla \varphi = 2(4 + 4) + 1 - 2(-2 - 8) = 16 + 1 + 20 = 37 l ⋅ ∇ φ = 2 ( 4 + 4 ) + 1 − 2 ( − 2 − 8 ) = 16 + 1 + 20 = 37
Answer: l ⋅ ∇ φ = 37 l \cdot \nabla \varphi = 37 l ⋅ ∇ φ = 37
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#340153 on Dec 2023