(a)
∇ ⋅ r = ∂ x ∂ x + ∂ y ∂ y + ∂ z ∂ z = 3 { \boldsymbol{\nabla} }\cdot\mathbf{r}
=\frac{\partial x}{\partial x}
+\frac{\partial y}{\partial y}
+\frac{\partial z}{\partial z}
=3 ∇ ⋅ r = ∂ x ∂ x + ∂ y ∂ y + ∂ z ∂ z = 3
(b)
∇ ( r 2 ) = ( ı ^ ∂ ∂ x + ȷ ^ ∂ ∂ y + k ^ ∂ ∂ z ) ( x 2 + y 2 + z 2 ) = 2 x ı ^ + 2 y ȷ ^ + 2 z k ^ = 2 r { \boldsymbol{\nabla} }(r^2)
=\left(\hat{\boldsymbol{\imath}}\frac{\partial }{\partial x}
+\hat{\boldsymbol{\jmath}}\frac{\partial }{\partial y}
+\hat{\mathbf{k}}\frac{\partial }{\partial z} \right)
\big(x^2+y^2+z^2\big)
=2x\,\hat{\boldsymbol{\imath}} +2y\,\hat{\boldsymbol{\jmath}} +2z\,\hat{\mathbf{k}} =2\mathbf{r} ∇ ( r 2 ) = ( ^ ∂ x ∂ + ^ ∂ y ∂ + k ^ ∂ z ∂ ) ( x 2 + y 2 + z 2 ) = 2 x ^ + 2 y ^ + 2 z k ^ = 2 r
(c) 다음과 같으므로
r × a = det [ ı ^ ȷ ^ k ^ x y z a 1 a 2 a 3 ] = ı ^ ( a 3 y − a 2 z ) + ȷ ^ ( a 1 z − a 3 x ) + k ^ ( a 2 x − a 1 y ) \mathbf{r}\times\mathbf{a} = \det\left[\begin{matrix}
\hat{\boldsymbol{\imath}} & \hat{\boldsymbol{\jmath}} & \hat{\mathbf{k}} \\
x & y & z \\
a_1 & a_2 & a_3
\end{matrix}\right]
=\hat{\boldsymbol{\imath}}\big(a_3y-a_2z\big)
+\hat{\boldsymbol{\jmath}}\big(a_1z-a_3x\big)
+\hat{\mathbf{k}}\big(a_2x-a_1y\big) r × a = det ^ x a 1 ^ y a 2 k ^ z a 3 = ^ ( a 3 y − a 2 z ) + ^ ( a 1 z − a 3 x ) + k ^ ( a 2 x − a 1 y )
다음을 얻는다.
∇ × ( r × a ) = det [ ı ^ ȷ ^ k ^ ∂ ∂ x ∂ ∂ y ∂ ∂ z a 3 y − a 2 z a 1 z − a 3 x a 2 x − a 1 y ] = − 2 a 1 ı ^ − 2 a 2 ȷ ^ − 2 a 3 k ^ = − 2 a \begin{aligned}
{ \boldsymbol{\nabla} }\times(\mathbf{r}\times\mathbf{a})
&= \det\left[\begin{matrix}\hat{\boldsymbol{\imath}}&\hat{\boldsymbol{\jmath}}&\hat{\mathbf{k}}\\[0.03in]
\frac{\partial }{\partial x}&
\frac{\partial }{\partial y}&
\frac{\partial }{\partial z}\\[0.03in]
a_3y-a_2z& a_1z-a_3x & a_2x-a_1y\end{matrix}\right]
=-2a_1\,\hat{\boldsymbol{\imath}} -2a_2\,\hat{\boldsymbol{\jmath}} -2a_3\,\hat{\mathbf{k}} \\
&=-2\mathbf{a} \end{aligned} ∇ × ( r × a ) = det ^ ∂ x ∂ a 3 y − a 2 z ^ ∂ y ∂ a 1 z − a 3 x k ^ ∂ z ∂ a 2 x − a 1 y = − 2 a 1 ^ − 2 a 2 ^ − 2 a 3 k ^ = − 2 a
(d) 다음과 같으므로
∇ ( r ) = ( ı ^ ∂ ∂ x + ȷ ^ ∂ ∂ y + k ^ ∂ ∂ z ) ( x 2 + y 2 + z 2 ) 1 / 2 = ı ^ x ( x 2 + y 2 + z 2 ) 1 / 2 + ȷ ^ y ( x 2 + y 2 + z 2 ) 1 / 2 + k ^ x ( x 2 + y 2 + z 2 ) 1 / 2 \begin{aligned}
{ \boldsymbol{\nabla} }(r) &= \left(\hat{\boldsymbol{\imath}}\frac{\partial }{\partial x}
+\hat{\boldsymbol{\jmath}}\frac{\partial }{\partial y}
+\hat{\mathbf{k}}\frac{\partial }{\partial z} \right) \big(x^2+y^2+z^2\big)^{1/2} \\
&=\hat{\boldsymbol{\imath}}\frac{x}{\big(x^2+y^2+z^2\big)^{1/2}}
+\hat{\boldsymbol{\jmath}}\frac{y}{\big(x^2+y^2+z^2\big)^{1/2}}
+\hat{\mathbf{k}}\frac{x}{\big(x^2+y^2+z^2\big)^{1/2}}
\end{aligned} ∇ ( r ) = ( ^ ∂ x ∂ + ^ ∂ y ∂ + k ^ ∂ z ∂ ) ( x 2 + y 2 + z 2 ) 1/2 = ^ ( x 2 + y 2 + z 2 ) 1/2 x + ^ ( x 2 + y 2 + z 2 ) 1/2 y + k ^ ( x 2 + y 2 + z 2 ) 1/2 x
다음을 얻는다.
∇ ⋅ ( ∇ ( r ) ) = ∂ ∂ x x ( x 2 + y 2 + z 2 ) 1 / 2 + ∂ ∂ y y ( x 2 + y 2 + z 2 ) 1 / 2 + ∂ ∂ z z ( x 2 + y 2 + z 2 ) 1 / 2 = 3 ( x 2 + y 2 + z 2 ) 1 / 2 − 1 2 2 x 2 + 2 y 2 + 2 z 2 ( x 2 + y 2 + z 2 ) 3 / 2 = 2 ( x 2 + y 2 + z 2 ) 1 / 2 = 2 r \begin{aligned}
{ \boldsymbol{\nabla} }\cdot\big({ \boldsymbol{\nabla} }(r)\big)
&= \frac{\partial }{\partial x} \frac{x}{\big(x^2+y^2+z^2\big)^{1/2}}
+\frac{\partial }{\partial y} \frac{y}{\big(x^2+y^2+z^2\big)^{1/2}}
+\frac{\partial }{\partial z} \frac{z}{\big(x^2+y^2+z^2\big)^{1/2}} \\
&= \frac{3}{\big(x^2+y^2+z^2\big)^{1/2}}
-\frac{1}{2}\frac{2x^2+2y^2+2z^2}{\big(x^2+y^2+z^2\big)^{3/2}}
=\frac{2}{\big(x^2+y^2+z^2\big)^{1/2}}
=\frac{2}{r}
\end{aligned} ∇ ⋅ ( ∇ ( r ) ) = ∂ x ∂ ( x 2 + y 2 + z 2 ) 1/2 x + ∂ y ∂ ( x 2 + y 2 + z 2 ) 1/2 y + ∂ z ∂ ( x 2 + y 2 + z 2 ) 1/2 z = ( x 2 + y 2 + z 2 ) 1/2 3 − 2 1 ( x 2 + y 2 + z 2 ) 3/2 2 x 2 + 2 y 2 + 2 z 2 = ( x 2 + y 2 + z 2 ) 1/2 2 = r 2
원본 교재 풀이(영어) (a)
∇ ⋅ r = ∂ x ∂ x + ∂ y ∂ y + ∂ z ∂ z = 3 { \boldsymbol{\nabla} }\cdot\mathbf{r}
=\frac{\partial x}{\partial x}
+\frac{\partial y}{\partial y}
+\frac{\partial z}{\partial z}
=3 ∇ ⋅ r = ∂ x ∂ x + ∂ y ∂ y + ∂ z ∂ z = 3
(b)
∇ ( r 2 ) = ( ı ^ ∂ ∂ x + ȷ ^ ∂ ∂ y + k ^ ∂ ∂ z ) ( x 2 + y 2 + z 2 ) = 2 x ı ^ + 2 y ȷ ^ + 2 z k ^ = 2 r { \boldsymbol{\nabla} }(r^2)
=\left(\hat{\boldsymbol{\imath}}\frac{\partial }{\partial x}
+\hat{\boldsymbol{\jmath}}\frac{\partial }{\partial y}
+\hat{\mathbf{k}}\frac{\partial }{\partial z} \right)
\big(x^2+y^2+z^2\big)
=2x\,\hat{\boldsymbol{\imath}} +2y\,\hat{\boldsymbol{\jmath}} +2z\,\hat{\mathbf{k}} =2\mathbf{r} ∇ ( r 2 ) = ( ^ ∂ x ∂ + ^ ∂ y ∂ + k ^ ∂ z ∂ ) ( x 2 + y 2 + z 2 ) = 2 x ^ + 2 y ^ + 2 z k ^ = 2 r
(c) Since
r × a = det [ ı ^ ȷ ^ k ^ x y z a 1 a 2 a 3 ] = ı ^ ( a 3 y − a 2 z ) + ȷ ^ ( a 1 z − a 3 x ) + k ^ ( a 2 x − a 1 y ) \mathbf{r}\times\mathbf{a} = \det\left[\begin{matrix}
\hat{\boldsymbol{\imath}} & \hat{\boldsymbol{\jmath}} & \hat{\mathbf{k}} \\
x & y & z \\
a_1 & a_2 & a_3
\end{matrix}\right]
=\hat{\boldsymbol{\imath}}\big(a_3y-a_2z\big)
+\hat{\boldsymbol{\jmath}}\big(a_1z-a_3x\big)
+\hat{\mathbf{k}}\big(a_2x-a_1y\big) r × a = det ^ x a 1 ^ y a 2 k ^ z a 3 = ^ ( a 3 y − a 2 z ) + ^ ( a 1 z − a 3 x ) + k ^ ( a 2 x − a 1 y )
we have
∇ × ( r × a ) = det [ ı ^ ȷ ^ k ^ ∂ ∂ x ∂ ∂ y ∂ ∂ z a 3 y − a 2 z a 1 z − a 3 x a 2 x − a 1 y ] = − 2 a 1 ı ^ − 2 a 2 ȷ ^ − 2 a 3 k ^ = − 2 a \begin{aligned}
{ \boldsymbol{\nabla} }\times(\mathbf{r}\times\mathbf{a})
&= \det\left[\begin{matrix}\hat{\boldsymbol{\imath}}&\hat{\boldsymbol{\jmath}}&\hat{\mathbf{k}}\\[0.03in]
\frac{\partial }{\partial x}&
\frac{\partial }{\partial y}&
\frac{\partial }{\partial z}\\[0.03in]
a_3y-a_2z& a_1z-a_3x & a_2x-a_1y\end{matrix}\right]
=-2a_1\,\hat{\boldsymbol{\imath}} -2a_2\,\hat{\boldsymbol{\jmath}} -2a_3\,\hat{\mathbf{k}} \\
&=-2\mathbf{a} \end{aligned} ∇ × ( r × a ) = det ^ ∂ x ∂ a 3 y − a 2 z ^ ∂ y ∂ a 1 z − a 3 x k ^ ∂ z ∂ a 2 x − a 1 y = − 2 a 1 ^ − 2 a 2 ^ − 2 a 3 k ^ = − 2 a
(d) Since
∇ ( r ) = ( ı ^ ∂ ∂ x + ȷ ^ ∂ ∂ y + k ^ ∂ ∂ z ) ( x 2 + y 2 + z 2 ) 1 / 2 = ı ^ x ( x 2 + y 2 + z 2 ) 1 / 2 + ȷ ^ y ( x 2 + y 2 + z 2 ) 1 / 2 + k ^ x ( x 2 + y 2 + z 2 ) 1 / 2 \begin{aligned}
{ \boldsymbol{\nabla} }(r) &= \left(\hat{\boldsymbol{\imath}}\frac{\partial }{\partial x}
+\hat{\boldsymbol{\jmath}}\frac{\partial }{\partial y}
+\hat{\mathbf{k}}\frac{\partial }{\partial z} \right) \big(x^2+y^2+z^2\big)^{1/2} \\
&=\hat{\boldsymbol{\imath}}\frac{x}{\big(x^2+y^2+z^2\big)^{1/2}}
+\hat{\boldsymbol{\jmath}}\frac{y}{\big(x^2+y^2+z^2\big)^{1/2}}
+\hat{\mathbf{k}}\frac{x}{\big(x^2+y^2+z^2\big)^{1/2}}
\end{aligned} ∇ ( r ) = ( ^ ∂ x ∂ + ^ ∂ y ∂ + k ^ ∂ z ∂ ) ( x 2 + y 2 + z 2 ) 1/2 = ^ ( x 2 + y 2 + z 2 ) 1/2 x + ^ ( x 2 + y 2 + z 2 ) 1/2 y + k ^ ( x 2 + y 2 + z 2 ) 1/2 x
we have
∇ ⋅ ( ∇ ( r ) ) = ∂ ∂ x x ( x 2 + y 2 + z 2 ) 1 / 2 + ∂ ∂ y y ( x 2 + y 2 + z 2 ) 1 / 2 + ∂ ∂ z z ( x 2 + y 2 + z 2 ) 1 / 2 = 3 ( x 2 + y 2 + z 2 ) 1 / 2 − 1 2 2 x 2 + 2 y 2 + 2 z 2 ( x 2 + y 2 + z 2 ) 3 / 2 = 2 ( x 2 + y 2 + z 2 ) 1 / 2 = 2 r \begin{aligned}
{ \boldsymbol{\nabla} }\cdot\big({ \boldsymbol{\nabla} }(r)\big)
&= \frac{\partial }{\partial x} \frac{x}{\big(x^2+y^2+z^2\big)^{1/2}}
+\frac{\partial }{\partial y} \frac{y}{\big(x^2+y^2+z^2\big)^{1/2}}
+\frac{\partial }{\partial z} \frac{z}{\big(x^2+y^2+z^2\big)^{1/2}} \\
&= \frac{3}{\big(x^2+y^2+z^2\big)^{1/2}}
-\frac{1}{2}\frac{2x^2+2y^2+2z^2}{\big(x^2+y^2+z^2\big)^{3/2}}
=\frac{2}{\big(x^2+y^2+z^2\big)^{1/2}}
=\frac{2}{r}
\end{aligned} ∇ ⋅ ( ∇ ( r ) ) = ∂ x ∂ ( x 2 + y 2 + z 2 ) 1/2 x + ∂ y ∂ ( x 2 + y 2 + z 2 ) 1/2 y + ∂ z ∂ ( x 2 + y 2 + z 2 ) 1/2 z = ( x 2 + y 2 + z 2 ) 1/2 3 − 2 1 ( x 2 + y 2 + z 2 ) 3/2 2 x 2 + 2 y 2 + 2 z 2 = ( x 2 + y 2 + z 2 ) 1/2 2 = r 2