ベクトル解析 gradの問題
問題
① \(\nabla r\) ② \(\nabla \displaystyle\frac{1}{r}\) ③ \(\nabla r^n\)
※\(r=\sqrt{x^2+y^2+z^2}\)
解答
1番
定義から考えていく。
\(\nabla r=\biggl(\displaystyle\frac{\partial r}{\partial x} , \displaystyle\frac{\partial r}{\partial y} , \displaystyle\frac{\partial r}{\partial z}\biggr)\)
\(\displaystyle\frac{\partial r}{\partial x}\)\(=\displaystyle\frac{\partial }{\partial x}(x^2+y^2+z^2)^{\frac{1}{2}}=\displaystyle\frac{1}{2}(x^2+y^2+z^2)^{-\frac{1}{2}}\cdot 2x=\displaystyle\frac{x}{\sqrt{x^2+y^2+z^2}}=\)\(\displaystyle\frac{x}{r}\)
同様に、 \(\displaystyle\frac{\partial r}{\partial y}=\displaystyle\frac{y}{r}\)および\(\displaystyle\frac{\partial r}{\partial z}=\displaystyle\frac{z}{r}\) なので
\(\nabla r=\biggl(\displaystyle\frac{\partial r}{\partial x} , \displaystyle\frac{\partial r}{\partial y} , \displaystyle\frac{\partial r}{\partial z}\biggr)=\biggl(\displaystyle\frac{x}{r} , \displaystyle\frac{y}{r} , \displaystyle\frac{z}{r}\biggr)=\displaystyle\frac{\boldsymbol{r}}{r}\)
※ \(\boldsymbol{r}=(x , y , z)\) である。
2番
\(\nabla \displaystyle\frac{1}{r}=\biggl[\displaystyle\frac{\partial }{\partial x}\biggl(\displaystyle\frac{1}{r}\biggr) , \displaystyle\frac{\partial }{\partial y}\biggl(\displaystyle\frac{1}{r}\biggr) , \displaystyle\frac{\partial }{\partial z}\biggl(\displaystyle\frac{1}{r}\biggr)\biggr]\)
\(\displaystyle\frac{\partial }{\partial x}\biggl(\displaystyle\frac{1}{r}\biggr)\)\(=\displaystyle\frac{\partial }{\partial x}(x^2+y^2+z^2)^{-\frac{1}{2}}=-\displaystyle\frac{1}{2}(x^2+y^2+z^2)^{-\frac{3}{2}}\cdot 2x\)
\(=-\displaystyle\frac{x}{(x^2+y^2+z^2)^{\frac{3}{2}}}=\)\(-\displaystyle\frac{x}{r^3}\)
同様に、 \(\displaystyle\frac{\partial }{\partial y}\biggl(\displaystyle\frac{1}{r}\biggr)=-\displaystyle\frac{y}{r^3}\)および\(\displaystyle\frac{\partial }{\partial z}\biggl(\displaystyle\frac{1}{r}\biggr)=-\displaystyle\frac{z}{r^3}\) なので
\(\nabla \displaystyle\frac{1}{r}=\biggl[\displaystyle\frac{\partial }{\partial x}\biggl(\displaystyle\frac{1}{r}\biggr) , \displaystyle\frac{\partial }{\partial y}\biggl(\displaystyle\frac{1}{r}\biggr) , \displaystyle\frac{\partial }{\partial z}\biggl(\displaystyle\frac{1}{r}\biggr)\biggr]=\biggl(-\displaystyle\frac{x}{r^3} , -\displaystyle\frac{y}{r^3} , -\displaystyle\frac{z}{r^3}\biggr)=-\displaystyle\frac{\boldsymbol{r}}{r^3}\)
3番
\(\nabla r^n=\biggl(\displaystyle\frac{\partial (r^n)}{\partial x} , \displaystyle\frac{\partial (r^n)}{\partial y} , \displaystyle\frac{\partial (r^n)}{\partial z}\biggr)\)
\(\displaystyle\frac{\partial (r^n)}{\partial x}\)\(=\displaystyle\frac{\partial }{\partial x}(x^2+y^2+z^2)^{\frac{n}{2}}=\displaystyle\frac{n}{2}(x^2+y^2+z^2)^{\frac{n}{2}-1}\cdot 2x=\)\(nxr^{n-2}\)
同様に、 \(\displaystyle\frac{\partial (r^n)}{\partial y}=ny r^{n-2}\)および\(\displaystyle\frac{\partial (r^n)}{\partial z}=nzr^{n-2}\) なので
\(\nabla r^n=\biggl(\displaystyle\frac{\partial r^n}{\partial x} , \displaystyle\frac{\partial r^n}{\partial y} , \displaystyle\frac{\partial r^n}{\partial z}\biggr)=(nxr^{n-2} , nyr^{n-2} ,nzr^{n-2})=nr^{n-2}\boldsymbol{r}\)