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202105241031 Homework 1 (Q1)

Let \boldsymbol{\lambda}\lambda be the separation vector from a fixed point (x',y',z') to the point (x,y,z), and let \lambda be its length. Show that

(a) \nabla (\lambda^2)=2\boldsymbol{\lambda}

(b) \nabla (1/\lambda )=-\hat{\boldsymbol{\lambda}}/\lambda^2

(c) What is the general formula for \nabla (\lambda^n).

Solution.

(a)

\begin{aligned} \boldsymbol{\lambda}\lambda & = (x-x',y-y',z-z') \\ \lambda & = \textrm{length } = \sqrt{(x-x')^2+(y-y')^2+(z-z')^2} \\ \lambda^2 & = (x-x')^2 + (y-y')^2 + (z-z')^2 \\ \nabla (\lambda^2)& =\Big( 2(x-x'), 2(y-y'), 2(z-z')\Big) \\ & = 2\boldsymbol{\lambda} \\ \end{aligned}

(b)

\begin{aligned} & \quad \nabla \bigg( \frac{1}{\lambda} \bigg) \\ & = \nabla \bigg( \Big( (x-x')^2 + (y-y')^2 + (z-z')^2 \Big)^{-\frac{1}{2}} \bigg) \\ & = -\frac{1}{2} \Big( (x-x')^2 + (y-y')^2 + (z-z')^2 \Big)^{-\frac{3}{2}} \cdot \nabla (\lambda^2) \\ & \stackrel{\textrm{(a)}}{=} -\frac{1}{2}(\lambda^2)^{-\frac{3}{2}}\cdot 2\boldsymbol{\lambda} \\ & = -\lambda^{-3}\cdot \lambda\hat{\boldsymbol{\lambda}} \\ & = -\frac{\hat{\boldsymbol{\lambda}}}{\lambda^2} \\ \end{aligned}

(c)

\begin{aligned} \nabla (\lambda^n) & = \nabla (\lambda^2)^{\frac{n}{2}} \\ & = \frac{n}{2}(\lambda^2)^{\frac{n}{2}-1}\cdot\nabla (\lambda^2) \\ & \stackrel{\textrm{(a)}}{=} \frac{n}{2}(\lambda^2)^{\frac{n-2}{2}}\cdot 2\boldsymbol{\lambda} \\ & = \frac{n}{2}(\lambda^2)^{\frac{n-2}{2}}\cdot 2\lambda\hat{\boldsymbol{\lambda}} \\ & = n\lambda^{n-1}\hat{\boldsymbol{\lambda}} \\ \end{aligned}

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