Differential Operators Exercise5(a)

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Differential Operators Exercise5(a)


1. Prove that

  1. \(\nabla\left(\frac{1}{r}\right)\)\(=-\frac{\bar{r}}{r^3}\)
  2. grad r= \(\frac{\mathbf{r}}{|\mathbf{r}|}\)r2=x2+y2+z

Solution:

⇒ \(|\mathbf{r}|=r=\sqrt{x^2+y^2+z^2} \Rightarrow r^2=x^2+y^2+z^2\)

∴ 2 r \(\frac{d \mathbf{r}}{d x}=2 x \Rightarrow \frac{d \mathbf{r}}{d t}=\frac{x}{r}\)

Similarly \(\frac{\partial r}{\partial y}=\frac{y}{r}, \frac{\partial r}{\partial z}=\frac{z}{r}\)

⇒ \(\nabla\left(\frac{1}{r}\right)=\sum \mathbf{i} \frac{\partial}{\partial x}\left(\frac{1}{r}\right)=-\frac{1}{r^2} \sum \mathbf{i} \frac{d \mathbf{r}}{d x}=-\frac{1}{r^2} \sum \frac{x \mathbf{i}}{r}=-\frac{(x \mathbf{i}+y \mathbf{j}+z \mathbf{k})}{r^3}=-\frac{\mathbf{r}}{r^3}\)

2.  If=\(\left(y \frac{\partial f}{\partial z}-z \frac{\partial f}{\partial y}\right) \mathbf{i}+\left(z \frac{\partial f}{\partial x}-x \frac{\partial f}{\partial z}\right) \mathbf{j}+\left(x \frac{\partial f}{\partial y}-y \frac{\partial f}{\partial x}\right) \mathbf{k}\)

prove that

  1. F.r=0 
  2. F. grad f=0

Solution:

1. \(\mathbf{F} . \mathbf{r}=\left[\left(y \frac{\partial f}{\partial z}-z \frac{\partial f}{\partial y}\right) \mathbf{i}+\left(z \frac{\partial f}{\partial x}-x \frac{\partial f}{\partial z}\right) \mathbf{j}+\left(x \frac{\partial f}{\partial y}-y \frac{\partial f}{\partial x}\right) \mathbf{k}\right]\)

= \(x\left(y \frac{\partial f}{\partial z}-z \frac{\partial f}{\partial y}\right)+y\left(z \frac{\partial f}{\partial x}-x \frac{\partial f}{\partial z}\right)+z\left(x \frac{\partial f}{\partial y}-y \frac{\partial f}{\partial x}\right)\)

∴ \(\mathbf{F} \cdot \mathbf{r}=0\)

2. \(\mathbf{F}=\sum \mathbf{i}\left(y \frac{\partial f}{\partial z}-z \frac{\partial f}{\partial y}\right) \text{grad} f=\nabla f=\sum \mathbf{i} \frac{\partial f}{\partial x}\)

⇒ \(\mathbf{F} \cdot \text{grad} f=\sum \frac{\partial f}{\partial x}\left(y \frac{\partial f}{\partial z}-z \frac{\partial f}{\partial y}\right)=0\)

Examples Of Differential Operator Solutions For Exercise 5(A)

3. Find grad f at the point (1,1,-2) where

  1. f= x3+y3+3xyz
  2. f= x2y+y2x+z2

Solution:

1. \(\text{grad} f=\nabla f=\mathbf{i} \frac{\partial f}{\partial x}+\mathbf{j} \frac{\partial f}{\partial y}+\mathbf{k} \frac{\partial f}{\partial z}\)

= \(\mathbf{i}\left(3 x^2+3 y z\right)+\mathbf{j}\left(3 y^2+3 z x\right)+\mathbf{k}(3 x y)\)

At (1,1,-2), \(\text{grad} f=-3 \bar{i}-3 \bar{j}+3 \bar{k}\)

2. grad f = \(\nabla f=\mathbf{i} \frac{\partial f}{\partial x}+\mathbf{j} \frac{\partial f}{\partial y}+\mathbf{k} \frac{\partial f}{\partial z}\)

= \(\mathbf{i}\left(2 x y+y^2\right)+\mathbf{j}\left(x^2+2 x y\right)+\mathbf{k}(2 z)\)

At (1,1,-2), \(\text{grad} f=3 \bar{i}+3 \bar{j}-4 \bar{k}\)

4. Find the directional derivative of

  1. Φ = xy+yz+zx at A in the direction of \(\overrightarrow{\mathrm{AB}}\), where A= (1,2,-1) ; B=(-1,2,3)
  2. Φ=xyz at (1,1,1) in the direction of the vector i+j+k

Solution:

⇒ \(\overrightarrow{\mathbf{A B}}=\mathbf{i}(-1-1)+\mathbf{j}(2-2)+\mathbf{k}(3+1)=-2 \mathbf{i}+4 \mathbf{k}\)

Unit vector along \(\mathbf{A B}, \mathbf{e}=\frac{-2}{2 \sqrt{5}}(\mathbf{i}-2 \mathbf{k})\)

∴ Directional derivative along \(\mathbf{A B}=\mathbf{e}. \nabla \phi\)

= \(-\frac{1}{\sqrt{5}}(\mathbf{i}-2 \mathbf{k}) \cdot[\mathbf{i}(y+z)+\mathbf{j}(z+x)+\mathbf{k}(x+y)]=\sqrt{5}\) at \(\mathbf{A}(1,2,-1)\).

5. Find the angle between the surfaces x2+y2+z2=9 and x2+y2-z=3 at (2,-1,2) [Hint: The angle between the surfaces is the angle between the normals]

Solution:

Given

x2+y2+z2=9 and x2+y2-z=3 at (2,-1,2)

Let f = \(x^2+y^2+z^2-9 ; \quad \phi=x^2+y^2-z-3\) then \(\nabla f=2 x \mathbf{i}+2 \boldsymbol{j}+2 z \mathbf{k} ; \quad \nabla \phi=2 x \mathbf{i}+2 \boldsymbol{y} \mathbf{j}-\mathbf{k}\)

At (2,-1,2), \(\nabla f=4 \mathbf{i}-2 \mathbf{j}+4 \mathbf{k}\)

⇒\(\nabla \phi=4 \mathbf{i}-2 \mathbf{j}-\mathbf{k}\)

Angle between the surfaces = Angle between the normals

= \(\text{Cos}^{-1} \frac{\nabla f . \nabla \phi}{|\nabla f||\nabla \phi|}=\text{Cos}^{-1} \frac{8}{3 \sqrt{21}}\)

Exercise 5(A) Calculus Differential Operator Explanation

6. Find the angle between the surfaces xy2z=3x+z2 and 3x2-y2+2z=1 at(1,-2,1)

Solution:

Given

y2z=3x+z2 and 3x2-y2+2z=1 at(1,-2,1)

f = \(x y^2 z-3 x-z^2 \text { and } \phi=3 x^2-y^2+2 z-1\)

⇒ \(\nabla f=\mathbf{i}\left[\left(y^2 z\right)-3\right]+\mathbf{j}(2 x y z)+\mathbf{k}\left(x y^2-2 z\right)\)

and \(\nabla \phi=\mathbf{i}(6 x)-\mathbf{j}(2 y)+2 \mathbf{k}\)

at (1,-2,1), \(\nabla f=\mathbf{i}-4 \mathbf{j}+2 \mathbf{k} \text { and } \nabla \phi=6 \mathbf{i}+4 \mathbf{j}+2 \mathbf{k}\)

∴ Angle between the surfaces = \(\cos ^{-1} \frac{\nabla f \cdot \nabla \phi}{|\nabla f||\nabla \phi|}\)

= \(\cos ^{-1}\left(\frac{6-16+4}{\sqrt{21} \sqrt{56}}\right)=\cos ^{-1}\left(\frac{-3}{7 \sqrt{6}}\right)\)

 

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