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Chapter 1: Vector Analysis Problem 1.1: Using the definitions in Eqs. 1.1 and 1.4, and appropriate diagrams, show that the dot product and cross product are distributive, a) when the three vectors are co-planar. b) in the general case.  Solution   Problem 1.2 Is the cross product associative? $$(\vec{A}\times \vec{B}) \times \vec{C} \overset{?}{=} \vec{A}\times (\vec{B} \times \vec{C})$$ If so, prove it; if not, provide a counterexample (the simpler the better). Solution   Problem 1.3 Find the angle between the body diagonals of a cube.  Solution Problem 1.4 Use the cross product to find the components of the unit vector $\hat{n}$ perpendicular to the shaded plane in Fig. 1.11. Solution   Problem 1.5 Prove the BAC-CAB rule by writing out both sides in component form. Solution Problem 1.6 Prove that $$[\vec{A}\times (\vec{B}\times \vec{C})]+[\vec{B}\times (\vec{C}\times \vec{A})]+[\vec{C}\times (\vec{A}\times \vec{B})] = 0$$ Under what conditions does $\vec{A}\times (\vec{B}\times \v
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Chapter1 Vector Analysis: Problem 1.18

  Problem 1.18 Calculate the CURL of the following vector functions: (a) $v_a = x^2\hat{i}+3xz^2\hat{j}-2xz\hat{k}$ (b) $v_b = xy\hat{i}+2yz\hat{j}+3zx\hat{k}$ (c) $v_c = y^2\hat{i}+(2xy+z^2)\hat{j}+2yz\hat{k}$ Solution: Here's the link on "How to calculate curl in different coordinate system?" (a) $v_a = x^2\hat{i}+3xz^2\hat{j}-2xz\hat{k}$ $$\vec{\nabla}\times \vec{v_a} = \begin{bmatrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ x^2 & 3xz^2 & (-2xz) \end{bmatrix}$$ $\vec{\nabla}\times \vec{v_a} = \big[\frac{\partial }{\partial y}(-2xz) - \frac{\partial}{\partial z}(3xz^2)\big]\hat{i}- \big[\frac{\partial }{\partial x}(-2xz) - \frac{\partial}{\partial z}(x^2)\big]\hat{j}+ \big[\frac{\partial }{\partial x}(3xz^2) - \frac{\partial}{\partial y}(x^2)\big]\hat{k}$ $$\vec{\nabla}\times \vec{v_a} = 6xz\hat{i}+2z\hat{j}+ 3z^2\hat{k}$$ (b) $v_b = xy\hat{i}+2yz\hat{j}+3zx\hat{k}
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Calculate CURL in different Coordinate Systems.

  How to calculate CURL in various Coordinate Systems?   Curl of a vector function $\vec{F} = F_u\hat{u}+F_v\hat{v}+F_w\hat{w}$ in a coordinate system where $\vec{dl} = f du\hat{u}+gdv\hat{v}+hdw\hat{w}$   is given by: $$\vec{\nabla}\times \vec{F} = \frac{1}{fgh} \begin{bmatrix} f\hat{u} & g\hat{v} & h\hat{w} \\ \frac{\partial}{\partial u} & \frac{\partial}{\partial v} & \frac{\partial}{\partial w} \\ fF_u & gF_v & hF_w \end{bmatrix}$$ In Cartesian Coordinate System: $$\vec{dl} = dx\hat{i}+dy\hat{j}+dz\hat{k}$$ $$f=1, g=1, h=1$$ $$\vec{\nabla}\times \vec{F} = \begin{bmatrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ F_x & F_y & F_z \end{bmatrix}$$ In Cylindrical Coordinate System: $$\vec{dl} = ds\hat{s}+sd\phi\hat{\phi}+dz\hat{k}$$ $$f=1, g=s, h=1$$ $$\vec{\nabla}\times \vec{F} = \frac{1}{s} \begin{bmatrix} \hat{s} & s\hat{\phi} & \hat{z} \\ \f
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Chapter 1 Vector Analysis: Problem 1.12

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 Problem 1.12 The height of a certain hill (in feet) is given by $$h(x, y) = 10(2xy − 3x2 − 4y2 − 18x + 28y + 12),$$ where $y$ is the distance (in miles) north, $x$ the distance east of South Hadley. (a) Where is the top of the hill located? (b) How high is the hill? (c) How steep is the slope (in feet per mile) at a point 1 mile north and one mile east of South Hadley? In what direction is the slope steepest, at that point? Here's is the (c) part. https://introductiontoelectrodynamics.blogspot.com/2021/09/problem-1_3.html If you have any doubt regarding the solution or you want solution of some problem which is not posted please let me know by commenting. This encourages me to answer more question because sometime it feels like all I am doing is just a waste. If it helps someone I will be happy to do it.
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Chapter 1 Vector Analysis: Problem 1.8

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 Problem 1.8 (a) Prove that the two-dimensional rotation matrix ($\bf Eq. 1.29$) preserves dot products. (That is, show that $\overline{A_y B_y} + \overline{A_z B_z} = A_y B_y + A_z B_z$ .) (b) What constraints must the elements ($R_{i j}$ ) of the three-dimensional rotation matrix ($\bf Eq. 1.30$) satisfy, in order to preserve the length of $\bf{A}$ (for all vectors $\vec{A}$)?   If you have any doubt regarding the solution or you want solution of some problem which is not posted please let me know by commenting. This encourages me to answer more question because sometime it feels like all I am doing is just a waste. If it helps someone I will be happy to do it.
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Chapter 1 Vector Analysis: Problem 1.16

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Problem 1.16 Sketch the vector function $$\vec{v} = \frac{\hat{r}}{r^2}$$ and compute its divergence. The answer may surprise you.. can you explain it? Solution:  Let's first sketch the given vector function. NOTE: Use this site to sketch vector function and play with it: https://c3d.libretexts.org/CalcPlot3D/index.html   Since it seems too complicated, let's see it from the top and see the field on the $x-y$ axis Now as we can see there is a net outward flux at the origin since the field is generated from the origin. So divergence at origin should be some positive value. We can't say anything about other coordinates as the field is expanding out but at the same time the magnitude is decreasing. Now compute the divergence:   $$\vec{v} = \frac{\hat{r}}{r^2}$$ $$\vec{\nabla}\cdot \vec{v} = \frac{1}{r^2\sin\theta}\big[\frac{\partial}{\partial r}(r^2\sin\theta v_r)+\frac{\partial}{\partial \theta}(r\sin\theta v_{\theta})+\frac{\partial}{\partial \phi}(rv_{\phi})\big]$$ $$\vec{\
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What is Divergence? How to calculate Divergence? What is the physical significance of Divergence?

 What is Divergence? How it looks? Divergence is a vector operator that is it acts on a vector to output a scalar just like a normal dot/scalar product and in simple terms it tells how much the vector divergences i.e whether the vector flux increases or decreases at a point(infinitesimal volume). First del operator in Cartesian coordinate is given by: $$\vec{\nabla} = \hat{i}\frac{\partial }{\partial x}+\hat{j}\frac{\partial }{\partial y}+\hat{k}\frac{\partial }{\partial z}$$  Since, divergence resembles the dot product in the way that it takes a vector and outputs a scalar. Divergence acting on an vector resembles dot product. Consider two vectors $\vec{A} = A_x\hat{i}+A_y\hat{j}+A_z\hat{k}$ and $\vec{B}=B_x\hat{i}+B_y\hat{j}+B_z\hat{k}$ . Their dot product is given by: $$\vec{A}\cdot \vec{B} = A_xB_x+A_yB_y+A_zB_z$$  Consider A vector function $\vec{v}=v_x\hat{i}+v_y\hat{j}+v_z\hat{k}$. The divergence of vector $\vec{v}$ is given be: $$\vec{\nabla}\cdot \vec{v} = \frac{\partial }{\p
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Chapter 1 Vector Analysis: Problem 1.15

Problem 1.15 Calculate the divergence of the following vector functions: (a) $v_a = x^2\hat{i}+3xz^2\hat{j}-2xz\hat{k}$ (b) $v_b = xy\hat{i}+2yz\hat{j}+3zx\hat{k}$ (c) $v_c = y^2\hat{i}+(2xy+z^2)\hat{j}+2yz\hat{k}$ Solution: Divergence of a vector function $\vec{F} = F_u\hat{u}+F_v\hat{v}+F_w\hat{w}$ in a coordinate system where $\vec{dl} = f du\hat{u}+gdv\hat{v}+hdw\hat{w}$  is given by: $$\vec{\nabla}\cdot \vec{F} = \frac{1}{fgh}\big[\frac{\partial}{\partial u}(ghF_u)+\frac{\partial}{\partial v}(fhF_v)+\frac{\partial}{\partial w}(fgF_w)\big]$$ In Cartesian Coordinate System: $$\vec{dl} = dx\hat{i}+dy\hat{j}+dz\hat{k}$$ $$\vec{\nabla}\cdot \vec{F} = \frac{\partial}{\partial x}(F_x)+\frac{\partial}{\partial y}(F_y)+\frac{\partial}{\partial z}(F_z)$$ In Cylindrical Coordinate System: $$\vec{dl} = ds\hat{s}+sd\phi\hat{\phi}+dz\hat{k}$$ $$\vec{\nabla}\cdot \vec{F} = \frac{1}{s}\big[\frac{\partial}{\partial s}(sF_s)+\frac{\partial}{\partial \phi}(F_{\phi})+\frac{\partial}{\partial z}(sF_
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Chapter 1 Vector Analysis: Problem 1.13

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        If you have any doubt regarding the solution or you want solution of some problem which is not posted please let me know by commenting. This encourages me to answer more question because sometime it feels like all I am doing is just a waste. If it helps someone I will be happy to do it.
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Chapter 1 Vector Analysis: Problem 1.12

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Problem 1.12 The height of a certain hill (in feet) is given by $$h(x, y) = 10(2xy − 3x^2 − 4y^2 − 18x + 28y + 12),$$ where $y$ is the distance (in miles) north, $x$ the distance east of South Hadley. (a) Where is the top of the hill located? (b) How high is the hill? (c) How steep is the slope (in feet per mile) at a point 1 mile north and one mile east of South Hadley? In what direction is the slope steepest, at that point? Solution: Before proceeding to the answer, I strongly encourage you to watch this amazing video to get better insight of gradient:  https://youtu.be/QQPz3eXXgQI OR if you are something who wants to learn by reading here's a great article  https://betterexplained.com/articles/vector-calculus-understanding-the-gradient/ . I hope you must have watched or/and read the article before proceeding. $$h(x,y) = 10(2xy-3x^2-4y^2-18x+28y+12)$$ First let's compute gradient of the function: $$\vec{\nabla} h = \frac{\partial h}{\partial x}\hat{i}+\frac{\partial h}{\par
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Chapter 1 Vector Analysis: Problem 1.11

Problem 1.11 Find the gradients of the following functions: (a) $f(x,y,z) = x^2+y^3+z^4$ (b) $f(x,y,z) = x^2y^3z^4$ (c) $f(x,y,z) = e^x\sin(y)\ln(z)$ Solution: Gradient of a function $F$ in a coordinate system where $\vec{dl} = f du\hat{u}+gdv\hat{v}+hdw\hat{w}$  is given by: $$\vec{\nabla}F = \frac{1}{f} \frac{\partial F}{\partial u}\hat{u}+\frac{1}{g}\frac{\partial F}{\partial v}\hat{v}+\frac{1}{h}\frac{\partial F}{\partial w}\hat{w}$$  In Cartesian Coordinate System: $$\vec{dl} = dx\hat{i}+dy\hat{j}+dz\hat{k}$$ $$\vec{\nabla} F = \frac{\partial F}{\partial x}\hat{i}+\frac{\partial F}{\partial y}\hat{j}+\frac{\partial F}{\partial z}\hat{k}$$ In Cylindrical Coordinate System: $$\vec{dl} = ds\hat{s}+sd\phi\hat{\phi}+dz\hat{k}$$$$\vec{\nabla} F = \frac{\partial F}{\partial s}\hat{s}+\frac{1}{s}\frac{\partial F}{\partial \phi}\hat{\phi}+\frac{\partial F}{\partial z}\hat{z}$$ In Spherical Coordinate System: $$\vec{dl} = dr\hat{r}+rd\theta\hat{\theta}+r\sin\theta d\phi\hat{\phi}$$$$\vec{\n
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Chapter 1 Vector Analysis: Problem 1.4

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Problem 1.4 Use the cross product to find the components of the unit vector $\hat{n}$ perpendicular to the shaded plane in Fig. 1.11.   Solution: We know that cross product of two vectors gives a vector which is perpendicular to both the initial vectors. Hence to find the vector that is perpendicular to a plane we need two non collinear vectors in the plane and take cross product of them.   Consider the two vectors as shown in the diagram,     $$\vec{r_1} = (0\hat{i}+2\hat{j}+0\hat{k})-(1\hat{i}+0\hat{j}+0\hat{k})$$ $$\vec{r_1} = -1\hat{i}+2\hat{j}+0\hat{k}$$   $$\vec{r_2} = (0\hat{i}+0\hat{j}+3\hat{k})-(1\hat{i}+0\hat{j}+0\hat{k})$$ $$\vec{r_2} = -1\hat{i}+0\hat{j}+3\hat{k}$$   $$\therefore \hat{n} = \frac{\vec{r_1}\times \vec{r_2}}{|\vec{r_1}\times \vec{r_2}|}$$  $$\vec{r_1}\times \vec{r_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ -1 & 2 & 0\\ -1 & 0 & 3\\ \end{vmatrix}$$ $$\vec{r_1}\times \vec{r_2} = 6\hat{i}+3\hat{j}+2\hat{k}$$ $$|\vec{r_1}\times \vec
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Chapter 1 Vector Analysis: Problem 1.3

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Problem 1.3 Find the angle between the body diagonals of a cube.     If you have any doubt regarding the solution or you want solution of some problem which is not posted please let me know by commenting. This encourages me to answer more question because sometime it feels like all I am doing is just a waste. If it helps someone I will be happy to do it.
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Chapter 1 Vector Analysis: Problem 1.5

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Problem 1.5 Prove the BAC-CAB rule by writing out both sides in component form.   If you have any doubt regarding the solution or you want solution of some problem which is not posted please let me know by commenting. This encourages me to answer more question because sometime it feels like all I am doing is just a waste. If it helps someone I will be happy to do it.
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Chapter 1 Vector Analysis: Problem 1.7

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      If you have any doubt regarding the solution or you want solution of some problem which is not posted please let me know by commenting. This encourages me to answer more question because sometime it feels like all I am doing is just a waste. If it helps someone I will be happy to do it.
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Chapter 1 Vector Analysis: Problem 1.6

Problem 1.6 Prove that $$[\vec{A}\times (\vec{B}\times \vec{C})]+[\vec{B}\times (\vec{C}\times \vec{A})]+[\vec{C}\times (\vec{A}\times \vec{B})] = 0$$ Under what conditions does $\vec{A}\times (\vec{B}\times \vec{C}) = (\vec{A}\times \vec{B})\times \vec{C}$? Solution: Note That: $$\vec{A}\times (\vec{B}\times \vec{C}) = (\vec{A}\cdot \vec{C})\vec{B} - (\vec{A}\cdot \vec{B})\vec{C}$$ Consider, $$[\vec{A}\times (\vec{B}\times \vec{C})]+[\vec{B}\times (\vec{C}\times \vec{A})]+[\vec{C}\times (\vec{A}\times \vec{B})]$$ $$ = [(\vec{A}\cdot \vec{C})\vec{B}-(\vec{A}\cdot \vec{B})\vec{C}] + [(\vec{B}\cdot \vec{A})\vec{C}-(\vec{B}\cdot \vec{C})\vec{A}]+ [(\vec{C}\cdot \vec{B})\vec{A}-(\vec{C}\cdot \vec{A})\vec{B}]$$ Using the commutative property of dot product i.e $\vec{A}\cdot \vec{B} = \vec{B}\cdot \vec{A}$   $$[\vec{A}\times (\vec{B}\times \vec{C})]+[\vec{B}\times (\vec{C}\times \vec{A})]+[\vec{C}\times (\vec{A}\times \vec{B})] = \vec{0}$$   Say $\vec{A}\times (\vec{B}\times \vec{C}) = (\vec
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Chapter 1 Vector Analysis: Problem 1.2

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Problem 1.2 Is the cross product associative? $$(\vec{A}\times \vec{B}) \times \vec{C} \overset{?}{=} \vec{A}\times (\vec{B} \times \vec{C})$$ If so, prove it; if not, provide a counterexample (the simpler the better). Solution: No, cross product is not associative. Counter Example Consider Three Non-Zero Vectors $\vec{A}$, $\vec{B}$ and $\vec{C}$. Just to keep the example simple, let $\vec{C}$ be perpendicular to the plane containing $\vec{A}$ and $\vec{B}$ $$\vec{A}\times \vec{B} = |A||B|\sin\theta \hat{n}$$ where $\theta$ is the angle between them and $\theta<90^{\circ}$ and $\hat{n}$ is the direction perpendicular to the plane containing $\vec{A}$ and $\vec{B}$ i.e the direction of $\vec{C}$. Hence, the angle between the cross product of $\vec{A}$ and $\vec{B}$ and that of $\vec{C}$ is $0$. $$(\vec{A}\times \vec{B})\times \vec{C} = |A||B|\sin\theta\hat{n} \times \vec{C} = \vec{0}$$ Now Consider the plane containing $\vec{B}$ and $\vec{C}$. $$\vec{A}\times (\vec{B} \times \vec{C}
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Chapter 1 Vector Analysis: Problem 1.1

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Problem 1.1: Using the definitions in Eqs. 1.1 and 1.4, and appropriate diagrams, show that the dot product and cross product are distributive, a) when the three vectors are co-planar. b) in the general case. Solution:  Eq$^n$ 1.1 $$\vec{A}\cdot \vec{B} = |A||B| \cos\theta$$where $\theta$ is the angle between the vectors $\vec{A}$ and $\vec{B}$ Eq$^n$ 1.4 $$\vec{A}\times \vec{B} = |A||B| \sin\theta \hat{n}$$where $\theta$ is the angle between the vectors $\vec{A}$ and $\vec{B}$, and $\hat{n}$ is a unit vector normal to both $\vec{A}$ and $\vec{B}$. We have to show that: $$\vec{A}\cdot (\vec{B}+\vec{C}) = \vec{A}\cdot \vec{B}+\vec{A}\cdot \vec{C}$$ $$\vec{A}\times (\vec{B}+\vec{C}) = \vec{A}\times \vec{B}+\vec{A}\times \vec{C}$$ $a) $ When $\vec{A}$, $\vec{B}$ and $\vec{C}$ are co-planar.  $\theta_1$ and $\theta_2$ are the angles the vector $\vec{B}$ and $\vec{C}$ makes with the $\vec{A}$ respectively. To get $\vec{B}+\vec{C}$ put the tail of $\vec{C}$ on the head of $\vec{B}$.   Let $
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