Mathematics II (BT-202) Unit 5 Notes - RGPV

How unit 5 is examined

Vector differentiation, the operators grad, div and curl, line integrals, and the three integral theorems; marks sit in Gauss/Green/Stokes verification (14 marks) and divergence-curl proofs (7 marks).

Differentiation of Vectors

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Definition. If $\vec r(t)$ is a vector function of a scalar $t$, then $\dfrac{d\vec r}{dt}=\lim_{\delta t\to0}\dfrac{\vec r(t+\delta t)-\vec r(t)}{\delta t}$.

Key points.

Differentiate component-wise: if $\vec r=x(t)\hat i+y(t)\hat j+z(t)\hat k$ then $\dfrac{d\vec r}{dt}=x'\hat i+y'\hat j+z'\hat k$.
$\dfrac{d\vec r}{dt}$ is tangent to the curve at that point, so it gives the direction of motion (velocity if $t$ is time).
$\dfrac{d}{dt}(\vec a\cdot\vec b)=\vec a\cdot\vec b'+\vec a'\cdot\vec b$ and $\dfrac{d}{dt}(\vec a\times\vec b)=\vec a\times\vec b'+\vec a'\times\vec b$ (keep the order in the cross product).
$\dfrac{d}{dt}(\phi\vec a)=\phi\vec a'+\phi'\vec a$, and a vector of constant magnitude satisfies $\vec a\cdot\vec a'=0$.

Scalar and vector point function

Definition. A scalar point function $\phi(x,y,z)$ assigns a number to each point of a region (temperature, density); a vector point function $\vec F(x,y,z)$ assigns a vector (velocity, force).

Key points.

A scalar field has only magnitude at each point, so it is written as one function $\phi(x,y,z)$.
A vector field has magnitude and direction, written $\vec F=F_1\hat i+F_2\hat j+F_3\hat k$ with $F_1,F_2,F_3$ scalar functions.
Level surfaces $\phi=\text{constant}$ describe a scalar field; grad turns a scalar field into a vector field.
The position vector $\vec r=x\hat i+y\hat j+z\hat k$ with $r=|\vec r|$ is the standard vector field used in proofs.

Gradient

Definition. For a scalar function $\phi$, $\nabla\phi=\operatorname{grad}\phi=\dfrac{\partial\phi}{\partial x}\hat i+\dfrac{\partial\phi}{\partial y}\hat j+\dfrac{\partial\phi}{\partial z}\hat k$, where $\nabla=\hat i\dfrac{\partial}{\partial x}+\hat j\dfrac{\partial}{\partial y}+\hat k\dfrac{\partial}{\partial z}$.

Key points.

The gradient of a scalar field is a vector field.
$\nabla(\phi\psi)=\phi\nabla\psi+\psi\nabla\phi$, and $\nabla f(r)=f'(r)\dfrac{\vec r}{r}$, so $\nabla r^n=nr^{n-2}\vec r$.
The gradient of a constant is zero.
If $\vec F=\nabla\phi$, then $\phi$ is called the scalar potential of $\vec F$.

Geometrical meaning of gradient

Definition. $\nabla\phi$ at a point is normal to the level surface $\phi=c$ through that point, and points in the direction of maximum increase of $\phi$.

Key points.

The unit normal to the surface $\phi=c$ is $\hat n=\dfrac{\nabla\phi}{|\nabla\phi|}$.
$|\nabla\phi|$ equals the greatest rate of change of $\phi$ at the point.
The angle between two surfaces is the angle between their normals: $\cos\theta=\dfrac{\nabla\phi_1\cdot\nabla\phi_2}{|\nabla\phi_1||\nabla\phi_2|}$.
Write each surface as $\phi=0$ before taking the gradient.

Example. $\phi_1=x^2+y^2+z^2-9$, $\phi_2=x^2+y^2-z-3$ at $(2,-1,2)$: $\nabla\phi_1=4\hat i-2\hat j+4\hat k$ (magnitude 6), $\nabla\phi_2=4\hat i-2\hat j-\hat k$ (magnitude $\sqrt{21}$), dot product $16+4-4=16$.

$\cos\theta=\dfrac{16}{6\sqrt{21}}=\dfrac{8}{3\sqrt{21}}$, so $\theta=\cos^{-1}\dfrac{8}{3\sqrt{21}}\approx54.4^\circ$.

Asked: [7 marks] (Jun 2022) Find the angle between the surfaces $x^2+y^2+z^2=9$ and $z=x^2+y^2-3$ at the point $(2,-1,2)$.

Directional Derivative

Definition. The directional derivative of $\phi$ at a point in the direction of a unit vector $\hat u$ is the rate of change of $\phi$ along $\hat u$: ==$D_{\hat u}\phi=\nabla\phi\cdot\hat u$==.

Key points.

The direction given must first be converted to a unit vector: $\hat u=\vec a/|\vec a|$.
$\nabla\phi$ is evaluated at the given point before the dot product is taken.
For a curve $\vec r(t)$, the direction is the tangent $d\vec r/dt$ at the given $t$.
The maximum value of $D_{\hat u}\phi$ is $|\nabla\phi|$, attained along $\nabla\phi$ itself.
The value is positive if $\phi$ increases along $\hat u$ and negative if it decreases.

Example 1. $f=e^{2x}\cos yz$ at $(0,0,0)$: $\nabla f=(2e^{2x}\cos yz,\,-ze^{2x}\sin yz,\,-ye^{2x}\sin yz)=2\hat i$. Tangent at $t=\pi/4$: $a\cos t\,\hat i-a\sin t\,\hat j+a\hat k=a\left(\tfrac1{\sqrt2},-\tfrac1{\sqrt2},1\right)$ with magnitude $a\sqrt2$, so $\hat u=\left(\tfrac12,-\tfrac12,\tfrac1{\sqrt2}\right)$.

$D_{\hat u}f=2\cdot\tfrac12=1$.

Example 2. $\phi=x^2yz+4xz^2$ at $(1,-2,-1)$ along $2\hat i-\hat j-2\hat k$: $\nabla\phi=(2xyz+4z^2)\hat i+x^2z\hat j+(x^2y+8xz)\hat k=8\hat i-\hat j-10\hat k$; $\hat u=\tfrac13(2,-1,-2)$.

$D_{\hat u}\phi=\dfrac{16+1+20}{3}=\dfrac{37}{3}$.

Answer frame. Open with "Directional derivative $=\nabla\phi\cdot\hat u$"; find $\nabla\phi$, substitute the point, make the direction a unit vector; dot them and box the value.

Pitfall: Forgetting to divide the direction vector by its magnitude.

Asked: [7 marks] (Jun 2023, Dec 2024) Find the directional derivative of $f=e^{2x}\cos yz$ at $(0,0,0)$ in the direction of the tangent to the curve $x=a\sin t,\ y=a\cos t,\ z=at$ at $t=\pi/4$; also of $\phi=x^2yz+4xz^2$ at $(1,-2,-1)$ in the direction of $2\bar i-\bar j-2\bar k$.

Divergence and Curl

Definition. For $\vec F=F_1\hat i+F_2\hat j+F_3\hat k$: ==$\operatorname{div}\vec F=\nabla\cdot\vec F=\dfrac{\partial F_1}{\partial x}+\dfrac{\partial F_2}{\partial y}+\dfrac{\partial F_3}{\partial z}$== (a scalar), and $\operatorname{curl}\vec F=\nabla\times\vec F=\begin{vmatrix}\hat i&\hat j&\hat k\\ \partial_x&\partial_y&\partial_z\\ F_1&F_2&F_3\end{vmatrix}$ (a vector).

Key points.

$\vec F$ is solenoidal if $\nabla\cdot\vec F=0$, meaning no net source and no net outflow at any point.
$\vec F$ is irrotational if $\nabla\times\vec F=\vec0$, and then $\vec F=\nabla\phi$ for a scalar potential $\phi$.
Curl measures rotation of the field (twice the angular velocity) and divergence measures outflow per unit volume.
$\nabla\cdot(\phi\vec A)=\nabla\phi\cdot\vec A+\phi\,\nabla\cdot\vec A$ and $\nabla\times(\phi\vec A)=\nabla\phi\times\vec A+\phi\,\nabla\times\vec A$.
For $\vec r$: $\nabla\cdot\vec r=3$, $\nabla\times\vec r=\vec0$, and $\nabla r^n=nr^{n-2}\vec r$.
$\nabla^2\phi=\nabla\cdot\nabla\phi=\phi_{xx}+\phi_{yy}+\phi_{zz}$ is the Laplacian, and curl of a gradient is always zero.

Proof A: $\nabla\cdot(r^n\vec r)=(n+3)r^n$. Using key point 4: $\nabla\cdot(r^n\vec r)=r^n(\nabla\cdot\vec r)+\vec r\cdot\nabla r^n=3r^n+\vec r\cdot nr^{n-2}\vec r=3r^n+nr^n=(n+3)r^n$. For $n=-3$ this is 0, so $r^{-3}\vec r=\vec r/r^3$ is solenoidal.

Proof B: $\nabla\times(r^n\vec r)=\vec0$. $\nabla\times(r^n\vec r)=\nabla r^n\times\vec r+r^n(\nabla\times\vec r)=nr^{n-2}(\vec r\times\vec r)+\vec0=\vec0$, since $\vec r\times\vec r=\vec0$ and $\nabla\times\vec r=\vec0$. Hence $r^n\vec r$ is irrotational.

Proof C: $\nabla^2f(r)=f''+\tfrac2rf'$. From $r^2=x^2+y^2+z^2$, $\partial r/\partial x=x/r$. Then $\dfrac{\partial f}{\partial x}=f'\dfrac xr$ and $\dfrac{\partial^2f}{\partial x^2}=f''\dfrac{x^2}{r^2}+f'\left(\dfrac1r-\dfrac{x^2}{r^3}\right)$. Adding the same for $y,z$ and using $x^2+y^2+z^2=r^2$: $\nabla^2f=f''+f'\left(\dfrac3r-\dfrac1r\right)$. So $\nabla^2f(r)=f''(r)+\dfrac2rf'(r)$.

Example (irrotational, potential). $\vec F=(x^2-yz,\,y^2-zx,\,z^2-xy)$. Curl: $\hat i(-x+x)-\hat j(-y+y)+\hat k(-z+z)=\vec0$. Put $\phi_x=x^2-yz$: $\phi=\tfrac{x^3}3-xyz+g(y,z)$; then $\phi_y=-xz+g_y=y^2-zx$ gives $g=\tfrac{y^3}3+h(z)$; $\phi_z=-xy+h'=z^2-xy$ gives $h=\tfrac{z^3}3$.

$\phi=\dfrac{x^3+y^3+z^3}{3}-xyz+C$.

Answer frame. For proofs: open with the definition ($\nabla\cdot\vec F=0$ or $\nabla\times\vec F=\vec0$); state the product identity, substitute $\nabla\cdot\vec r=3$ or $\nabla r^n=nr^{n-2}\vec r$, simplify, close with the conclusion. For the potential: show the curl determinant is zero, integrate the three partials in turn, close with $\phi$.

Pitfall: Writing $\nabla r^n=nr^{n-1}$ without the vector factor; it is $nr^{n-2}\vec r$.

Asked: [7 marks] (Jun 2022, Nov 2022) Prove that $r^n\bar r$ is solenoidal if $n=-3$; show that $\vec r/r^3$ is solenoidal. Asked: [7 marks] (Dec 2023, Jun 2025) Prove that $\operatorname{curl}(r^n\bar r)=\bar0$; prove that $r^n\bar r$ is irrotational. Asked: [7 marks] (Nov 2022) Show that $(x^2-yz)\hat i+(y^2-zx)\hat j+(z^2-xy)\hat k$ is irrotational; find its scalar potential. Asked: [7 marks] (Jun 2023) Prove that $\nabla^2f(r)=f''(r)+\frac2rf'(r)$.

Line, Surface and Volume Integral

Definition. The line integral of $\vec F$ along a curve $C$ is $\int_C\vec F\cdot d\vec r$; it equals the work done by force $\vec F$ in moving a particle along $C$.

Key points.

For a parametric curve, $\int_C\vec F\cdot d\vec r=\int_{t_1}^{t_2}\vec F\cdot\dfrac{d\vec r}{dt}\,dt$, after writing $\vec F$ in terms of $t$.
The surface integral $\iint_S\vec F\cdot\hat n\,dS$ is the flux of $\vec F$ through $S$.
The volume integral is $\iiint_V\phi\,dV$ (or of a vector, component-wise).
A line integral reverses sign when the direction of travel is reversed, and over a closed curve it is the circulation.

Example. $\vec F=z\hat i+x\hat j+y\hat k$, $\vec r=\cos t\,\hat i+\sin t\,\hat j-t\hat k$, $0\le t\le2\pi$. Then $\vec F=-t\hat i+\cos t\,\hat j+\sin t\,\hat k$ and $d\vec r/dt=-\sin t\,\hat i+\cos t\,\hat j-\hat k$, so $\vec F\cdot d\vec r/dt=t\sin t+\cos^2t-\sin t$. Integrating: $[-t\cos t+\sin t]_0^{2\pi}=-2\pi$, $\int\cos^2t=\pi$, $\int\sin t=0$.

Work done $=-2\pi+\pi=-\pi$.

Asked: [7 marks] (Jun 2025) Find the work done by $\bar F=z\bar i+x\bar j+y\bar k$ moving a particle along $\bar r=\cos t\,\bar i+\sin t\,\bar j-t\bar k$ from $t=0$ to $t=2\pi$.

Gauss Divergence, Stokes and Green theorems

Definition. ==Gauss: $\iint_S\vec F\cdot\hat n\,dS=\iiint_V\nabla\cdot\vec F\,dV$ for a closed surface $S$ enclosing $V$.== Stokes: $\oint_C\vec F\cdot d\vec r=\iint_S(\nabla\times\vec F)\cdot\hat n\,dS$. Green: $\oint_C(M\,dx+N\,dy)=\iint_R\left(\dfrac{\partial N}{\partial x}-\dfrac{\partial M}{\partial y}\right)dx\,dy$.

Key points.

Gauss converts a closed-surface flux into a volume integral; its surface integral must be computed over all six faces of a cube.
Stokes converts a circulation around a closed curve into flux of the curl through a surface bounded by it.
Green is the plane case of Stokes, with $\hat n=\hat k$ and $\vec F=M\hat i+N\hat j$.
The curve $C$ is traversed counterclockwise so the region lies on the left.
To verify a theorem, evaluate both sides separately and show they are equal.
Area by Green: $A=\tfrac12\oint_C(x\,dy-y\,dx)$.

Example 1 (Gauss). $\vec F=x^2\hat i+y^2\hat j+z^2\hat k$ over the box $0\le x\le a,\ 0\le y\le b,\ 0\le z\le c$. Volume side: $\nabla\cdot\vec F=2(x+y+z)$, and $\int_0^a\!\int_0^b\!\int_0^c2(x+y+z)\,dz\,dy\,dx=abc(a+b+c)$. Surface side: on $x=a$ flux is $a^2bc$ and on $x=0$ it is 0; likewise $ab^2c$ on $y=b$ and $abc^2$ on $z=c$, others 0.

Sum $=abc(a+b+c)$, so LHS = RHS. (For $\vec F=x^3\hat i+y^3\hat j+z^3\hat k$ on the cube of side $a$: $\nabla\cdot\vec F=3(x^2+y^2+z^2)$, volume integral $=3a^5$; three faces at $x,y,z=a$ each give $a^3\cdot a^2=a^5$, total $3a^5$.)

Example 2 (Green). $M=3x^2-8y^2$, $N=4y-6xy$ over the triangle $x=0,y=0,x+y=1$. Double integral: $N_x-M_y=-6y+16y=10y$; $\int_0^1\!\int_0^{1-x}10y\,dy\,dx=\int_0^15(1-x)^2dx=\tfrac53$.

Side of C	Path	Value
$C_1$	$y=0$, $x:0\to1$	$\int3x^2dx=1$
$C_2$	$y=1-x$, $x:1\to0$	$\tfrac83$
$C_3$	$x=0$, $y:1\to0$	$\int_1^04y\,dy=-2$

Line integral $=1+\tfrac83-2=\tfrac53$, equal to the double integral, so Green is verified.

(Variant: $M=xy+y^2,\ N=x^2$ between $y=x$ and $y=x^2$: $N_x-M_y=x-2y$, double integral $\int_0^1\!\int_{x^2}^x(x-2y)\,dy\,dx=-\tfrac1{20}$; line integral $\tfrac{19}{20}$ along $y=x^2$ plus $-1$ along $y=x$ back $=-\tfrac1{20}$.)

Example 3 (Stokes). $\vec F=(x^2-y^2)\hat i+2xy\hat j$ over the rectangle $0\le x\le a,\ 0\le y\le b$ in the $xy$-plane, $\hat n=\hat k$. Surface side: $\nabla\times\vec F=(2y+2y)\hat k=4y\hat k$, and $\int_0^a\!\int_0^b4y\,dy\,dx=2ab^2$. Line side, counterclockwise, using $\vec F\cdot d\vec r=(x^2-y^2)dx+2xy\,dy$:

Edge	Path	Value
bottom	$y=0$, $x:0\to a$	$\tfrac{a^3}3$
right	$x=a$, $y:0\to b$	$\int2ay\,dy=ab^2$
top	$y=b$, $x:a\to0$	$-\tfrac{a^3}3+ab^2$
left	$x=0$	$0$

Sum $=2ab^2$, equal to the surface integral, so Stokes is verified.

Example 4 (Area). Curves $y=x$, $y=1/x$, $y=x/4$ meet at $(0,0)$, $(1,1)$, $(2,\tfrac12)$. Counterclockwise: along $y=x/4$ to $(2,\tfrac12)$, along $y=1/x$ back to $(1,1)$, along $y=x$ to the origin. On lines through the origin $x\,dy-y\,dx=0$. On $y=1/x$, $dy=-dx/x^2$ gives $x\,dy-y\,dx=-\dfrac{2}{x}dx$, taken from $x=2$ to $1$.

Area $=\tfrac12\int_2^1\left(-\dfrac2x\right)dx=\ln2$.

Answer frame. Open by stating the theorem in full; find the curl, divergence or $N_x-M_y$ and evaluate that side first; then split the boundary into its edges or faces in a table, sum them, and close with "LHS = RHS, hence verified". For area, state $A=\tfrac12\oint(x\,dy-y\,dx)$, find intersection points, go counterclockwise.

Pitfall: Traversing the edges clockwise, which flips the sign of the line integral.

Asked: [14 marks] (Jun 2022, Dec 2023) Verify Gauss divergence theorem for $\bar F=x^2\bar i+y^2\bar j+z^2\bar k$ over the cube $x=0,a$; $y=0,b$; $z=0,c$; also for $\bar F=x^3\bar i+y^3\bar j+z^3\bar k$ over the cube $0\le x,y,z\le a$. Asked: [14 marks] (Nov 2022, Dec 2024) Verify Green's theorem for $\oint_C[(3x^2-8y^2)dx+(4y-6xy)dy]$ over the region bounded by $x=0$, $y=0$, $x+y=1$; also for $\int_C[(xy+y^2)dx+x^2dy]$ bounded by $y=x$ and $y=x^2$. Asked: [14 marks] (Jun 2025) Verify Stokes theorem for $\bar F=(x^2-y^2)\bar i+2xy\bar j$ over the box bounded by $x=0,x=a,y=0,y=b$. Asked: [7 marks] (Jun 2023) Using Green's theorem, find the area of the region in the first quadrant bounded by $y=x$, $y=1/x$, $y=x/4$.

Last-minute revision

$\nabla\phi=(\phi_x,\phi_y,\phi_z)$ is normal to $\phi=c$; $\cos\theta=\dfrac{\nabla\phi_1\cdot\nabla\phi_2}{|\nabla\phi_1||\nabla\phi_2|}$ for the angle between surfaces.
Directional derivative $=\nabla\phi\cdot\hat u$ with $\hat u$ a unit vector; maximum value $|\nabla\phi|$.
Solenoidal: $\nabla\cdot\vec F=0$; irrotational: $\nabla\times\vec F=\vec0$, and then $\vec F=\nabla\phi$.
$\nabla\cdot\vec r=3$, $\nabla\times\vec r=\vec0$, $\nabla r^n=nr^{n-2}\vec r$.
$\nabla\cdot(r^n\vec r)=(n+3)r^n$, so $\vec r/r^3$ is solenoidal; $r^n\vec r$ is always irrotational.
$\nabla^2f(r)=f''+\tfrac2rf'$.
Work $=\int\vec F\cdot d\vec r=\int\vec F\cdot\dfrac{d\vec r}{dt}dt$; asked answer $-\pi$.
Gauss box $a,b,c$ with $F=(x^2,y^2,z^2)$: both sides $abc(a+b+c)$; cube $x^3$ case: $3a^5$.
Green triangle problem: both sides $\tfrac53$; Green $y=x,y=x^2$ problem: $-\tfrac1{20}$.
Stokes rectangle: both sides $2ab^2$.
Green area $=\tfrac12\oint(x\,dy-y\,dx)$; asked area $=\ln2$.
Angle between surfaces at $(2,-1,2)$: $\cos^{-1}\dfrac{8}{3\sqrt{21}}$; directional derivatives asked: $1$ and $\tfrac{37}3$.

Memory hooks

Gauss = Get the volume: flux out of a closed surface equals divergence inside.
Stokes = Spin: circulation on the edge equals curl through the surface.
Green = Flat Stokes: use $N_x-M_y$ and walk counterclockwise.
Div is a dot (scalar answer); curl is a cross (vector answer).
Gradient stands on the surface (normal); take the unit vector before the dot.

Coverage checklist

Differentiation of Vectors: no past question; definition and rules covered.
Scalar and vector point function: no past question; covered.
Gradient: no past question; covered.
Geometrical meaning of gradient: angle between surfaces at $(2,-1,2)$ (Jun 2022).
Directional Derivative: $e^{2x}\cos yz$ along tangent, $x^2yz+4xz^2$ (Jun 2023, Dec 2024).
Divergence and Curl: $r^n\bar r$ solenoidal, curl $r^n\bar r$, irrotational with potential, $\nabla^2f(r)$ (Jun 2022, Nov 2022, Jun 2023, Dec 2023, Jun 2025).
Line Integral, Surface Integral and Volume Integral: work done along helix (Jun 2025).
Gauss Divergence, Stokes and Green theorems: Gauss cube, Green triangle and $y=x,y=x^2$, Stokes rectangle, Green area (Jun 2022, Nov 2022, Jun 2023, Dec 2023, Dec 2024, Jun 2025).

Account

Mathematics II (BT-202) - Unit 5 Short Notes

How unit 5 is examined

Differentiation of Vectors

Scalar and vector point function

Gradient

Geometrical meaning of gradient

Directional Derivative

Divergence and Curl

Line, Surface and Volume Integral

Gauss Divergence, Stokes and Green theorems

Last-minute revision

Memory hooks

Coverage checklist