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BT-102 · Mathematics I/Quick Revision Short Notes

Mathematics I (BT-102) - Unit 1 Short Notes

How unit 1 is examined

Differential calculus of one and several variables: Rolle and mean value theorems, Taylor series, partial derivatives, maxima and minima, Lagrange multipliers. Every topic is unasked in the tagged list, but untagged papers show verify-the-theorem, Taylor, partial-derivative and extremum problems worth 7 marks each.

Rolle’s theorem

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Definition. If $f$ is continuous on $[a,b]$, differentiable on $(a,b)$ and $f(a)=f(b)$, then there is at least one $c\in(a,b)$ with $f'(c)=0$.

Key points.

  1. The three conditions are continuity on the closed interval, differentiability on the open interval, and equal end values.
  2. Geometrically, some point of the curve has a horizontal tangent.
  3. To verify: check the three conditions, solve $f'(c)=0$, and show $c$ lies inside $(a,b)$.

==If $f(a)=f(b)$ and $f$ is continuous and differentiable, then $f'(c)=0$ for some $c$ in $(a,b)$.==

Example. $f=x^2+2x$ on $[-2,0]$: $f(-2)=f(0)=0$, $f'=2x+2=0$ gives $c=-1\in(-2,0)$. Verified. For $f=\sin x/e^x$ on $[0,\pi]$: $f(0)=f(\pi)=0$, $f'=(\cos x-\sin x)/e^x=0$ gives $\tan c=1$, so $c=\pi/4$.

Asked: [7 marks] (Jun 2023) State Rolle's theorem hence verify for $f(x)=x^2+2x$ defined in the interval $[-2,0]$. Asked: [7 marks] (Jun 2022) Verify Rolle's theorem for the function $f(x)=\frac{\sin x}{e^x}$.

Mean Value theorems

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Definition. Lagrange: if $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then $f'(c)=\dfrac{f(b)-f(a)}{b-a}$ for some $c\in(a,b)$. Cauchy: for $f,g$ continuous on $[a,b]$ and differentiable on $(a,b)$, with $g'\neq0$, $\dfrac{f'(c)}{g'(c)}=\dfrac{f(b)-f(a)}{g(b)-g(a)}$.

Key points.

  1. Lagrange's theorem says some tangent is parallel to the chord joining the end points.
  2. Rolle's theorem is the special case $f(a)=f(b)$.
  3. Cauchy's theorem reduces to Lagrange's when $g(x)=x$.
  4. For inequalities, apply the theorem to a suitable $f$ and bound $f'(c)$ over the interval.

==Lagrange's mean value theorem: $f(b)-f(a)=(b-a)f'(c)$ for some $c$ between $a$ and $b$.==

Example. $f=x^2+2x$ on $[-2,0]$: $\frac{f(0)-f(-2)}{2}=0=2c+2$, so $c=-1$. Cauchy with $e^x,e^{-x}$: $\frac{e^c}{-e^{-c}}=-e^{2c}$ and $\frac{e^b-e^a}{e^{-b}-e^{-a}}=-e^{a+b}$, so $c=\frac{a+b}{2}$. Cauchy with $\sin x,\cos x$ on $[0,\pi/2]$: $\frac{\cos c}{-\sin c}=\frac{1-0}{0-1}=-1$, so $\cot c=1$, $c=\pi/4$. Inequality: take $f=\cos^{-1}x$ on $[\frac12,\frac35]$. Then $\cos^{-1}\frac35-\frac\pi3=-\frac{1}{10\sqrt{1-c^2}}$. For $c\in(\frac12,\frac35)$, $\frac45<\sqrt{1-c^2}<\frac{\sqrt3}{2}$, so the subtracted term lies between $\frac18$ and $\frac{1}{5\sqrt3}$, giving $\frac\pi3-\frac{1}{5\sqrt3}>\cos^{-1}\frac35>\frac\pi3-\frac18$.

Asked: [7 marks] (Dec 2023) State Lagrange's theorem hence verify for $f(x)=x^2+2x$ defined in the interval $[-2,0]$. Asked: [7 marks] (Nov 2022) Prove that $\frac{\pi}{3}-\frac{1}{5\sqrt3}>\cos^{-1}\frac35>\frac\pi3-\frac18$ using Lagrange's mean value theorem. Asked: [7 marks] (Nov 2022) Find $C$ of Cauchy's Mean value theorem on $[a,b]$ for $f(x)=e^x$ and $g(x)=e^{-x}$, $(a,b>0)$. Asked: [7 marks] (Jun 2022) Verify Cauchy's Mean value theorem for the function $\sin x$ and $\cos x$ in $[0,\frac\pi2]$.

Expansion of functions by Mc. Laurin’s and Taylor’s for one variable

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Definition. Taylor's series expands $f$ about $x=a$ in powers of $(x-a)$; Maclaurin's series is the case $a=0$.

Formula. $$f(x)=f(a)+(x-a)f'(a)+\frac{(x-a)^2}{2!}f''(a)+\frac{(x-a)^3}{3!}f'''(a)+\cdots$$

Key points.

  1. Maclaurin's series is $f(x)=f(0)+xf'(0)+\frac{x^2}{2!}f''(0)+\cdots$.
  2. Write the derivatives of $f$, evaluate each at $a$, and substitute into the formula.
  3. The expansion is valid when the remainder term tends to zero.

==Taylor's series: $f(x)=\sum \frac{f^{(n)}(a)}{n!}(x-a)^n$.==

Example. $f=\sin x$ about $\pi/2$: $f,f',f'',f'''$ at $\pi/2$ are $1,0,-1,0$, so $$\sin x=1-\frac{(x-\pi/2)^2}{2!}+\frac{(x-\pi/2)^4}{4!}-\cdots$$

Asked: [7 marks] (Dec 2024) Find the Taylor's expansion of $y=\sin x$ about point $x=\pi/2$.

Taylor’s theorem for function of two variables

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Definition. About $(a,b)$, with $h=x-a$ and $k=y-b$, $f(x,y)=f(a,b)+\left(h\partial_x+k\partial_y\right)f+\frac{1}{2!}\left(h\partial_x+k\partial_y\right)^2f+\cdots$, all derivatives taken at $(a,b)$.

Key points.

  1. The second-order term is $\frac{1}{2}\left(h^2f_{xx}+2hk\,f_{xy}+k^2f_{yy}\right)$.
  2. About $(0,0)$ this is Maclaurin's series in $x$ and $y$, and $h=x$, $k=y$.
  3. For a product like $e^x\cos y$, multiplying the one-variable series is faster than differentiating.

==Taylor's theorem for two variables expands $f(x,y)$ in powers of $h=x-a$ and $k=y-b$ using the operator $(h\partial_x+k\partial_y)^n/n!$.==

Example. $e^x\cos y=(1+x+\frac{x^2}{2}+\frac{x^3}{6})(1-\frac{y^2}{2})=1+x+\frac{x^2}{2}-\frac{y^2}{2}+\frac{x^3}{6}-\frac{xy^2}{2}+\cdots$ $e^x\log(1+y)=(1+x+\frac{x^2}{2})(y-\frac{y^2}{2}+\frac{y^3}{3})=y+xy-\frac{y^2}{2}+\frac{x^2y}{2}-\frac{xy^2}{2}+\frac{y^3}{3}+\cdots$

Asked: [7 marks] (Dec 2023) Find the first six terms of the expansions of the function $e^x\cos y$ in a Taylor series in the neighbourhood of the point $(0,0)$. Asked: [7 marks] (Jun 2023) Find the first six terms of the expansions of the function $e^x\log(1+y)$ in a Taylor series in the neighbourhood of the point $(0,0)$.

Partial Differentiation

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Definition. The partial derivative $\partial f/\partial x$ is the derivative of $f(x,y)$ with respect to $x$ with $y$ held constant.

Key points.

  1. Chain rule: if $u=f(x,y)$ with $x,y$ functions of $t$, then $\frac{du}{dt}=\frac{\partial u}{\partial x}\frac{dx}{dt}+\frac{\partial u}{\partial y}\frac{dy}{dt}$.
  2. Euler's theorem: if $f$ is homogeneous of degree $n$, then $xf_x+yf_y=nf$.
  3. If $u=\tan^{-1}(\cdot)$ of a homogeneous expression, apply Euler to $z=\tan u$, of degree $n$, which gives $xu_x+yu_y=n\sin u\cos u$.
  4. Mixed derivatives are equal, $u_{xy}=u_{yx}$, when they are continuous.

==Euler's theorem: for a homogeneous function of degree $n$, $x\frac{\partial f}{\partial x}+y\frac{\partial f}{\partial y}=nf$.==

Example. $u=x^2+y^2$, $x=a\cos t$, $y=b\sin t$: $\frac{du}{dt}=2x(-a\sin t)+2y(b\cos t)=$ $(b^2-a^2)\sin 2t$. $u=\tan^{-1}\frac{x^3+y^3}{x-y}$: $\tan u$ has degree 2, so $xu_x+yu_y=2\sin u\cos u=$ $\sin 2u$. $u=x^2\tan^{-1}\frac yx-y^2\tan^{-1}\frac xy$: $u_x=2x\tan^{-1}\frac yx-y$, so $u_{xy}=$ $\frac{x^2-y^2}{x^2+y^2}$. $u=u(r,s)$ with $r=\frac1x-\frac1y$, $s=\frac1x-\frac1z$: $x^2u_x=-(u_r+u_s)$, $y^2u_y=u_r$, $z^2u_z=u_s$, so the sum is $0$.

Asked: [7 marks] (Jun 2025) If $u=\tan^{-1}\frac{x^3+y^3}{x-y}$, then prove that $x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}=\sin 2u$. Asked: [7 marks] (Dec 2023) If $u=u\left(\frac{y-x}{xy},\frac{z-x}{xz}\right)$ find the value of $x^2u_x+y^2u_y+z^2u_z$. Asked: [7 marks] (Jun 2023) If $u=x^2\tan^{-1}\frac yx-y^2\tan^{-1}\frac xy$, find the value of $\frac{\partial^2u}{\partial x\partial y}$. Asked: [7 marks] (Dec 2023, Jun 2023) Find $\frac{du}{dt}$ if $u=x^2+y^2$, $x=a\cos t$, $y=b\sin t$.

Maxima & Minima (two and three variables)

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Definition. A stationary point of $f(x,y)$ satisfies $f_x=0$, $f_y=0$; it is a maximum, minimum or saddle according to the second-order test.

Steps.

  1. Solve $f_x=0$, $f_y=0$ for all stationary points.
  2. Find $r=f_{xx}$, $s=f_{xy}$, $t=f_{yy}$ at each point.
  3. If $rt-s^2>0$: minimum when $r>0$, maximum when $r<0$. If $rt-s^2<0$: saddle, no extremum. If $rt-s^2=0$: doubtful.

<mark>At a stationary point, $rt-s^2>0$ with $r<0$ gives a maximum and with $r>0$ a minimum.</mark>

Example. $f=x^3+y^3-3axy$: points $(0,0)$ and $(a,a)$. At $(0,0)$, $rt-s^2=-9a^2<0$, saddle. At $(a,a)$, $rt-s^2=27a^2>0$, $r=6a>0$, minimum $-a^3$. $f=x^3+3xy^2-3x^2-3y^2+4$: points $(0,0),(2,0),(1,\pm1)$. Maximum $f(0,0)=4$, minimum $f(2,0)=0$, $(1,\pm1)$ are saddles. $f=x^3+y^3-63(x+y)+12xy$: points $(3,3),(-7,-7),(5,-1),(-1,5)$. Minimum $-216$ at $(3,3)$, maximum $784$ at $(-7,-7)$, the other two are saddles. Three variables: box with sides $2x,2y,2z$ in the sphere $x^2+y^2+z^2=a^2$ has volume $8xyz$, maximum when $x=y=z=a/\sqrt3$, a cube, volume $\frac{8a^3}{3\sqrt3}$ (proved with a multiplier below).

Asked: [7 marks] (Dec 2024) Find the points where the function $x^3+y^3-3axy$ has maximum or minimum value. Asked: [7 marks] (Nov 2022) Find the minimum and maximum value of $f(x,y)=x^3+3xy^2-3x^2-3y^2+4$. Asked: [7 marks] (Dec 2023) Estimate the extreme values of the function $x^3+y^3-63(x+y)+12xy$. Asked: [7 marks] (Dec 2023, Jun 2025) Show that a rectangular solid of maximum volume that can be inscribed in a given sphere is a cube.

Method of Lagranges Multipliers

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Definition. To find the extreme values of $f(x,y,z)$ subject to $\phi(x,y,z)=0$, form $F=f+\lambda\phi$ and solve $F_x=F_y=F_z=0$ together with $\phi=0$.

Key points.

  1. Eliminate $\lambda$ from the equations to get relations between $x,y,z$, then use the constraint.
  2. The multiplier $\lambda$ is an unknown constant, not a variable of the problem.
  3. Taking $\log f$ first often simplifies products of powers.

==Lagrange's method: extremise $F=f+\lambda\phi$ and solve $F_x=F_y=F_z=0$ with $\phi=0$.==

Example. Volume $xyz$ with $x^2+y^2+z^2=a^2$: $yz=-2\lambda x$, $zx=-2\lambda y$, $xy=-2\lambda z$, so $x^2=y^2=z^2=a^2/3$: a cube. $x^2yz^3$ with $2x+y+3z=a$: taking logs, $\frac2x+2\lambda=0$, $\frac1y+\lambda=0$, $\frac3z+3\lambda=0$ give $x=y=z=-\frac1\lambda$, so $6x=a$ and the value is $(a/6)^6=a^6/46656$. It is a maximum for positive variables. $400xyz^2$ on $x^2+y^2+z^2=1$: $x^2=y^2=\frac14$, $z^2=\frac12$, so the highest temperature is $400\cdot\frac18=$ $50$. Nearest point on $x^2+4xy+6y^2=140$: minimise $x^2+y^2$; then $x^2+y^2=-140\lambda$ with $2\lambda^2+7\lambda+1=0$, so the least value is $35(7-\sqrt{41})$ and distance $\approx4.57$.

Asked: [7 marks] (Jun 2022) Find the minimum value of $x^2yz^3$, subject to the condition $2x+y+3z=a$. Asked: [7 marks] (Jun 2023) The temperature $u(x,y,z)$ at any point in space is $u=400xyz^2$ find the highest temperature on surface of the sphere $x^2+y^2+z^2=1$. Asked: [7 marks] (Jun 2023) Find shortest distance from the origin to the curve $x^2+4xy+6y^2=140$. Asked: [7 marks] (Jun 2025) Prove that a rectangular solid of maximum volume within a sphere is a cube.

Last-minute revision

  • Rolle: $f(a)=f(b)\Rightarrow f'(c)=0$; for $x^2+2x$ on $[-2,0]$, $c=-1$.
  • Lagrange: $f'(c)=\frac{f(b)-f(a)}{b-a}$; Cauchy: $\frac{f'(c)}{g'(c)}=\frac{f(b)-f(a)}{g(b)-g(a)}$.
  • Cauchy with $e^x,e^{-x}$ gives $c=\frac{a+b}{2}$; with $\sin x,\cos x$ gives $c=\frac\pi4$.
  • Taylor: $f(x)=\sum\frac{f^{(n)}(a)}{n!}(x-a)^n$; Maclaurin is $a=0$.
  • $e^x\cos y=1+x+\frac{x^2}{2}-\frac{y^2}{2}+\frac{x^3}{6}-\frac{xy^2}{2}+\cdots$
  • Euler: $xf_x+yf_y=nf$; chain rule $\frac{du}{dt}=u_xx'+u_yy'$.
  • Extremum test: $rt-s^2>0$ with $r<0$ maximum, $r>0$ minimum; $<0$ saddle.
  • $x^3+y^3-3axy$ has a saddle at $(0,0)$ and a minimum $-a^3$ at $(a,a)$.
  • Lagrange multipliers: $F=f+\lambda\phi$; solid in a sphere is a cube; $400xyz^2$ gives $50$.

Memory hooks

  • Rolle is Lagrange with equal ends: a flat tangent.
  • Cauchy is Lagrange for two functions: divide their derivatives.
  • $rt-s^2$: positive means an extremum, negative means a saddle, and $r$ gives the sign.
  • Homogeneous of degree $n$ means Euler: $x,y$ multiply the partials and give $n f$.

Coverage checklist

  • Rolle’s theorem: Jun 2023 verify $x^2+2x$; Jun 2022 $\sin x/e^x$.
  • Mean Value theorems: Dec 2023 Lagrange verify; Nov 2022 inequality; Nov 2022 Cauchy $e^x,e^{-x}$; Jun 2022 Cauchy $\sin x,\cos x$.
  • Expansion of functions by Mc. Laurin’s and Taylor’s for one variable: Dec 2024 $\sin x$ about $\pi/2$.
  • Taylor’s theorem for function of two variables: Dec 2023 $e^x\cos y$; Jun 2023 $e^x\log(1+y)$.
  • Partial Differentiation: Jun 2025 $\tan^{-1}$; Dec 2023 $u_x$ sum; Jun 2023 $u_{xy}$; Dec 2023 and Jun 2023 $du/dt$.
  • Maxima & Minima (two and three variables): Dec 2024, Nov 2022, Dec 2023 extrema; Dec 2023 solid in sphere.
  • Method of Lagranges Multipliers: Jun 2022 $x^2yz^3$; Jun 2023 $400xyz^2$; Jun 2023 shortest distance; Jun 2025 solid in sphere.
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