Applied Calculus – Comprehensive Exam Study Notes

Exam Context & Administrative Details

• Course: Applied Calculus
  • Code: BUM2123
  • Faculty: Industrial Sciences & Technology
  • Session/Semester: 2018/2019 Semester I
  • Date/Duration: 2 Jan 2019 – 3 hours
  • Programmes Sitting: $\text{BAA, BCG, BEE, BEP, BFF, BFM, BHA, BHM, BKC, BMA, BMM, BPS, BSB, BSK, BSP, BTC, BTF, BTP, FBAA, FBMM}$
• Paper Structure: 9 questions, 5 printed pages (+ appendix)
• Instructions:
  • Answer all questions & start each on new page.
  • Show every calculation & assumption.
  • Formula sheet provided (trig identities, polar/cylindrical/spherical conversions, multiple‐integral Jacobians, surface‐area formulas, second‐derivative test, etc.)

Q1 – Chain-Rule Differentiation (5 marks)

• Given composite variable dependencies:
  $w = x e^{y}, \quad x = \ln t^{2}, \quad y = \sin t$
• Objective: find $\dfrac{dw}{dt}$ using chain rule.
• Chain Rule in multivariable form:
  $\frac{dw}{dt}=\frac{\partial w}{\partial x}\frac{dx}{dt}+\frac{\partial w}{\partial y}\frac{dy}{dt}$
• Derivative components:
  • $\frac{\partial w}{\partial x}=e^{y}$
  • $\frac{\partial w}{\partial y}=x e^{y}$
  • $\frac{dx}{dt}=\frac{d}{dt}\big(\ln t^{2}\big)=\frac{2}{t}$
  • $\frac{dy}{dt}=\cos t$
• Substitution yields answer (paper’s numeric simplification not shown here):
  $\frac{dw}{dt}=e^{\sin t}\left(\frac{2}{t}+x\cos t\right)$ where $x=\ln t^{2}$.
• Pedagogical notes:
  • Emphasises multivariable chain rule; crucial in physics (e.g.
    thermodynamic state functions) & engineering (automatic differentiation of nested formulas).

Q2 – Implicit Differentiation of $F(x,y,z)=0$ (8 marks)

• Function: $F=2x^{2}y^{2}-y^{2}+xz-\sin(2x-5z)+\cos(6y)=0$
• Task: compute $\dfrac{\partial z}{\partial x}$ and $\dfrac{\partial z}{\partial y}$ (implicitly) and then evaluate at
  $\bigl(x,y,z\bigr)=\left(\tfrac{5\pi}{12}, \tfrac{5\pi}{12}, \tfrac{5\pi}{2}\right)$.
• General implicit‐differentiation formula for $z$:
  $\frac{\partial z}{\partial x}=-\frac{F<em>{x}}{F</em>{z}}, \qquad \frac{\partial z}{\partial y}=-\frac{F<em>{y}}{F</em>{z}}$ (holding other variables constant).
• Compute partials:
  \begin{aligned}
  F{x}&=4xy^{2}+z-2\cos(2x-5z)\cdot(2) \ &=4xy^{2}+z-4\cos(2x-5z)\[4pt] F{y}&=4x^{2}y-2y-6\sin(6y)\[4pt]
  F_{z}&=x+5\sin(2x-5z)
  \end{aligned}
• Substitute numerical coordinates (paper’s worked value $\approx1.3579$ supplied for $\partial z/\partial x$; symmetrical method for $\partial z/\partial y$).
• Key concept: gradient vector $\nabla F=(F<em>{x},F</em>{y},F_{z})$ is orthogonal to implicit surface; directional rates follow from gradient ratios.

Q3 – Relative Extrema via Second-Derivative Test (15 marks)

• Function:
  $f(x,y)=4x^{4}-2x^{2}y^{2}+x^{2}y- y^{3}-8x^{2}+12y$
• Steps:
  1. Find critical points by solving $f<em>{x}=0,\;f</em>{y}=0$.
  2. Build Hessian: H=\begin{bmatrix}f{xx} & f{xy}\f{yx} & f{yy}\end{bmatrix}; compute $D=\det H$ at each point.
  3. Classification:
  • $D>0, f_{xx}>0$ → local min.
  • $D>0, f_{xx}<0$ → local max.
  • $D<0$ → saddle.
• Supplied results:
  • $\bigl(0,2\bigr)$ → $D=64, f_{xx}=-16$ ⇒ relative maximum.
  • $\bigl(2,2\bigr)$ & $(-2,2)$ → $D=-128$ ⇒ saddle points.
• Practical relevance: design optimisation (finding maxima/minima of multivariable cost); economic surfaces; machine‐learning loss landscapes.

Q4 – Cardioid $r=3+3\cos\theta$ (Polar) (9 marks)

(i) Sketch (2 marks)

• Cardioid symmetric about polar axis ($\theta=0$ line).
• Maximum radius $r_{\max}=6$ at $\theta=0$; cusp at origin when $\theta=\pi$.
  (ii) Area via single polar integral (7 marks)
• Formula: $A=\dfrac{1}{2}\displaystyle\int_{\alpha}^{\beta}r^{2}\,d\theta$.
• Full cardioid: $\alpha=-\pi,\;\beta=\pi$ (or $0\to2\pi$).
• Computation leads to paper’s answer $A\approx 42.4115$ (units$^{2}$).
  • Exact symbolic value: $A=\frac{27\pi}{2}$.
• Concept: single‐integral polar area avoids double‐integral set-up; crucial in antenna lobe plots, orbital paths etc.

Q5 – Double Integral over Region Bounded by Two Parabolas (14 marks)

• Curves:
  $y=2-x^{2}, \qquad y=-x^{2}+4$
• Region R: vertical “lens” intersection (sketch & shading worth 3 marks).
• Limits: solve equality $2-x^{2}=-x^{2}+4\implies x=\pm\sqrt{2}$.
• Integral: $\displaystyle\iint_{R}y\,dA$.
• Typical evaluation: integrate with $x$ outer; yields answer $\frac{125}{6}\approx20.8333$.
• Learning point: choosing order $dydx$ vs $dxdy$ can simplify algebra; visual sketching is pivotal.

Q6 – Common Region of Circle & Cardioid in First Quadrant (16 marks)

• Circle: $r=2\sin\theta$ (radius 2, centred on $y$-axis).
• Cardioid: $r=-2+2\sin\theta$ (left-shifted, cusp on negative $x$).
  (i) Combined sketch (3 marks): first‐quadrant snippet.
  (ii) Intersection angle (2 marks):
  $2\sin\theta=-2+2\sin\theta\;\Rightarrow\;\sin\theta=1 \Rightarrow \theta=\frac{\pi}{2}$.
• Reported answer: $\theta=\dfrac{\pi}{6}$? (Paper lists “$1\sin\theta/2$” then $\theta=\pi/6$; interpretation: likely other intersection solving $2\sin\theta = -2+2\sin\theta$ + domain restrictions → cross-check in class.)
  (iii) Area via polar double integral (11 marks): integrate $\dfrac12 r^{2}$ from lower curve to upper, $\theta$ bounds $0\to\theta_{\text{int}}$; answer in sheet $A=\dfrac{7\pi}{6}-\dfrac{2}{3}\approx0.2011$.

Q7 – Triple Integral Volume in First Octant beneath Plane (9 marks)

• Region: bounded by coordinate planes $x=0,y=0,z=0$ and plane
  $6x+6y+3z=6 \;\Rightarrow\; z=2-2x-2y$.
• First octant ⇒ $x,y,z\ge0$; projection onto $xy$ plane is triangle with intercepts $x=y=1$.
• Volume:
  $V=\int<em>{0}^{1}\int</em>{0}^{1-x}\int_{0}^{2-2x-2y}dz\,dy\,dx$.
• Paper’s answer: $V=1$.
• Interpretation: tetrahedral volume verification (classic exam favourite; showcases transforming planes to limits).

Q8 – Spherical-Coordinate Evaluation of a Symmetric Solid (12 marks)

• Rectangular description: $G={(x,y,z):x^{2}+y^{2}+z^{2}\le4,\;x,y,z\ge0}$ (first-octant quarter-ball of radius 2).
• Given integral $\displaystyle\iiint_{G}(4-x^{2}-y^{2}-z^{2})\,dV$.
• Spherical substitution:
  $x=\rho\sin\phi\cos\theta,\;y=\rho\sin\phi\sin\theta,\;z=\rho\cos\phi$.
  Jacobian $\rho^{2}\sin\phi$.
• Limits: $\rho:0\to2, \theta:0\to\frac{\pi}{2}, \phi:0\to\frac{\pi}{2}$.
• Integral in spherical:
  $\int<em>{0}^{\pi/2}\int</em>{0}^{\pi/2}\int_{0}^{2}(4-\rho^{2})\rho^{2}\sin\phi\,d\rho\,d\phi\,d\theta$.
• Evaluate:
  • Radial: $\int<em>{0}^{2}(4\rho^{2}-\rho^{4})d\rho=\Bigl(\tfrac{4\rho^{3}}{3}-\tfrac{\rho^{5}}{5}\Bigr)</em>{0}^{2}=\tfrac{32}{3}-\tfrac{32}{5}=\tfrac{64}{15}$.
  • Angular product: $\int<em>{0}^{\pi/2}\sin\phi\,d\phi=1,$ and $\int</em>{0}^{\pi/2}d\theta=\tfrac{\pi}{2}$.
  • Result: $V=\frac{64}{15}\cdot\frac{\pi}{2}=\frac{32\pi}{15}\approx6.702$ (paper shows symbolic triple integral template; final answer not printed).
• Key learning: spherical coordinates simplify radially symmetric integrands; especially 3-D probability densities, gravitational potentials.

Q9 – Surface Area of Paraboloid $z=4-x^{2}-y^{2}$ in First Octant (12 marks)

• First octant rules: $x\ge0,y\ge0,z\ge0$; projection onto $xy$ plane is quarter of disk $x^{2}+y^{2}\le4$ (radius 2).
• Surface-area formula (from appendix):
  $A=\iint_{R}\sqrt{1+\left(\frac{\partial z}{\partial x}\right)^{2}+\left(\frac{\partial z}{\partial y}\right)^{2}}\,dA.$
• Gradients: $z<em>{x}=-2x, z</em>{y}=-2y$ → integrand $\sqrt{1+4x^{2}+4y^{2}}$.
• Switch to polar: $x=r\cos\theta,y=r\sin\theta, r:0\to2, \theta:0\to\tfrac{\pi}{2}$.
  $A=\int<em>{0}^{\pi/2}\int</em>{0}^{2}\sqrt{1+4r^{2}}\,r\,dr\,d\theta.$
• Evaluate:
  • Inner: set $u=1+4r^{2} \Rightarrow du=8rdr$.
  • After substitution & limits $u:1\to17$, get $\tfrac{1}{8}\int_{1}^{17}u^{1/2}du=\tfrac{1}{8}\cdot\tfrac{2}{3}(17^{3/2}-1)$.
  • Multiply by angular $\tfrac{\pi}{2}$.
  • Paper’s numeric: $A=\dfrac{\pi}{24}\bigl(17\sqrt{17}-1\bigr)\approx9.0442$.
• Applications: manufacturing cost of parabolic dishes; surface coating.

Appendix Highlights – Key Formulae & Identities

• Trig identities (excerpts):
  $\sin^{2}x+\cos^{2}x=1, \quad \sin2x=2\sin x\cos x, \quad \cos2x=\cos^{2}x-\sin^{2}x$.
• Polar conversions: $x=r\cos\theta,\;y=r\sin\theta,\;r^{2}=x^{2}+y^{2},\;\tan\theta=\frac{y}{x}$.
• Polar area: $A=\tfrac12\int_{\alpha}^{\beta}r^{2}\,d\theta$.
• Multiple‐integral Jacobians:
  • Cylindrical: $dV=r\,dr\,d\theta\,dz$.
  • Spherical: $dV=\rho^{2}\sin\phi\,d\rho\,d\phi\,d\theta$.
• Second-derivative test in $\mathbb R^{2}$:
  $D=f<em>{xx}f</em>{yy}-f_{xy}^{2}$ (see classification earlier).
• Surface area of graph $z=f(x,y)$:
  $A=\iint<em>{R}\sqrt{1+f</em>{x}^{2}+f_{y}^{2}}\,dA$.

Conceptual Connections & Practical/Philosophical Notes

• Differentiation rules extend to automatic differentiation in machine learning, enabling gradient‐based optimisation.
• Implicit‐surface methods underpin level‐set modelling in computer graphics and fluid interfaces.
• Polar & spherical integrals illustrate value of choosing coordinates matching symmetry—key ethical lesson: proper modelling reduces computational waste (energy, time).
• Optimisation (Q3) governs allocation of scarce resources; need awareness of assumptions in modelling human systems.
• Surface‐area calculations (Q9) directly feed sustainable material‐usage forecasts in engineering.