FAD1014: MATHEMATICS II — Interleaved Mastery Problem Set

5-Day Intensive Study Plan
Topics: Geometry (Circle, Parabola, Ellipse, Hyperbola) + Series Expansion + Taylor-Maclaurin + Trigonometric Integration

How to Use This Set

Each problem intentionally mixes multiple concepts. Don't compartmentalize—let the ideas flow together. This builds the neural connections you need for exam conditions when you don't know which technique to use.

Recommended approach:

Day 1: Problems 1-3
Day 2: Problems 4-6
Day 3: Problems 7-9
Day 4: Problems 10-12
Day 5: Problems 13-15 + review

Topic Interconnection Map

graph TD
    A[Interleaved Math II Problems] --> B[Geometry]
    A --> C[Series & Taylor-Maclaurin]
    A --> D[Trigonometric Integration]
    B --> B1[Circle]
    B --> B2[Parabola]
    B --> B3[Ellipse]
    B --> B4[Hyperbola]
    C --> C1[Binomial Series]
    C --> C2[Maclaurin Series]
    C --> C3[Taylor Series]
    C --> C4[Power Series]
    D --> D1[Power Reduction]
    D --> D2[Product-to-Sum]
    D --> D3[Trig Substitution]
    B1 --> E[Cross-Topic Problems]
    C2 --> E
    D1 --> E
    B2 --> E
    C1 --> E
    D3 --> E
    B3 --> E
    D2 --> E
    B4 --> E
    C3 --> E

The Mastery Problems

Problem 1: The Oscillating Circle [Geometry + Series + Integration]

A particle moves along a path defined by the parametric equations: $$x(t) = \cos t, \quad y(t) = \sin 2t$$

(a) Show that at $t = \frac{\pi}{4}$, the particle lies on the circle $x^2 + y^2 = 1$.

(b) Find the Maclaurin series for $y(t) = \sin 2t$ up to the $t^5$ term. Use this to approximate $y(0.1)$.

(c) The area swept out by the position vector from $t = 0$ to $t = \frac{\pi}{4}$ is given by: $$A = \frac{1}{2} \int_0^{\pi/4} (x \frac{dy}{dt} - y \frac{dx}{dt}) , dt$$

Evaluate this integral using trigonometric identities (power reduction formulas).

Problem 2: The Parabolic Mirror [Geometry + Series Expansion]

A parabolic mirror has equation $(x-2)^2 = 8(y-1)$.

(a) Find the focus and directrix of this parabola.

(b) A light ray parallel to the axis of symmetry hits the mirror at point $P$ where $x = 4$. Find the coordinates of $P$ and show that the reflected ray passes through the focus.

(c) Expand $\sqrt{1 + \frac{x-2}{4}}$ as a binomial series up to the $(x-2)^3$ term. Use this to approximate the $y$-coordinate of the mirror when $x = 2.5$.

(d) The arc length of a parabola from $x = a$ to $x = b$ involves $\int \sqrt{1 + (\frac{dy}{dx})^2} , dx$. For this parabola, set up (but do not evaluate) the integral for arc length from $x = 2$ to $x = 4$, then use the trigonometric substitution $x - 2 = 2\tan\theta$ to transform it.

Problem 3: Elliptical Orbits [Geometry + Maclaurin + Trigonometric Integrals]

A planet orbits in an ellipse with equation $\frac{x^2}{9} + \frac{y^2}{4} = 1$.

(a) Find the coordinates of the foci and vertices.

(b) The area of an ellipse is $\pi ab$. Verify this by evaluating: $$A = 4\int_0^3 y , dx = 4\int_0^3 2\sqrt{1 - \frac{x^2}{9}} , dx$$

Use the substitution $x = 3\sin\theta$ (a trigonometric substitution) and show all steps.

(c) The orbital period $T$ involves the integral $\int_0^{2\pi} \frac{d\theta}{(1 + e\cos\theta)^2}$ where $e$ is eccentricity. For this ellipse, $e = \frac{\sqrt{5}}{3}$. Use the power series expansion of $(1 + u)^{-2}$ with $u = e\cos\theta$ to write the first three non-zero terms of the integrand.

(d) Evaluate $\int_0^{2\pi} \cos^2\theta , d\theta$ using power reduction formulas—you'll need this for part (c).

Problem 4: The Hyperbolic Cooling Tower [Geometry + Integration + Series]

A cooling tower has the shape of a hyperbola with equation $\frac{x^2}{16} - \frac{y^2}{9} = 1$.

(a) Find the equations of the asymptotes and the coordinates of the vertices.

(b) The volume of revolution formed by rotating the right branch of this hyperbola from $y = 0$ to $y = 3$ about the y-axis is: $$V = \pi \int_0^3 x^2 , dy$$

Substitute the hyperbola equation and evaluate this integral.

(c) The slope of the tangent line at any point on the hyperbola is $\frac{dy}{dx} = \frac{9x}{16y}$. Find the angle $\alpha$ that the tangent makes with the horizontal at point $(5, \frac{9}{4})$ using $\tan\alpha = \frac{dy}{dx}$.

(d) Use the Maclaurin series for $\arctan u$ to approximate this angle $\alpha$ in radians (convert $\frac{dy}{dx}$ to a decimal, then use series with $u = $ that value minus 1, adjusting appropriately).

Problem 5: Taylor's Trigonometric Challenge [Taylor Series + Integration]

Consider $f(x) = \sin x \cdot \cos^2 x$.

(a) Find the Taylor series for $f(x)$ about $x = 0$ (Maclaurin series) up to the $x^5$ term using:

The series for $\sin x = x - \frac{x^3}{6} + \frac{x^5}{120} - \cdots$
The series for $\cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24} - \cdots$

(b) Use your series from (a) to approximate $\int_0^{0.5} \sin x \cos^2 x , dx$ by integrating term by term.

(c) Verify your answer to (b) by evaluating the integral exactly using trigonometric integration techniques:

Let $u = \cos x$, or
Use identity $\sin x \cos^2 x = \frac{1}{4}\sin x + \frac{1}{4}\sin 3x$ (product-to-sum)

Compare: Is your series approximation within 1% of the exact value?

Problem 6: The Intersecting Curves [Geometry + Series]

A circle $C: x^2 + y^2 = 4$ and a parabola $P: y^2 = 3x$ intersect at points $A$ and $B$ in the first and fourth quadrants.

(a) Find the coordinates of $A$ and $B$.

(b) Find the angle between the tangent lines to the circle and parabola at point $A$.

For the circle: tangent is perpendicular to radius
For the parabola: use implicit differentiation

(c) Expand $\sqrt{4 - x^2}$ as a binomial series up to the $x^6$ term. This represents the upper semicircle.

(d) Use your series from (c) to approximate the area of the circle in the first quadrant: $$A \approx \int_0^1 \left(2 - \frac{x^2}{4} - \frac{x^4}{64} - \frac{x^6}{512}\right) dx$$ Compare to the exact area $\pi$.

Problem 7: The Reflecting Property [Geometry + Trigonometric Integrals]

An ellipse has foci at $F_1(-3, 0)$ and $F_2(3, 0)$, and passes through point $(4, 2.4)$.

(a) Find the equation of this ellipse.

(b) The reflecting property states that a ray from one focus reflects to the other. If a ray from $F_1$ hits the ellipse at $(4, 2.4)$, show that it reflects toward $F_2$ by verifying that the angle of incidence equals the angle of reflection.

(c) The perimeter of the ellipse involves the complete elliptic integral: $$P = 4a \int_0^{\pi/2} \sqrt{1 - e^2\sin^2\theta} , d\theta$$

Expand the integrand using binomial series $(1 + u)^{1/2}$ with $u = -e^2\sin^2\theta$, keeping terms up to $\sin^4\theta$. Then use power reduction to express $\sin^2\theta$ and $\sin^4\theta$ in terms of $\cos 2\theta$ and $\cos 4\theta$.

Problem 8: The Gateway Arch [Geometry + Maclaurin + Integration]

The Gateway Arch in St. Louis has the shape of an inverted catenary, but we can approximate it with a parabola: $y = 192 - \frac{x^2}{31.25}$ (in meters), from $x = -77.5$ to $x = 77.5$.

(a) Find the height and base width of this parabolic arch.

(b) Find the Maclaurin series for $\sqrt{1 + t^2}$ up to the $t^4$ term. You'll need this for arc length.

(c) The arc length of the arch is: $$L = 2\int_0^{77.5} \sqrt{1 + \left(\frac{dy}{dx}\right)^2} , dx$$

With $\frac{dy}{dx} = -\frac{2x}{31.25}$, set up the integral. Then use the substitution $t = \frac{2x}{31.25}$ and your series from (b) to approximate the arc length (integrate term by term).

Problem 9: The Lissajous Figure [Parametric + Series + Trigonometry]

A Lissajous figure is defined by $x = \sin 3t$, $y = \cos 2t$.

(a) Show that this curve lies on the surface of a cylinder. What is the equation of this cylinder?

(b) Find the Maclaurin series for $x(t) = \sin 3t$ up to the $t^5$ term.

(c) The curve passes through the origin when $t = \frac{\pi}{4}$. Find the slope $\frac{dy}{dx}$ at this point using parametric differentiation.

(d) The area enclosed by this curve can be found using: $$A = \frac{1}{2} \int_0^{2\pi} (x \frac{dy}{dt} - y \frac{dx}{dt}) , dt$$

Evaluate this using trigonometric identities and product-to-sum formulas.

Problem 10: The Whispering Gallery [Geometry + Integration + Series]

A whispering gallery has elliptical ceiling with equation $\frac{x^2}{25} + \frac{y^2}{9} = 1$ (in meters), where $y \geq 0$.

(a) If one person stands at focus $F_1$ and whispers, where should another person stand to hear clearly? Find both foci.

(b) The surface area of the ceiling (for painting) is: $$S = 2\pi \int_{-5}^{5} y \sqrt{1 + \left(\frac{dy}{dx}\right)^2} , dx$$

First find $\frac{dy}{dx}$ by implicit differentiation, then simplify the integrand.

(c) Use the trigonometric substitution $x = 5\sin\theta$ to transform the integral from (b). The integral becomes: $$S = 18\pi \int_{-\pi/2}^{\pi/2} \sqrt{1 - \frac{16}{25}\sin^2\theta} \cos\theta , d\theta$$

(d) Expand $\sqrt{1 - \frac{16}{25}\sin^2\theta}$ using binomial series and evaluate the first three terms of the surface area approximation.

Problem 11: The Conic Lens [Geometry + Maclaurin + Trig Integration]

A lens has one side as a parabola $y = x^2$ and the other as a circle $x^2 + y^2 = 2$ (for $y \geq 0$).

(a) Find where these curves intersect.

(b) Find the volume of the lens (between the surfaces) from $x = -1$ to $x = 1$: $$V = \pi \int_{-1}^{1} \left[(\sqrt{2-x^2})^2 - (x^2)^2\right] dx$$

(c) The upper surface of the lens (the circular part) can be written as $y = \sqrt{2}\sqrt{1 - \frac{x^2}{2}}$. Expand this as a binomial series up to $x^4$.

(d) Use your series from (c) to approximate: $$\int_0^1 \sqrt{2-x^2} , dx$$

Then verify using trigonometric substitution $x = \sqrt{2}\sin\theta$.

Problem 12: The Satellite Dish [Geometry + Series]

A satellite dish is a paraboloid formed by rotating $y = \frac{x^2}{16}$ about the y-axis.

(a) Find the focus of this parabola. Why is this location important for the receiver?

(b) The surface area of the dish (from $y = 0$ to $y = 4$) is: $$S = 2\pi \int_0^8 x \sqrt{1 + \left(\frac{dy}{dx}\right)^2} , dx$$

Evaluate this integral using trigonometric substitution or by simplifying the radical.

(c) If the dish extends to $y = h$, the radius is $r = 4\sqrt{h}$. Use binomial approximation to show that for small $h$, the surface area $S \approx 8\pi h$.

Problem 13: The Pendulum Period [Maclaurin + Trigonometric Integrals]

The period of a simple pendulum with amplitude $\theta_0$ is: $$T = 4\sqrt{\frac{L}{g}} \int_0^{\pi/2} \frac{d\phi}{\sqrt{1 - k^2\sin^2\phi}}$$ where $k = \sin(\theta_0/2)$.

(a) Expand $(1 - k^2\sin^2\phi)^{-1/2}$ using binomial series up to the $k^4$ term.

(b) Use power reduction formulas to express:

$\sin^2\phi$ in terms of $\cos 2\phi$
$\sin^4\phi$ in terms of $\cos 2\phi$ and $\cos 4\phi$

(c) Evaluate $\int_0^{\pi/2} \cos(2n\phi) , d\phi$ for $n = 0, 1, 2$.

(d) Combine (a), (b), and (c) to show: $$T \approx 2\pi\sqrt{\frac{L}{g}}\left(1 + \frac{1}{4}k^2 + \frac{9}{64}k^4\right)$$

For small angles, $T \approx 2\pi\sqrt{L/g}$ (the familiar formula).

Problem 14: The Bernoulli Spiral [Geometry + Taylor + Integration]

The spiral $r = e^{\theta}$ (in polar coordinates) can be analyzed using conic sections at each point.

(a) Convert to Cartesian: $x = e^{\theta}\cos\theta$, $y = e^{\theta}\sin\theta$. Find $\frac{dy}{dx}$ in terms of $\theta$.

(b) Find the Taylor series for $\tan\theta$ about $\theta = 0$ up to the $\theta^3$ term. Use this to find the slope at $\theta = 0$.

(c) The arc length of the spiral from $\theta = 0$ to $\theta = \pi$ is: $$L = \sqrt{2} \int_0^{\pi} e^{\theta} , d\theta$$

Evaluate this.

(d) The area swept out is $A = \frac{1}{2}\int_0^{\pi} r^2 , d\theta$. Evaluate and compare to the area of a circle with radius $r(\pi) = e^{\pi}$.

Problem 15: The Grand Finale [All Topics Interleaved]

A particle moves along an ellipse $\frac{x^2}{4} + y^2 = 1$ with parametric equations $x = 2\cos t$, $y = \sin t$.

(a) [Geometry] Find the foci and verify that $x^2 + y^2$ varies as the particle moves. At what point is the particle closest to the origin?

(b) [Maclaurin] The distance from origin is $d(t) = \sqrt{4\cos^2 t + \sin^2 t} = \sqrt{1 + 3\cos^2 t}$. Find the Maclaurin series for $d(t)$ up to $t^4$ using:

Series for $\cos t = 1 - \frac{t^2}{2} + \frac{t^4}{24} - \cdots$
Binomial expansion of $\sqrt{1 + u}$ with $u = 3\cos^2 t - 3$

(c) [Trigonometric Integration] The area enclosed by the ellipse is $A = 4\int_0^1 y , dx$. Using parametric substitution, show: $$A = 8\int_0^{\pi/2} \sin^2 t , dt$$

Evaluate using the power reduction formula $\sin^2 t = \frac{1 - \cos 2t}{2}$.

(d) [Series Verification] Expand $\sqrt{1 - \frac{x^2}{4}}$ as a binomial series and integrate term by term from 0 to 2 to verify the area is $2\pi$.

(e) [Conic Section Property] The sum of distances from any point on the ellipse to the two foci is constant and equals $2a$. Verify this at point $(0, 1)$.

Trigonometric Integration Decision Tree

graph TD
    A[Trig Integral Strategy] --> B{Contains sqrt?}
    B -->|sqrt(a^2-x^2)| C[x = a sin theta]
    B -->|sqrt(a^2+x^2)| D[x = a tan theta]
    B -->|sqrt(x^2-a^2)| E[x = a sec theta]
    B -->|No sqrt| F{Powers of sin/cos?}
    F -->|Both even| G[Power Reduction<br/>sin^2 x = 1/2(1-cos2x)]
    F -->|One odd| H[u-substitution]
    F -->|Product sinAcosB| I[Product-to-Sum]
    C --> J[Reduce to powers<br/>of sin/cos then integrate]
    D --> J
    E --> J
    G --> J
    H --> J
    I --> J

Summary of Techniques Used

Problem	Geometry	Series	Taylor/Mac	Trig Integrals
1	Circle	Maclaurin	✓	Power reduction
2	Parabola	Binomial	—	Trig sub
3	Ellipse	Power series	—	Trig sub, power reduction
4	Hyperbola	Maclaurin	✓	—
5	—	—	Maclaurin	Product-to-sum
6	Circle + Parabola	Binomial	—	—
7	Ellipse	Binomial	—	Product-to-sum
8	Parabola	Maclaurin	✓	—
9	Cylinder	Maclaurin	✓	Product-to-sum
10	Ellipse	Binomial	—	Trig sub
11	Parabola + Circle	Binomial	—	Trig sub
12	Parabola	Binomial	—	Trig sub
13	—	Binomial	—	Power reduction
14	Spiral	Taylor	✓	—
15	Ellipse	Binomial + Mac	✓	Power reduction

Key Formulas Reference

Geometry

Circle: $(x-h)^2 + (y-k)^2 = r^2$
Parabola: $(x-h)^2 = 4a(y-k)$ or $(y-k)^2 = 4a(x-h)$
Ellipse: $\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$, foci at $c^2 = a^2 - b^2$
Hyperbola: $\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1$, asymptotes $y = \pm\frac{b}{a}x$

Series

Binomial: $(1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \cdots$
Maclaurin: $f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \cdots$
Taylor: $f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \cdots$

Trigonometric Integration

Power reduction: $\sin^2 x = \frac{1-\cos 2x}{2}$, $\cos^2 x = \frac{1+\cos 2x}{2}$
Product-to-sum: $\sin A \cos B = \frac{1}{2}[\sin(A+B) + \sin(A-B)]$
Substitution patterns:
- $\sqrt{a^2 - x^2}$: use $x = a\sin\theta$
- $\sqrt{a^2 + x^2}$: use $x = a\tan\theta$
- $\sqrt{x^2 - a^2}$: use $x = a\sec\theta$

Related Resources

#mathematics #interleaved-practice #mastery #geometry #series #taylor-maclaurin #integration #fad1014