Model Answer

STEP 1 - Given Data

• Integrals: $\int_1^{\infty} \frac{1}{x^2},dx$ and $\int_1^{\infty} \frac{1}{x},dx$
• Function: $f(x,y) = \dfrac{xy}{x^2+y^2}$; limit as $(x,y)\to(0,0)$
• Acceleration: $a(t) = 6t^2 + t$
• Initial velocity: $v(0) = 4$ m/s
• Initial displacement: $s(0) = 5$ cm

Note on units: $v(0)$ is given in m/s while $s(0)$ is given in cm. This is a stated inconsistency in the problem. Following standard textbook treatment, I use the numeric values as given (the answer is presented symbolically; strict unit conversion would require converting one to the other, but the problem intends the plain numbers).

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STEP 2 - Solution

Part 1: Convergence and Divergence [2 marks]

(a) $\int_1^{\infty} \frac{1}{x^2},dx$ converges

$$\int_1^{\infty} x^{-2},dx = \lim_{b\to\infty}\left[-\frac{1}{x}\right]1^{b} = \lim{b\to\infty}\left(-\frac{1}{b}+1\right) = 1$$

Finite limit, so the integral converges to $1$.

(b) $\int_1^{\infty} \frac{1}{x},dx$ diverges

$$\int_1^{\infty}\frac{1}{x},dx = \lim_{b\to\infty}\big[\ln x\big]1^{b} = \lim{b\to\infty}\ln b = \infty$$

The limit is infinite, so the integral diverges.

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Part 2: Limit of $f(x,y)=\dfrac{xy}{x^2+y^2}$ [3 marks]

Approach along paths $y = mx$:

$$\lim_{x\to0}\frac{x(mx)}{x^2+m^2x^2} = \lim_{x\to0}\frac{mx^2}{x^2(1+m^2)} = \frac{m}{1+m^2}$$

The value depends on $m$:

• $m=0$ (along $y=0$): limit $=0$
• $m=1$ (along $y=x$): limit $=\dfrac{1}{2}$

Since different paths give different values, the limit does not exist.

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Part 3: Position Function $s(t)$ [5 marks]

Step 1: Velocity

$$v(t) = \int(6t^2+t),dt = 2t^3 + \frac{t^2}{2} + C_1$$

Apply $v(0)=4$: $;C_1 = 4$

$$v(t) = 2t^3 + \frac{t^2}{2} + 4$$

Step 2: Position

$$s(t) = \int\left(2t^3+\frac{t^2}{2}+4\right)dt = \frac{t^4}{2} + \frac{t^3}{6} + 4t + C_2$$

Apply $s(0)=5$: $;C_2 = 5$

$$\boxed{s(t) = \frac{t^4}{2} + \frac{t^3}{6} + 4t + 5}$$

Given Data

Part 1: Double integral $$\int_{3}^{2}\int_{0}^{\pi/2}(y + y^2\cos x),dx,dy$$

• Outer limits (for $y$): from $3$ to $2$
• Inner limits (for $x$): from $0$ to $\pi/2$

Part 2: Find Maclaurin series for $\cos x$ and prove it represents $\cos x$ for all $x$.

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Part 1: Evaluate the Double Integral

The limits are literally $\int_3^2$, so I will evaluate as written.

Step 1: Inner Integral (with respect to $x$)

Treat $y$ as constant: $$I_1 = \int_0^{\pi/2}(y + y^2\cos x),dx = \left[yx + y^2\sin x\right]_0^{\pi/2}$$

$$= \left(y\cdot\frac{\pi}{2} + y^2\sin\frac{\pi}{2}\right) - (0 + 0) = \frac{\pi y}{2} + y^2$$

Step 2: Outer Integral (with respect to $y$, from 3 to 2)

$$I = \int_3^2\left(\frac{\pi y}{2} + y^2\right)dy = \left[\frac{\pi y^2}{4} + \frac{y^3}{3}\right]_3^2$$

Evaluate at upper limit $y=2$: $$\frac{\pi(4)}{4} + \frac{8}{3} = \pi + \frac{8}{3}$$

Evaluate at lower limit $y=3$: $$\frac{\pi(9)}{4} + \frac{27}{3} = \frac{9\pi}{4} + 9$$

Subtract: $$I = \left(\pi + \frac{8}{3}\right) - \left(\frac{9\pi}{4} + 9\right)$$

$$= \pi - \frac{9\pi}{4} + \frac{8}{3} - 9 = \frac{4\pi - 9\pi}{4} + \frac{8 - 27}{3}$$

$$= -\frac{5\pi}{4} - \frac{19}{3}$$

$$\boxed{\int_{3}^{2}\int_{0}^{\pi/2}(y+y^2\cos x),dx,dy = -\frac{5\pi}{4} - \frac{19}{3}}$$

If the intended (standard) orientation is $\int_2^3$, the answer is $+\left(\dfrac{5\pi}{4}+\dfrac{19}{3}\right)$.

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Part 2: Maclaurin's Series for $\cos x$

Finding the Series

The Maclaurin series is: $$f(x) = \sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}x^n$$

Let $f(x)=\cos x$:

Odd derivatives vanish at $0$; even derivatives satisfy $f^{(2n)}(0)=(-1)^n$. Hence:

$$\cos x = \sum_{n=0}^{\infty}\frac{(-1)^n}{(2n)!}x^{2n} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots$$

Proof of Convergence to $\cos x$ for all $x$

By Taylor's theorem, $\cos x = P_n(x) + R_n(x)$ where the Lagrange remainder is: $$R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!}x^{n+1}, \quad c \text{ between } 0 \text{ and } x$$

Every derivative of $\cos x$ is $\pm\sin x$ or $\pm\cos x$, so: $$\left|f^{(n+1)}(c)\right| \le 1 \quad \text{for all } c$$

Therefore: $$|R_n(x)| \le \frac{|x|^{n+1}}{(n+1)!}$$

For any fixed $x$, the standard limit holds: $$\lim_{n\to\infty}\frac{|x|^{n+1}}{(n+1)!} = 0$$

(These are the terms of the convergent series for $e^{|x|}$, so they tend to $0$.)

By the squeeze theorem, $\lim_{n\to\infty}|R_n(x)| = 0$ for every real $x$.

Conclusion: Since the remainder tends to zero for all $x \in \mathbb{R}$, the Maclaurin series converges to $\cos x$ for all real $x$: $$\cos x = \sum_{n=0}^{\infty}\frac{(-1)^n}{(2n)!}x^{2n}, \quad \forall x \in \mathbb{R} \qquad \blacksquare$$

Continuity of a Function at a Point and Continuity of f(x) = √(1-x²) on [-1, 1]

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Part 1: Definition of Continuity at a Point x = a

A function f(x) is said to be continuous at a point x = a if the following three conditions are all satisfied:

1. f(a) is defined (the function has a value at x = a)
2. lim(x→a) f(x) exists (the left-hand limit equals the right-hand limit)
3. lim(x→a) f(x) = f(a) (the limit equals the function value)

If any one of these conditions fails, the function is said to be discontinuous at x = a.

Note: For continuity on a closed interval [a, b], we require:

• Continuity at every interior point (two-sided)
• Right-continuity at the left endpoint: lim(x→a⁺) f(x) = f(a)
• Left-continuity at the right endpoint: lim(x→b⁻) f(x) = f(b)

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Part 2: Show f(x) = √(1 - x²) is Continuous on [-1, 1]

Note: The interval is [-1, 1] (the question contains a typo writing [1, -1]).

Step 1: Domain Check

For f(x) = √(1 - x²) to be defined, we need:

$$1 - x^2 \geq 0 \implies x^2 \leq 1 \implies -1 \leq x \leq 1$$

So the domain of f(x) is exactly [-1, 1]. The function is defined at every point in this interval.

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Step 2: Continuity at an Interior Point x = a, where -1 < a < 1

We need to show: lim(x→a) f(x) = f(a)

$$\lim_{x \to a} f(x) = \lim_{x \to a} \sqrt{1 - x^2}$$

Since the square root function is continuous and (1 - x²) is a polynomial (hence continuous everywhere), the composition is continuous. Therefore:

$$\lim_{x \to a} \sqrt{1 - x^2} = \sqrt{1 - a^2} = f(a)$$

So f is continuous at every interior point a ∈ (-1, 1).

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Step 3: Right-Continuity at the Left Endpoint x = -1

We check the right-hand limit:

$$\lim_{x \to -1^+} f(x) = \lim_{x \to -1^+} \sqrt{1 - x^2}$$

Substituting x → -1:

$$= \sqrt{1 - (-1)^2} = \sqrt{1 - 1} = \sqrt{0} = 0$$

Also:

$$f(-1) = \sqrt{1 - (-1)^2} = \sqrt{0} = 0$$

Therefore:

$$\lim_{x \to -1^+} f(x) = 0 = f(-1) \checkmark$$

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Step 4: Left-Continuity at the Right Endpoint x = 1

We check the left-hand limit:

$$\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} \sqrt{1 - x^2}$$

Substituting x → 1:

$$= \sqrt{1 - (1)^2} = \sqrt{1 - 1} = \sqrt{0} = 0$$

Also:

$$f(1) = \sqrt{1 - 1^2} = \sqrt{0} = 0$$

Therefore:

$$\lim_{x \to 1^-} f(x) = 0 = f(1) \checkmark$$

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Conclusion

Since:

• f(x) = √(1 - x²) is defined at every point in [-1, 1],
• f is continuous at every interior point in (-1, 1),
• f is right-continuous at x = -1,
• f is left-continuous at x = 1,

Therefore, f(x) = √(1 - x²) is continuous on the closed interval [-1, 1]. $\blacksquare$

Volume of Solid Rotated About the Y-Axis

Step 1 - Extract: Given Data

• Region bounded by curves: $y = x$ and $y = x^2$
• Axis of rotation: the $y$-axis

All data present.

Step 2 - Solve

Find Intersection Points

Set the curves equal:

$$x = x^2 \implies x^2 - x = 0 \implies x(x-1) = 0$$

$$x = 0 \quad \text{and} \quad x = 1$$

Determine Upper Curve on $[0,1]$

At $x = 0.5$: $y = x = 0.5$ and $y = x^2 = 0.25$.

So $y = x$ is above $y = x^2$ on $(0,1)$.

Method: Cylindrical Shells

Rotating about the $y$-axis with the region described in terms of $x$, use the shell method:

$$V = 2\pi \int_a^b x,[f(x) - g(x)],dx$$

with $f(x) = x$, $g(x) = x^2$, $a = 0$, $b = 1$.

$$V = 2\pi \int_0^1 x,(x - x^2),dx = 2\pi \int_0^1 (x^2 - x^3),dx$$

Evaluate

$$V = 2\pi \left[ \frac{x^3}{3} - \frac{x^4}{4} \right]_0^1 = 2\pi \left( \frac{1}{3} - \frac{1}{4} \right)$$

$$V = 2\pi \cdot \frac{4 - 3}{12} = 2\pi \cdot \frac{1}{12}$$

$$\boxed{V = \frac{\pi}{6} \text{ cubic units}}$$

Verification (Washer Cross-check)

Using washers in $y$ (from $y=x^2$, $x=\sqrt{y}$ outer; from $y=x$, $x=y$ inner), over $y \in [0,1]$:

$$V = \pi \int_0^1 \left[(\sqrt{y})^2 - y^2\right]dy = \pi \int_0^1 (y - y^2),dy = \pi\left(\frac{1}{2} - \frac{1}{3}\right) = \frac{\pi}{6}$$

Both methods agree.

Result

$$V = \frac{\pi}{6} \approx 0.5236 \text{ cubic units}$$

Vector Perpendicular to a Plane Through Three Points

Given Data

• $P(1, 4, 6)$
• $Q(-2, 5, -1)$
• $R(1, -1, 1)$

Concept

A normal vector to the plane is the cross product of two vectors lying in the plane: $$\vec{n} = \overrightarrow{PQ} \times \overrightarrow{PR}$$

Step 1: Vectors in the Plane

$$\overrightarrow{PQ} = Q - P = (-2-1,\ 5-4,\ -1-6) = (-3,\ 1,\ -7)$$

$$\overrightarrow{PR} = R - P = (1-1,\ -1-4,\ 1-6) = (0,\ -5,\ -5)$$

Step 2: Cross Product

$$\vec{n} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \ -3 & 1 & -7 \ 0 & -5 & -5 \end{vmatrix}$$

i-component: $$(1)(-5) - (-7)(-5) = -5 - 35 = -40$$

j-component: $$-\big[(-3)(-5) - (-7)(0)\big] = -[15 - 0] = -15$$

k-component: $$(-3)(-5) - (1)(0) = 15 - 0 = 15$$

Result

$$\boxed{\vec{n} = -40,\vec{i} - 15,\vec{j} + 15,\vec{k} = (-40,\ -15,\ 15)}$$

Verification (dot products should be zero):

$$\vec{n} \cdot \overrightarrow{PQ} = (-40)(-3) + (-15)(1) + (15)(-7) = 120 - 15 - 105 = 0 \checkmark$$

$$\vec{n} \cdot \overrightarrow{PR} = (-40)(0) + (-15)(-5) + (15)(-5) = 0 + 75 - 75 = 0 \checkmark$$

Both dot products vanish, confirming $\vec{n}$ is perpendicular to the plane.

(Any nonzero scalar multiple, e.g. dividing by $-5$ to get $(8, 3, -3)$, is also a valid answer.)

Local Extrema and Saddle Points of $f(x,y) = x^4 + y^4 - 4xy + 1$

Step 1: Given Data

Function: $f(x,y) = x^4 + y^4 - 4xy + 1$

Step 2: First-Order Partial Derivatives (Critical Points)

$$ f_x = 4x^3 - 4y = 0 ;\Rightarrow; y = x^3 \tag{1} $$

$$ f_y = 4y^3 - 4x = 0 ;\Rightarrow; x = y^3 \tag{2} $$

Substitute (1) into (2):

$$x = (x^3)^3 = x^9 ;\Rightarrow; x^9 - x = 0 ;\Rightarrow; x(x^8 - 1) = 0$$

So $x = 0$ or $x^8 = 1 \Rightarrow x = \pm 1$.

• $x = 0 \Rightarrow y = 0$ → $(0,0)$
• $x = 1 \Rightarrow y = 1$ → $(1,1)$
• $x = -1 \Rightarrow y = -1$ → $(-1,-1)$

Step 3: Second-Derivative Test

$$f_{xx} = 12x^2, \quad f_{yy} = 12y^2, \quad f_{xy} = -4$$

$$D = f_{xx}f_{yy} - (f_{xy})^2 = 144x^2y^2 - 16$$

At $(0,0)$: $$D = 144(0)(0) - 16 = -16 < 0 ;\Rightarrow; \textbf{saddle point}$$

At $(1,1)$: $$D = 144(1)(1) - 16 = 128 > 0, \quad f_{xx} = 12 > 0 ;\Rightarrow; \textbf{local minimum}$$ $$f(1,1) = 1 + 1 - 4 + 1 = -1$$

At $(-1,-1)$: $$D = 144(1)(1) - 16 = 128 > 0, \quad f_{xx} = 12 > 0 ;\Rightarrow; \textbf{local minimum}$$ $$f(-1,-1) = 1 + 1 - 4 + 1 = -1$$

Summary

Conclusion: No local maximum. Local minimum value $-1$ at $(1,1)$ and $(-1,-1)$. Saddle point at $(0,0)$.