Taylor Series for f(x) = 1/x at a = 2

Given data

• Function: $f(x) = \dfrac{1}{x}$
• Center: $a = 2$

Step 1: Derivatives of f(x) = 1/x

$$f(x) = x^{-1}$$ $$f'(x) = -x^{-2}$$ $$f''(x) = 2x^{-3}$$ $$f'''(x) = -6x^{-4}$$

General pattern: $$f^{(n)}(x) = (-1)^n , n! , x^{-(n+1)}$$

Step 2: Evaluate at a = 2

$$f^{(n)}(2) = (-1)^n , n! , 2^{-(n+1)} = \frac{(-1)^n , n!}{2^{n+1}}$$

Values:

• $f(2) = \dfrac{1}{2}$
• $f'(2) = -\dfrac{1}{4}$
• $f''(2) = \dfrac{2}{8} = \dfrac{1}{4}$ (coefficient term $\frac{f''(2)}{2!} = \frac{1}{8}$)
• $f'''(2) = -\dfrac{6}{16} = -\dfrac{3}{8}$

Step 3: Taylor series

$$f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(2)}{n!}(x-2)^n = \sum_{n=0}^{\infty} \frac{(-1)^n , n!}{n! , 2^{n+1}}(x-2)^n$$

$$\boxed{\frac{1}{x} = \sum_{n=0}^{\infty} \frac{(-1)^n}{2^{n+1}}(x-2)^n = \frac{1}{2} - \frac{1}{4}(x-2) + \frac{1}{8}(x-2)^2 - \frac{1}{16}(x-2)^3 + \cdots}$$

Convergence Analysis

This is a geometric series. Write: $$\frac{1}{x} = \frac{1}{2 + (x-2)} = \frac{1}{2}\cdot \frac{1}{1 + \frac{x-2}{2}} = \frac{1}{2}\sum_{n=0}^\infty \left(-\frac{x-2}{2}\right)^n$$

which matches the series above.

Ratio Test: $$\left|\frac{a_{n+1}}{a_n}\right| = \left|\frac{(x-2)^{n+1}/2^{n+2}}{(x-2)^n/2^{n+1}}\right| = \frac{|x-2|}{2}$$

Convergence requires: $$\frac{|x-2|}{2} < 1 \implies |x-2| < 2 \implies 0 < x < 4$$

Endpoint check: At $x=0$ and $x=4$, $\left|\frac{x-2}{2}\right| = 1$, so terms do not tend to 0; the series diverges at both endpoints.

Conclusion:

Because a geometric series converges to $\dfrac{1}{1-r}$ (here $\frac{1}{2}\cdot\frac{1}{1+(x-2)/2} = \frac{1}{x}$) whenever it converges, the sum equals $\dfrac{1}{x}$ throughout the interval of convergence:

$$\text{The series converges to } \frac{1}{x} \text{ for } 0 < x < 4.$$

• Function: $y = 3 + \sin x$ - Interval: $[0, 2\pi]$ $$y' = \cos x$$ $$y'' = -\sin x$$ Concave up requires $y'' 0$: $$-\sin x 0 \implies \sin x < 0$$ On $[0, 2\pi]$, $\sin x < 0$ for $x \in (\pi, 2\pi)$. Concave down requires $y'' < 0$: ...

Given data

• Integral to evaluate: $\displaystyle \int \sqrt{4 - x^2}, dx$
• Form: $\sqrt{a^2 - x^2}$ with $a^2 = 4 \Rightarrow a = 2$

Solution

Step 1: Substitution

Let $x = 2\sin\theta \Rightarrow dx = 2\cos\theta, d\theta$.

Then: $$\sqrt{4 - x^2} = \sqrt{4 - 4\sin^2\theta} = \sqrt{4\cos^2\theta} = 2\cos\theta$$

Step 2: Substitute

$$\int \sqrt{4 - x^2}, dx = \int (2\cos\theta)(2\cos\theta), d\theta = \int 4\cos^2\theta, d\theta$$

Step 3: Power-reduction identity $\cos^2\theta = \dfrac{1+\cos 2\theta}{2}$

$$= 4\int \frac{1+\cos 2\theta}{2}, d\theta = 2\int (1 + \cos 2\theta), d\theta$$

$$= 2\left(\theta + \frac{\sin 2\theta}{2}\right) + C = 2\theta + \sin 2\theta + C$$

Step 4: Expand $\sin 2\theta = 2\sin\theta\cos\theta$

$$= 2\theta + 2\sin\theta\cos\theta + C$$

Step 5: Back-substitute

From $x = 2\sin\theta$:

• $\sin\theta = \dfrac{x}{2}$, so $\theta = \sin^{-1}\dfrac{x}{2}$
• $\cos\theta = \sqrt{1 - \dfrac{x^2}{4}} = \dfrac{\sqrt{4 - x^2}}{2}$

Therefore: $$2\theta = 2\sin^{-1}\frac{x}{2}$$ $$2\sin\theta\cos\theta = 2 \cdot \frac{x}{2} \cdot \frac{\sqrt{4 - x^2}}{2} = \frac{x\sqrt{4 - x^2}}{2}$$

Final Answer

$$\boxed{\int \sqrt{4 - x^2}, dx = \frac{x\sqrt{4 - x^2}}{2} + 2\sin^{-1}\left(\frac{x}{2}\right) + C}$$

Verification (by differentiation)

Let $F(x) = \dfrac{x\sqrt{4-x^2}}{2} + 2\sin^{-1}\frac{x}{2}$.

$$\frac{d}{dx}\left[\frac{x\sqrt{4-x^2}}{2}\right] = \frac{1}{2}\left(\sqrt{4-x^2} + x\cdot\frac{-x}{\sqrt{4-x^2}}\right) = \frac{1}{2}\cdot\frac{(4-x^2)-x^2}{\sqrt{4-x^2}} = \frac{4-2x^2}{2\sqrt{4-x^2}}$$

$$\frac{d}{dx}\left[2\sin^{-1}\frac{x}{2}\right] = 2\cdot\frac{1/2}{\sqrt{1-x^2/4}} = \frac{1}{\sqrt{(4-x^2)/4}} = \frac{2}{\sqrt{4-x^2}}$$

Sum: $$\frac{4-2x^2}{2\sqrt{4-x^2}} + \frac{2}{\sqrt{4-x^2}} = \frac{4-2x^2 + 4}{2\sqrt{4-x^2}} = \frac{8-2x^2}{2\sqrt{4-x^2}} = \frac{2(4-x^2)}{2\sqrt{4-x^2}} = \sqrt{4-x^2}\ \checkmark$$

The result is confirmed.

• Series: $\displaystyle\sum{n=1}^{\infty}\frac{1}{n^2+1}$ - Associated function: $f(x)=\dfrac{1}{x^2+1}$, tested on $[1,\infty)$ Let $f(x)$ be a function that is continuous, positive, and monotonically decreasing on $[1,\infty)$, and su...

STEP 1 - EXTRACT

Given data:

• Curve (implicit relation): $x^2 + y^2 = 25$ (a circle of radius 5 centered at origin)
• Point of interest: $(3, -4)$

Check that the point lies on the circle: $3^2 + (-4)^2 = 9 + 16 = 25$ ✓

All required data present.

STEP 2 - SOLVE

Definition of Implicit Differentiation

Implicit differentiation is the process of finding the derivative $\dfrac{dy}{dx}$ when $y$ is defined implicitly as a function of $x$ through an equation $F(x, y) = 0$, rather than explicitly as $y = f(x)$.

Procedure:

• Differentiate both sides of the equation with respect to $x$.
• Treat $y$ as a function of $x$, applying the chain rule to every term containing $y$ (i.e., $\frac{d}{dx}[f(y)] = f'(y)\frac{dy}{dx}$).
• Algebraically solve the resulting equation for $\dfrac{dy}{dx}$.

Finding the Slope

Step 1: Differentiate both sides with respect to $x$

$$\frac{d}{dx}(x^2 + y^2) = \frac{d}{dx}(25)$$

$$2x + 2y\frac{dy}{dx} = 0$$

Step 2: Solve for $\dfrac{dy}{dx}$

$$2y\frac{dy}{dx} = -2x \implies \frac{dy}{dx} = -\frac{x}{y}$$

Step 3: Substitute the point $(3, -4)$

$$\frac{dy}{dx}\bigg|_{(3,-4)} = -\frac{3}{-4} = \frac{3}{4}$$

Final Result

$$\boxed{\text{Slope} = \frac{3}{4} = 0.75}$$

Taylor Series for f(x) = 1/x at a = 2

Given data

• Function: $f(x) = \dfrac{1}{x}$
• Center: $a = 2$

Step 1: Derivatives of f(x) = 1/x

$$f(x) = x^{-1}$$ $$f'(x) = -x^{-2}$$ $$f''(x) = 2x^{-3}$$ $$f'''(x) = -6x^{-4}$$

General pattern: $$f^{(n)}(x) = (-1)^n , n! , x^{-(n+1)}$$

Step 2: Evaluate at a = 2

$$f^{(n)}(2) = (-1)^n , n! , 2^{-(n+1)} = \frac{(-1)^n , n!}{2^{n+1}}$$

Values:

• $f(2) = \dfrac{1}{2}$
• $f'(2) = -\dfrac{1}{4}$
• $f''(2) = \dfrac{2}{8} = \dfrac{1}{4}$ (coefficient term $\frac{f''(2)}{2!} = \frac{1}{8}$)
• $f'''(2) = -\dfrac{6}{16} = -\dfrac{3}{8}$

Step 3: Taylor series

$$f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(2)}{n!}(x-2)^n = \sum_{n=0}^{\infty} \frac{(-1)^n , n!}{n! , 2^{n+1}}(x-2)^n$$

$$\boxed{\frac{1}{x} = \sum_{n=0}^{\infty} \frac{(-1)^n}{2^{n+1}}(x-2)^n = \frac{1}{2} - \frac{1}{4}(x-2) + \frac{1}{8}(x-2)^2 - \frac{1}{16}(x-2)^3 + \cdots}$$

Convergence Analysis

This is a geometric series. Write: $$\frac{1}{x} = \frac{1}{2 + (x-2)} = \frac{1}{2}\cdot \frac{1}{1 + \frac{x-2}{2}} = \frac{1}{2}\sum_{n=0}^\infty \left(-\frac{x-2}{2}\right)^n$$

which matches the series above.

Ratio Test: $$\left|\frac{a_{n+1}}{a_n}\right| = \left|\frac{(x-2)^{n+1}/2^{n+2}}{(x-2)^n/2^{n+1}}\right| = \frac{|x-2|}{2}$$

Convergence requires: $$\frac{|x-2|}{2} < 1 \implies |x-2| < 2 \implies 0 < x < 4$$

Endpoint check: At $x=0$ and $x=4$, $\left|\frac{x-2}{2}\right| = 1$, so terms do not tend to 0; the series diverges at both endpoints.

Conclusion:

Because a geometric series converges to $\dfrac{1}{1-r}$ (here $\frac{1}{2}\cdot\frac{1}{1+(x-2)/2} = \frac{1}{x}$) whenever it converges, the sum equals $\dfrac{1}{x}$ throughout the interval of convergence:

$$\text{The series converges to } \frac{1}{x} \text{ for } 0 < x < 4.$$

• Function: $y = 3 + \sin x$ - Interval: $[0, 2\pi]$ $$y' = \cos x$$ $$y'' = -\sin x$$ Concave up requires $y'' 0$: $$-\sin x 0 \implies \sin x < 0$$ On $[0, 2\pi]$, $\sin x < 0$ for $x \in (\pi, 2\pi)$. Concave down requires $y'' < 0$: ...

Given data

• Integral to evaluate: $\displaystyle \int \sqrt{4 - x^2}, dx$
• Form: $\sqrt{a^2 - x^2}$ with $a^2 = 4 \Rightarrow a = 2$

Solution

Step 1: Substitution

Let $x = 2\sin\theta \Rightarrow dx = 2\cos\theta, d\theta$.

Then: $$\sqrt{4 - x^2} = \sqrt{4 - 4\sin^2\theta} = \sqrt{4\cos^2\theta} = 2\cos\theta$$

Step 2: Substitute

$$\int \sqrt{4 - x^2}, dx = \int (2\cos\theta)(2\cos\theta), d\theta = \int 4\cos^2\theta, d\theta$$

Step 3: Power-reduction identity $\cos^2\theta = \dfrac{1+\cos 2\theta}{2}$

$$= 4\int \frac{1+\cos 2\theta}{2}, d\theta = 2\int (1 + \cos 2\theta), d\theta$$

$$= 2\left(\theta + \frac{\sin 2\theta}{2}\right) + C = 2\theta + \sin 2\theta + C$$

Step 4: Expand $\sin 2\theta = 2\sin\theta\cos\theta$

$$= 2\theta + 2\sin\theta\cos\theta + C$$

Step 5: Back-substitute

From $x = 2\sin\theta$:

• $\sin\theta = \dfrac{x}{2}$, so $\theta = \sin^{-1}\dfrac{x}{2}$
• $\cos\theta = \sqrt{1 - \dfrac{x^2}{4}} = \dfrac{\sqrt{4 - x^2}}{2}$

Therefore: $$2\theta = 2\sin^{-1}\frac{x}{2}$$ $$2\sin\theta\cos\theta = 2 \cdot \frac{x}{2} \cdot \frac{\sqrt{4 - x^2}}{2} = \frac{x\sqrt{4 - x^2}}{2}$$

Final Answer

$$\boxed{\int \sqrt{4 - x^2}, dx = \frac{x\sqrt{4 - x^2}}{2} + 2\sin^{-1}\left(\frac{x}{2}\right) + C}$$

Verification (by differentiation)

Let $F(x) = \dfrac{x\sqrt{4-x^2}}{2} + 2\sin^{-1}\frac{x}{2}$.

$$\frac{d}{dx}\left[\frac{x\sqrt{4-x^2}}{2}\right] = \frac{1}{2}\left(\sqrt{4-x^2} + x\cdot\frac{-x}{\sqrt{4-x^2}}\right) = \frac{1}{2}\cdot\frac{(4-x^2)-x^2}{\sqrt{4-x^2}} = \frac{4-2x^2}{2\sqrt{4-x^2}}$$

$$\frac{d}{dx}\left[2\sin^{-1}\frac{x}{2}\right] = 2\cdot\frac{1/2}{\sqrt{1-x^2/4}} = \frac{1}{\sqrt{(4-x^2)/4}} = \frac{2}{\sqrt{4-x^2}}$$

Sum: $$\frac{4-2x^2}{2\sqrt{4-x^2}} + \frac{2}{\sqrt{4-x^2}} = \frac{4-2x^2 + 4}{2\sqrt{4-x^2}} = \frac{8-2x^2}{2\sqrt{4-x^2}} = \frac{2(4-x^2)}{2\sqrt{4-x^2}} = \sqrt{4-x^2}\ \checkmark$$

The result is confirmed.

• Series: $\displaystyle\sum{n=1}^{\infty}\frac{1}{n^2+1}$ - Associated function: $f(x)=\dfrac{1}{x^2+1}$, tested on $[1,\infty)$ Let $f(x)$ be a function that is continuous, positive, and monotonically decreasing on $[1,\infty)$, and su...

STEP 1 - EXTRACT

Given data:

• Curve (implicit relation): $x^2 + y^2 = 25$ (a circle of radius 5 centered at origin)
• Point of interest: $(3, -4)$

Check that the point lies on the circle: $3^2 + (-4)^2 = 9 + 16 = 25$ ✓

All required data present.

STEP 2 - SOLVE

Definition of Implicit Differentiation

Implicit differentiation is the process of finding the derivative $\dfrac{dy}{dx}$ when $y$ is defined implicitly as a function of $x$ through an equation $F(x, y) = 0$, rather than explicitly as $y = f(x)$.

Procedure:

• Differentiate both sides of the equation with respect to $x$.
• Treat $y$ as a function of $x$, applying the chain rule to every term containing $y$ (i.e., $\frac{d}{dx}[f(y)] = f'(y)\frac{dy}{dx}$).
• Algebraically solve the resulting equation for $\dfrac{dy}{dx}$.

Finding the Slope

Step 1: Differentiate both sides with respect to $x$

$$\frac{d}{dx}(x^2 + y^2) = \frac{d}{dx}(25)$$

$$2x + 2y\frac{dy}{dx} = 0$$

Step 2: Solve for $\dfrac{dy}{dx}$

$$2y\frac{dy}{dx} = -2x \implies \frac{dy}{dx} = -\frac{x}{y}$$

Step 3: Substitute the point $(3, -4)$

$$\frac{dy}{dx}\bigg|_{(3,-4)} = -\frac{3}{-4} = \frac{3}{4}$$

Final Result

$$\boxed{\text{Slope} = \frac{3}{4} = 0.75}$$

Model Answer: Asymptotes of $f(x) = x - \sqrt{x^2 + 16}$

Given Data

• Function: $f(x) = x - \sqrt{x^2 + 16}$
• Domain check: $x^2 + 16 > 0$ for all real $x$, so domain is all of $\mathbb{R}$.

(a) Definitions

Horizontal Asymptote: A line $y = L$ is a horizontal asymptote of $y = f(x)$ if $$\lim_{x \to +\infty} f(x) = L \quad \text{or} \quad \lim_{x \to -\infty} f(x) = L.$$

Vertical Asymptote: A line $x = a$ is a vertical asymptote of $y = f(x)$ if $$\lim_{x \to a^{+}} f(x) = \pm\infty \quad \text{or} \quad \lim_{x \to a^{-}} f(x) = \pm\infty.$$

(An oblique/slant asymptote is a line $y = mx + c$, $m \neq 0$, that the curve approaches at infinity.)

(b) Finding Asymptotes

Vertical Asymptotes

Since $x^2 + 16 \geq 16 > 0$ for all $x$, the function is defined and finite everywhere. There is no finite value of $x$ where $f(x)$ becomes unbounded.

No vertical asymptotes.

Behaviour as $x \to +\infty$

Rationalize: $$f(x) = x - \sqrt{x^2 + 16} = \frac{(x - \sqrt{x^2+16})(x + \sqrt{x^2+16})}{x + \sqrt{x^2+16}} = \frac{x^2 - (x^2 + 16)}{x + \sqrt{x^2 + 16}} = \frac{-16}{x + \sqrt{x^2 + 16}}$$

As $x \to +\infty$, denominator $\to +\infty$, so $$\lim_{x \to +\infty} f(x) = 0.$$

Horizontal asymptote: $y = 0$ (to the right).

Behaviour as $x \to -\infty$

Using the rationalized form, as $x \to -\infty$ we have $\sqrt{x^2+16} \to +\infty$ while $x \to -\infty$. Note $\sqrt{x^2+16} \approx |x| = -x$, so $$x + \sqrt{x^2 + 16} \to 0^{+}, \qquad \frac{-16}{x + \sqrt{x^2+16}} \to -\infty.$$

So $\lim_{x \to -\infty} f(x) = -\infty$; no horizontal asymptote on the left.

Check for oblique asymptote as $x \to -\infty$: $$m = \lim_{x \to -\infty}\frac{f(x)}{x} = \lim_{x \to -\infty}\frac{x - \sqrt{x^2+16}}{x}.$$ With $x < 0$, $\sqrt{x^2+16} = -x\sqrt{1 + 16/x^2}$, so $$f(x) = x - (-x)\sqrt{1 + \tfrac{16}{x^2}} = x + x\sqrt{1 + \tfrac{16}{x^2}}.$$ $$\frac{f(x)}{x} = 1 + \sqrt{1 + \tfrac{16}{x^2}} \to 1 + 1 = 2 = m.$$ $$c = \lim_{x \to -\infty}\big(f(x) - 2x\big) = \lim_{x \to -\infty}\big(-x + x\sqrt{1+\tfrac{16}{x^2}}\big) = \lim_{x \to -\infty} x\Big(\sqrt{1+\tfrac{16}{x^2}} - 1\Big).$$ Expand: $\sqrt{1+\tfrac{16}{x^2}} - 1 \approx \tfrac{8}{x^2}$, so $x \cdot \tfrac{8}{x^2} = \tfrac{8}{x} \to 0$. Thus $c = 0$, giving the oblique asymptote $y = 2x$ as $x \to -\infty$.

Final Answer

• Vertical asymptote: none.
• Horizontal asymptote: $y = 0$ (as $x \to +\infty$).
• Oblique asymptote: $y = 2x$ (as $x \to -\infty$).

Definition: A function $f$ is continuous on a closed interval $[a,b]$ if it is continuous at every interior point and continuous from the right at $a$ and from the left at $b$. Given: $f(x) = 1 - \sqrt{1 - x^2}$, interval $[-1, 1]$. Step...

Model Answer

1. Initial Value Problem (1 mark)

An Initial Value Problem (IVP) is a differential equation together with the value of the unknown function (and possibly its derivatives) specified at a single point, called the initial condition.

General form: $$\frac{dy}{dx} = f(x,y), \qquad y(x_0) = y_0$$

The initial condition pins down the arbitrary constant in the general solution, giving a unique particular solution.

2. Solution of the IVP (4 marks)

Given: $x\dfrac{dy}{dx} - y = x^2$, with $y(2) = 5$.

Step 1: Standard form. Divide by $x$: $$\frac{dy}{dx} - \frac{1}{x},y = x$$ This is linear with $P(x) = -\dfrac{1}{x}$, $Q(x) = x$.

Step 2: Integrating factor. $$\mu(x) = e^{\int -\frac{1}{x},dx} = e^{-\ln x} = \frac{1}{x}$$

Step 3: Multiply through by $\mu$. $$\frac{1}{x}\frac{dy}{dx} - \frac{y}{x^2} = 1 \quad\Longrightarrow\quad \frac{d}{dx}!\left(\frac{y}{x}\right) = 1$$

Step 4: Integrate. $$\frac{y}{x} = x + C \quad\Longrightarrow\quad y = x^2 + Cx$$

Step 5: Apply $y(2) = 5$. $$5 = (2)^2 + C(2) = 4 + 2C \implies 2C = 1 \implies C = \frac{1}{2}$$

Particular solution: $$\boxed{,y = x^2 + \frac{x}{2},}$$

Check: $y(2) = 4 + 1 = 5$ ✓

3. Evaluate the Limit (5 marks)

Given: $$\lim_{x\to\infty} \frac{3x^2 - 5x + 2}{5x^2 + 8x + 7}$$

Divide numerator and denominator by the highest power $x^2$: $$= \lim_{x\to\infty} \frac{3 - \dfrac{5}{x} + \dfrac{2}{x^2}}{5 + \dfrac{8}{x} + \dfrac{7}{x^2}}$$

As $x \to \infty$, all terms $\dfrac{5}{x}, \dfrac{2}{x^2}, \dfrac{8}{x}, \dfrac{7}{x^2} \to 0$: $$= \frac{3 - 0 + 0}{5 + 0 + 0} = \frac{3}{5}$$

$$\boxed{,\lim_{x\to\infty} \frac{3x^2 - 5x + 2}{5x^2 + 8x + 7} = \frac{3}{5},}$$

• Integrand: $x \sin x$ - Limits: from $0$ to $\pi$ Using $\int u , dv = uv - \int v , du$. Let: - $u = x \Rightarrow du = dx$ - $dv = \sin x , dx \Rightarrow v = -\cos x$ $$\int0^{\pi} x \sin x , dx = \Big[-x\cos x\Big]0^{\pi} - \in...

Volume of Solid of Revolution about the y-axis

STEP 1 - EXTRACT: Given data

• Curve 1: $y = x$
• Curve 2: $y = x^2$
• Axis of rotation: the y-axis
• Region: bounded between the two curves

All data needed is present.

STEP 2 - SOLVE

Step 1: Intersection points

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

So $x = 0$ or $x = 1$, giving intersection points $(0,0)$ and $(1,1)$.

Step 2: Determine top curve on $[0,1]$

For $0 < x < 1$, since $x > x^2$, the line $y = x$ lies above the parabola $y = x^2$.

Step 3: Shell method (rotation about y-axis)

$$V = 2\pi \int_a^b (\text{radius})(\text{height}), dx$$

• Radius $= x$
• Height $= x - x^2$
• Limits: $x = 0$ to $x = 1$

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

Step 4: Evaluate

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

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

$$V = 2\pi \left(\frac{1}{3} - \frac{1}{4}\right) = 2\pi \cdot \frac{1}{12} = \frac{\pi}{6}$$

Verification via disk/washer method (integrating in $y$):

For a horizontal strip at height $y$ (with $0 \le y \le 1$):

• $y = x \Rightarrow x = y$ (inner boundary, smaller x)
• $y = x^2 \Rightarrow x = \sqrt{y}$ (outer boundary, larger x)

Outer radius $= \sqrt{y}$, inner radius $= y$:

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

$$V = \pi \left[\frac{y^2}{2} - \frac{y^3}{3}\right]_0^1 = \pi\left(\frac{1}{2} - \frac{1}{3}\right) = \pi \cdot \frac{1}{6} = \frac{\pi}{6}$$

Both methods agree.

Final Answer:

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

Given Data

• Function: $f(x,y) = \sqrt{x}, y^3 + x^4 y$
• Evaluation point: $(-4, 1)$

Step 1: Compute $f_x$

Treat $y$ as constant. Write $\sqrt{x} = x^{1/2}$.

$$f_x = \frac{1}{2}x^{-1/2},y^3 + 4x^3 y = \frac{y^3}{2\sqrt{x}} + 4x^3 y$$

Step 2: Compute $f_y$

Treat $x$ as constant.

$$f_y = 3\sqrt{x},y^2 + x^4$$

Step 3: Compute $f_{xy}$

Differentiate $f_x$ with respect to $y$:

$$f_{xy} = \frac{\partial}{\partial y}\left(\frac{y^3}{2\sqrt{x}} + 4x^3 y\right) = \frac{3y^2}{2\sqrt{x}} + 4x^3$$

Step 4: Evaluate at $(-4, 1)$

Domain note: Since $x = -4 < 0$, $\sqrt{x} = \sqrt{-4}$ is not real. The function (and its derivatives) are not defined at $(-4,1)$ in the real domain. Proceeding formally using $\sqrt{-4} = 2i$:

$f_x(-4,1)$: $$f_x = \frac{1^3}{2\sqrt{-4}} + 4(-4)^3(1) = \frac{1}{2(2i)} - 256 = \frac{1}{4i} - 256 = -\frac{i}{4} - 256$$

So $f_x(-4,1) = -256 - \dfrac{i}{4}$.

$f_y(-4,1)$: $$f_y = 3\sqrt{-4}(1)^2 + (-4)^4 = 3(2i) + 256 = 256 + 6i$$

$f_{xy}(-4,1)$: $$f_{xy} = \frac{3(1)^2}{2\sqrt{-4}} + 4(-4)^3 = \frac{3}{4i} - 256 = -\frac{3i}{4} - 256$$

So $f_{xy}(-4,1) = -256 - \dfrac{3i}{4}$.

Summary

Conclusion: The derivative expressions are correct. However, because $\sqrt{-4}$ is not real, the function is not defined at $(-4,1)$ over the reals. The point is almost certainly a typo for $(4,1)$, which would give real values:

• $f_x(4,1) = \frac{1}{4} + 256 = 256.25$
• $f_y(4,1) = 6 + 256 = 262$
• $f_{xy}(4,1) = \frac{3}{4} + 256 = 256.75$

The derivative formulas and formal evaluations above are correct, and $\sqrt{-4} = 2i$ has been simplified to give explicit complex values.

Verification of Mean Value Theorem

Step 1 - Given Data

• Function: $f(x) = x^2 + 3x + 1$
• Interval: $[a, b] = [-1, 1]$, so $a = -1$, $b = 1$

Step 2 - Solve

MVT Statement

If $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then there exists $c \in (a,b)$ such that: $$f'(c) = \frac{f(b) - f(a)}{b - a}$$

Verify Conditions

• $f(x) = x^2 + 3x + 1$ is a polynomial, hence continuous on $[-1, 1]$.
• Being a polynomial, it is differentiable on $(-1, 1)$.

Both conditions hold, so MVT is applicable.

Compute f(a) and f(b)

$$f(-1) = (-1)^2 + 3(-1) + 1 = 1 - 3 + 1 = -1$$ $$f(1) = (1)^2 + 3(1) + 1 = 1 + 3 + 1 = 5$$

Average Rate of Change

$$\frac{f(1) - f(-1)}{1 - (-1)} = \frac{5 - (-1)}{2} = \frac{6}{2} = 3$$

Derivative

$$f'(x) = 2x + 3$$

Solve f'(c) = 3

$$2c + 3 = 3 \implies 2c = 0 \implies c = 0$$

Check

$c = 0 \in (-1, 1)$ ✓ and $f'(0) = 2(0) + 3 = 3$ ✓

Conclusion

There exists $c = 0 \in (-1, 1)$ such that $$f'(0) = 3 = \frac{f(1) - f(-1)}{1 - (-1)}.$$ Hence the Mean Value Theorem is verified.

Part 1: - Function: $f(x,y,z) = x^3 - xy^2 + z$ - Point: $P(1,0,0)$ - Direction: $\vec{v} = 2\vec{i} - \vec{j} + \vec{k}$ Part 2: - Region bounded by $y = \sqrt{x}$, line $y = 1$, and $x = 4$ - Axis of revolution: $y = 1$ All data presen...

Model Answer: Area Between Curves and Volume of Revolution

Part 1: Definition of Area Between Two Curves (1 mark)

The area between two curves is the region bounded between two curves in a plane over a common interval. If $y = f(x)$ and $y = g(x)$ with $f(x) \ge g(x)$ on $[a,b]$, then:

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

Part 2: Area Enclosed by $y = 2 - x^2$ and $y = -x$ (3 marks)

Step 1: Intersection points

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

So $x = -1$ and $x = 2$.

Step 2: Upper curve

At $x = 0$: parabola gives $2$, line gives $0$. So parabola is above on $[-1, 2]$.

Step 3: Integrate

$$A = \int_{-1}^{2} \big[(2 - x^2) - (-x)\big],dx = \int_{-1}^{2} (2 + x - x^2),dx$$

$$= \left[2x + \frac{x^2}{2} - \frac{x^3}{3}\right]_{-1}^{2}$$

At $x = 2$: $;4 + 2 - \frac{8}{3} = 6 - \frac{8}{3} = \frac{10}{3}$

At $x = -1$: $;-2 + \frac{1}{2} + \frac{1}{3} = -2 + \frac{5}{6} = -\frac{7}{6}$

$$A = \frac{10}{3} - \left(-\frac{7}{6}\right) = \frac{20}{6} + \frac{7}{6} = \frac{27}{6} = \frac{9}{2} \text{ square units}$$

Part 3: Definition of Volume Integral

A volume integral is the integral of a function over a three-dimensional region $V$, giving the volume (or accumulated quantity) of the solid:

$$V = \iiint_V dV$$

For a solid of revolution using the washer method:

$$V = \pi \int_a^b \Big([R]^2 - [r]^2\Big),dx$$

Part 4: Volume of Solid Revolved about $y = 1$ (6 marks)

Given: curve $y^2 = x$ (i.e. $y = \sqrt{x}$), lines $y = 1$ and $x = 4$; revolve about $y = 1$.

Step 1: Identify the region

The region lies between the curve $y = \sqrt{x}$ (upper) and the line $y = 1$ (lower), from where they meet to $x = 4$.

• $y = \sqrt{x}$ meets $y = 1$ at $x = 1$.
• Right boundary $x = 4$, where $y = \sqrt{4} = 2$.

So $x$ runs from $1$ to $4$, with $\sqrt{x} \ge 1$ on this interval.

Step 2: Disk method about $y = 1$

Revolving about the horizontal line $y = 1$ with vertical strips gives disks of radius

$$R(x) = \sqrt{x} - 1$$

(The region touches $y = 1$ so the inner radius is zero.)

$$V = \pi \int_1^4 (\sqrt{x} - 1)^2 , dx$$

Step 3: Expand and integrate

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

$$V = \pi \int_1^4 \left(x - 2x^{1/2} + 1\right) dx = \pi \left[\frac{x^2}{2} - \frac{4}{3}x^{3/2} + x\right]_1^4$$

At $x = 4$: $;\frac{16}{2} - \frac{4}{3}(8) + 4 = 8 - \frac{32}{3} + 4 = 12 - \frac{32}{3} = \frac{4}{3}$

At $x = 1$: $;\frac{1}{2} - \frac{4}{3} + 1 = \frac{3}{2} - \frac{4}{3} = \frac{9 - 8}{6} = \frac{1}{6}$

$$V = \pi\left(\frac{4}{3} - \frac{1}{6}\right) = \pi\left(\frac{8}{6} - \frac{1}{6}\right) = \frac{7\pi}{6} \text{ cubic units}$$

Common mistake: setting the solid up as horizontal washers with outer radius $4$ and inner radius $y^2$, that is $\int_1^2 (16 - y^4),dy$. That set-up does not represent revolving the given region about $y = 1$. The region bounded by $y^2 = x$, $y = 1$ and $x = 4$ revolved about $y = 1$ gives disks of radius $\sqrt{x} - 1$, yielding $\frac{7\pi}{6}$, not $\frac{49\pi}{5}$.

Final Answers:

• Area $= \dfrac{9}{2}$ square units
• Volume $= \dfrac{7\pi}{6}$ cubic units

• Function: $f(x,y,z) = x^3 - xy^2 - z$ - Point: $P(1, 1, 0)$ - Direction vector: $\mathbf{v} = 2\mathbf{i} - 3\mathbf{j} + 6\mathbf{k}$ The directional derivative is $D{\mathbf{u}}f = \nabla f \cdot \hat{\mathbf{u}}$, where $\hat{\mathb...