Model Answer

Given Data

Part 1: $y = \dfrac{\tan^{-1} x}{\sqrt{x}}$

Part 2: Curves $y = -x$ and $x = y^2 + 3y$

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(a) Derivative

Let $u = \tan^{-1} x$ and $v = \sqrt{x} = x^{1/2}$.

$$\frac{du}{dx} = \frac{1}{1+x^2}, \qquad \frac{dv}{dx} = \frac{1}{2\sqrt{x}}$$

Quotient rule: $\dfrac{dy}{dx} = \dfrac{v,\frac{du}{dx} - u,\frac{dv}{dx}}{v^2}$

$$\frac{dy}{dx} = \frac{\sqrt{x}\cdot\dfrac{1}{1+x^2} - \tan^{-1}x \cdot \dfrac{1}{2\sqrt{x}}}{x}$$

Numerator over common denominator $2\sqrt{x}(1+x^2)$:

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

$$\boxed{\frac{dy}{dx} = \frac{2x - (1+x^2)\tan^{-1}x}{2x^{3/2}(1+x^2)}}$$

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(b) Area bounded by $y = -x$ and $x = y^2 + 3y$

Line: $x = -y$. Parabola: $x = y^2 + 3y$.

Intersections: Set $-y = y^2 + 3y$

$$0 = y^2 + 4y = y(y+4) \implies y = 0 \text{ or } y = -4$$

Points: $(0,0)$ and $(4,-4)$.

Rightmost curve: At $y = -2$: line gives $x = 2$; parabola gives $x = 4-6 = -2$. Line is to the right.

Integrate in $y$:

$$A = \int_{-4}^{0}\left[(-y) - (y^2+3y)\right]dy = \int_{-4}^{0}\left(-y^2 - 4y\right)dy$$

$$= \left[-\frac{y^3}{3} - 2y^2\right]_{-4}^{0}$$

At $y=0$: $0$. At $y=-4$: $-\dfrac{-64}{3} - 2(16) = \dfrac{64}{3} - 32 = \dfrac{-32}{3}$.

$$A = 0 - \left(-\frac{32}{3}\right) = \frac{32}{3}$$

$$\boxed{A = \frac{32}{3} \text{ square units} \approx 10.67}$$

Both parts verified.

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.

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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: $y(2) = 4 + 1 = 5$ ✓

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

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

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.

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: $(-4, 1)$

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Step 1: Compute $f_x$

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

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

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Step 2: Compute $f_y$

Treat $x$ as constant.

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

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Step 3: Compute $f_{xy}$

Differentiate $f_x$ with respect to $y$:

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

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

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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.

Continuity Test at $x = 2$

Given data

$$f(x) = \begin{cases} \dfrac{x^2 - 2x}{x - 2}, & x \ne 2 \[2mm] 1, & x = 2 \end{cases}$$

Point to test: $x = 2$.

Conditions for continuity at $x = 2$

A function $f$ is continuous at $x = a$ if:

1. $f(a)$ is defined,
2. $\lim_{x \to a} f(x)$ exists,
3. $\lim_{x \to a} f(x) = f(a)$.

Step 1: Value of the function

$$f(2) = 1 \quad \text{(defined)}$$

Step 2: Limit as $x \to 2$

For $x \ne 2$:

$$f(x) = \frac{x^2 - 2x}{x - 2} = \frac{x(x - 2)}{x - 2} = x$$

Left-hand limit: $$\lim_{x \to 2^-} f(x) = 2$$

Right-hand limit: $$\lim_{x \to 2^+} f(x) = 2$$

Since LHL = RHL, the limit exists: $$\lim_{x \to 2} f(x) = 2$$

Step 3: Compare

$$\lim_{x \to 2} f(x) = 2, \qquad f(2) = 1$$

Since $$\lim_{x \to 2} f(x) = 2 \ne 1 = f(2),$$

the third condition fails.

Conclusion

The function is discontinuous at $x = 2$.

Because the two-sided limit exists (equals $2$) but does not equal the function value $f(2) = 1$, this is a removable discontinuity. Redefining $f(2) = 2$ would make the function continuous. (It is not a jump discontinuity, since the left and right limits are equal.)