110 marksNumericalIntroductionAnswer

If $\vec{a}=(4,0,3)$ and $\vec{b}=(-2,1,5)$ , find $|\vec{a}|$ , $3\vec{b}$ , $\vec{a}+\vec{b}$ and $2\vec{a}+5\vec{b}$ .

Estimate the value of $\lim_{x \to 0} \frac{\sqrt{x^2 + 9} - 3}{x^2}$ [5+5]

If $\vec{a}=(4,0,3)$ and $\vec{b}=(-2,1,5)$, find $|\vec{a}|$, $3\vec{b}$, $\vec{a}+\vec{b}$ and $2\vec{a}+5\vec{b}$.

Estimate the value of $\lim_{x \to 0} \frac{\sqrt{x^2 + 9} - 3}{x^2}$ [5+5]

Given data: - $\vec{a} = (4, 0, 3)$ - $\vec{b} = (-2, 1, 5)$ - Limit: $\displaystyle \lim{x \to 0} \frac{\sqrt{x^2 + 9} - 3}{x^2}$ Required: $\vec{a}$ , $3\vec{b}$ , $\vec{a}+\vec{b}$ , $2\vec{a}+5\vec{b}$ , and the limit value. All data pre...

Given data: - $\vec{a} = (4, 0, 3)$ - $\vec{b} = (-2, 1, 5)$ - Limit: $\displaystyle \lim{x \to 0} \frac{\sqrt{x^2 + 9} - 3}{x^2}$ Required: $\vec{a}$, $3\vec{b}$, $\vec{a}+\vec{b}$, $2\vec{a}+5\vec{b}$, and the limit value. All data pre...

210 marksNumericalLinear mathematical modelAnswer

As dry air moves upward, it expands and cools. If the ground temperature is $20°C$ and the temperature at height of 1 km is $10°C$ , express the temperature $T$ (in $°C$ ) as a function of the height $h$ (in kilometer), assuming that linear model is appropriate.

(a) Draw a graph of the function in part (b). What does the slope represent?

(c) What is the temperature at a height of 2.5 km?

[5+5]

As dry air moves upward, it expands and cools. If the ground temperature is $20°C$ and the temperature at height of 1 km is $10°C$, express the temperature $T$ (in $°C$) as a function of the height $h$ (in kilometer), assuming that linear model is appropriate.

(a) Draw a graph of the function in part (b). What does the slope represent?

(c) What is the temperature at a height of 2.5 km?

[5+5]

Temperature as a Linear Function of Height

STEP 1 - Given Data

Ground level: $h = 0$ km, $T = 20^{\circ}C$ → point $(0, 20)$
At $h = 1$ km: $T = 10^{\circ}C$ → point $(1, 10)$
Linear model assumed: $T = mh + b$
Required: temperature at $h = 2.5$ km

STEP 2 - Solution

Building the Linear Model

Slope:

$m = \frac{10 - 20}{1 - 0} = \frac{-10}{1} = -10$

Intercept (from point $(0,20)$ ):

$b = 20$

Therefore:

$\boxed{T(h) = 20 - 10h}$

with $T$ in $^{\circ}C$ and $h$ in km.

Part (a): Graph and Slope Meaning

The graph is a straight line through $(0,20)$ with negative slope $-10$ :

T (°C)
 20 |* (0,20)
    |  \
 10 |    * (1,10)
    |      \
  0 |________*___(2,0)_____ h (km)
    |          \
 -5 |            * (2.5,-5)
    +----+----+----+----+---
    0    1    2   2.5   3

Key points: $(0,20),\ (1,10),\ (2,0),\ (2.5,-5)$ .

Slope interpretation: The slope $m = -10\ ^{\circ}C/\text{km}$ is the rate of change of temperature with height. For each 1 km rise in altitude, the temperature falls by $10^{\circ}C$ (the lapse rate).

Part (c): Temperature at $h = 2.5$ km

$T(2.5) = 20 - 10(2.5) = 20 - 25 = -5$

$\boxed{T(2.5) = -5^{\circ}C}$

Final Results

$T(h) = 20 - 10h$
Slope $= -10^{\circ}C/\text{km}$ (rate of temperature decrease per km)
$T(2.5\text{ km}) = -5^{\circ}C$

Temperature as a Linear Function of Height

STEP 1 - Given Data

Ground level: $h = 0$ km, $T = 20^{\circ}C$ → point $(0, 20)$
At $h = 1$ km: $T = 10^{\circ}C$ → point $(1, 10)$
Linear model assumed: $T = mh + b$
Required: temperature at $h = 2.5$ km

STEP 2 - Solution

Building the Linear Model

Slope:

$$m = \frac{10 - 20}{1 - 0} = \frac{-10}{1} = -10$$

Intercept (from point $(0,20)$):

$$b = 20$$

Therefore:

$$\boxed{T(h) = 20 - 10h}$$

with $T$ in $^{\circ}C$ and $h$ in km.

Part (a): Graph and Slope Meaning

The graph is a straight line through $(0,20)$ with negative slope $-10$:

T (°C)
 20 |* (0,20)
    |  \
 10 |    * (1,10)
    |      \
  0 |________*___(2,0)_____ h (km)
    |          \
 -5 |            * (2.5,-5)
    +----+----+----+----+---
    0    1    2   2.5   3

Key points: $(0,20),\ (1,10),\ (2,0),\ (2.5,-5)$.

Slope interpretation: The slope $m = -10\ ^{\circ}C/\text{km}$ is the rate of change of temperature with height. For each 1 km rise in altitude, the temperature falls by $10^{\circ}C$ (the lapse rate).

Part (c): Temperature at $h = 2.5$ km

$$T(2.5) = 20 - 10(2.5) = 20 - 25 = -5$$

$$\boxed{T(2.5) = -5^{\circ}C}$$

Final Results

$T(h) = 20 - 10h$
Slope $= -10^{\circ}C/\text{km}$ (rate of temperature decrease per km)
$T(2.5\text{ km}) = -5^{\circ}C$

310 marksNumericalArea of surface of revolutionAnswer

The area of the parabola $y = x^2$ from (1,1) to (2,4) is rotated about the y-axis. Find the area of the resulting surface. Find the solution of the equation $y^2 dy = x^2 dx$ that satisfies the initial condition $y(0) = 2$ . [5+5]

The area of the parabola $y = x^2$ from (1,1) to (2,4) is rotated about the y-axis. Find the area of the resulting surface. Find the solution of the equation $y^2 dy = x^2 dx$ that satisfies the initial condition $y(0) = 2$. [5+5]

(a) Surface Area of Revolution of $y = x^2$ about the Y-axis - Curve: $y = x^2$ - Segment from $(1,1)$ to $(2,4)$ - Axis of rotation: y-axis For rotation about the y-axis: $$S = 2\pi \int x , ds, \quad ds = \sqrt{1 + \left(\frac{dx}{dy}...

45 marksNumericalDefinite integralAnswer

Integrate $\int_0^1 x^2 \sqrt{x^3 + 1} dx$ [5]

Definite integral to evaluate: $\int0^1 x^2\sqrt{x^3+1}\, dx$ Lower limit $x=0$ , upper limit $x=1$ . Let $u = x^3 + 1$ . $\frac{du}{dx} = 3x^2 \implies du = 3x^2\, dx \implies x^2\, dx = \frac{du}{3}$ -

Definite integral to evaluate: $$\int0^1 x^2\sqrt{x^3+1}, dx$$ Lower limit $x=0$, upper limit $x=1$. Let $u = x^3 + 1$. $$\frac{du}{dx} = 3x^2 \implies du = 3x^2, dx \implies x^2, dx = \frac{du}{3}$$ -

55 marksTaylor's and Maclaurin's seriesAnswer

Find the Maclaurin series expansion of $f(x) = e^x$ at $x = 0$ . [5]

Find the Maclaurin series expansion of $f(x) = e^x$ at $x = 0$. [5]

The Maclaurin series is a special case of the Taylor series expanded about x = 0. It is given by: $f(x) = f(0) + f'(0)\cdot x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots = \sum{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}x^n$ --- Si...

The Maclaurin series is a special case of the Taylor series expanded about x = 0. It is given by: $$f(x) = f(0) + f'(0)\cdot x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots = \sum{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}x^n$$ --- Si...

65 marksNumericalCurve sketchingAnswer

Find where the function $f(x) = 3x^4 - 4x^3 - 12x^2 + 5$ is increasing and where it is decreasing. [5]

$f(x) = 3x^4 - 4x^3 - 12x^2 + 5$ $f'(x) = 12x^3 - 12x^2 - 24x$ $12x^3 - 12x^2 - 24x = 0$ $12x(x^2 - x - 2) = 0$ $12x(x-2)(x+1) = 0$ $x = -1,\ 0,\ 2$ Interval Test $12x$ $(x-2)$ $(x+1)$ $f'(x)$ Behaviour --------------------- ...

$f(x) = 3x^4 - 4x^3 - 12x^2 + 5$ $$f'(x) = 12x^3 - 12x^2 - 24x$$ $$12x^3 - 12x^2 - 24x = 0$$ $$12x(x^2 - x - 2) = 0$$ $$12x(x-2)(x+1) = 0$$ $$x = -1,\ 0,\ 2$$ Interval Test $12x$ $(x-2)$ $(x+1)$ $f'(x)$ Behaviour --------------------- ...

75 marksNumericalReview of derivativeAnswer

Find y' if $x^3 + y^3 = 6xy$ . [5]

Find y' if $x^3 + y^3 = 6xy$. [5]

Equation (folium of Descartes): $x^3 + y^3 = 6xy$ Find: $y' = \dfrac{dy}{dx}$ Differentiate both sides with respect to $x$ : $\frac{d}{dx}(x^3) + \frac{d}{dx}(y^3) = \frac{d}{dx}(6xy)$ Left side (chain rule on $y^3$ ): $$3x^2 + 3y^2,y'

Equation (folium of Descartes): $x^3 + y^3 = 6xy$ Find: $y' = \dfrac{dy}{dx}$ Differentiate both sides with respect to $x$: $$\frac{d}{dx}(x^3) + \frac{d}{dx}(y^3) = \frac{d}{dx}(6xy)$$ Left side (chain rule on $y^3$): $$3x^2 + 3y^2,y'

85 marksNumericalRange and domain of functions and their GrAnswer

Sketch the graph and find the domain and range of the function $f(x) = 2x - 1$ . [5]

Sketch the graph and find the domain and range of the function $f(x) = 2x - 1$. [5]

$f(x) = 2x - 1, \quad \text{i.e., } y = 2x - 1$ This is a linear function (degree 1 polynomial) with slope $m = 2$ and y-intercept $-1$ . --- A polynomial function has no restrictions (no division by zero, no even roots, no logs). It is...

$$f(x) = 2x - 1, \quad \text{i.e., } y = 2x - 1$$ This is a linear function (degree 1 polynomial) with slope $m = 2$ and y-intercept $-1$. --- A polynomial function has no restrictions (no division by zero, no even roots, no logs). It is...

95 marksNumericalConvergence tests and power seriesAnswer

Determine whether the series converges or diverges $\sum_{n=1}^{\infty} \frac{n^2}{5n^2+4}$ [5]

Convergence/Divergence of $\sum_{n=1}^{\infty} \dfrac{n^2}{5n^2+4}$

STEP 1 - Given Data

Series: $\displaystyle\sum_{n=1}^{\infty} a_n$ where $a_n = \dfrac{n^2}{5n^2+4}$
Task: determine convergence or divergence.

STEP 2 - Solve

Method: nth-Term Divergence Test.

If $\displaystyle\lim_{n\to\infty} a_n \neq 0$ (or does not exist), then $\sum a_n$ diverges.

Compute the limit of the general term. Divide numerator and denominator by $n^2$ :

\lim_{n\to\infty} \frac{n^2}{5n^2+4} = \lim_{n\to\infty} \frac{1}{5 + \dfrac{4}{n^2}}.

As $n \to \infty$ , $\dfrac{4}{n^2} \to 0$ , hence

\lim_{n\to\infty} a_n = \frac{1}{5+0} = \frac{1}{5}.

Apply the test.

\lim_{n\to\infty} a_n = \frac{1}{5} \neq 0.

Since the necessary condition for convergence ( $\lim a_n = 0$ ) fails, the series cannot converge.

Conclusion

\boxed{\sum_{n=1}^{\infty} \frac{n^2}{5n^2+4} \text{ diverges (by the nth-term / Divergence Test).}}

Convergence/Divergence of $\sum_{n=1}^{\infty} \dfrac{n^2}{5n^2+4}$

STEP 1 - Given Data

Series: $\displaystyle\sum_{n=1}^{\infty} a_n$ where $a_n = \dfrac{n^2}{5n^2+4}$
Task: determine convergence or divergence.

STEP 2 - Solve

Method: nth-Term Divergence Test.

If $\displaystyle\lim_{n\to\infty} a_n \neq 0$ (or does not exist), then $\sum a_n$ diverges.

Compute the limit of the general term. Divide numerator and denominator by $n^2$:

$$ \lim_{n\to\infty} \frac{n^2}{5n^2+4} = \lim_{n\to\infty} \frac{1}{5 + \dfrac{4}{n^2}}. $$

As $n \to \infty$, $\dfrac{4}{n^2} \to 0$, hence

$$ \lim_{n\to\infty} a_n = \frac{1}{5+0} = \frac{1}{5}. $$

Apply the test.

$$ \lim_{n\to\infty} a_n = \frac{1}{5} \neq 0. $$

Since the necessary condition for convergence ($\lim a_n = 0$) fails, the series cannot converge.

Conclusion

$$ \boxed{\sum_{n=1}^{\infty} \frac{n^2}{5n^2+4} \text{ diverges (by the nth-term / Divergence Test).}} $$

105 marksNumericalPartial derivativesAnswer

If $f(x,y) = x^3 + x^2y^3 - 2y^2$ , find $f_x(2,1)$ and $f_y(2,1)$ . [5]

If $f(x,y) = x^3 + x^2y^3 - 2y^2$, find $f_x(2,1)$ and $f_y(2,1)$. [5]

$f(x, y) = x^3 + x^2y^3 - 2y^2$ Required: $fx(2,1)$ and $fy(2,1)$ . --- Differentiate with respect to $x$ , treating $y$ as constant: - $\frac{\partial}{\partial x}(x^3) = 3x^2$ - $\frac{\partial}{\partial x}(x^2 y^3) = 2x y^3$ -

$$f(x, y) = x^3 + x^2y^3 - 2y^2$$ Required: $fx(2,1)$ and $fy(2,1)$. --- Differentiate with respect to $x$, treating $y$ as constant: - $\frac{\partial}{\partial x}(x^3) = 3x^2$ - $\frac{\partial}{\partial x}(x^2 y^3) = 2x y^3$ -

115 marksNumericalContinuityAnswer

Show that the function $f(x) = x^2 + \sqrt{7-x}$ is continuous at $x=4$ . [5]

Show that the function $f(x) = x^2 + \sqrt{7-x}$ is continuous at $x=4$. [5]

Function: $f(x) = x^2 + \sqrt{7 - x}$ - Point: $x = 4$ $f(x)$ is continuous at $x = a$ if all three hold: 1. $f(a)$ is defined 2. $\lim{x \to a} f(x)$ exists 3. $\lim{x \to a} f(x) = f(a)$ Here $a = 4$ . --- $$f(4) = 4^2 + \sqrt{7 - 4} ...

Function: $f(x) = x^2 + \sqrt{7 - x}$ - Point: $x = 4$ $f(x)$ is continuous at $x = a$ if all three hold: 1. $f(a)$ is defined 2. $\lim{x \to a} f(x)$ exists 3. $\lim{x \to a} f(x) = f(a)$ Here $a = 4$. --- $$f(4) = 4^2 + \sqrt{7 - 4} ...

125 marksNumericalIntroduction to first order equations SepaAnswer

Show that $y = x - \frac{1}{x}$ is a solution of the differential equation $xy' + y = 2x$ . [5]

Show that $y = x - \frac{1}{x}$ is a solution of the differential equation $xy' + y = 2x$. [5]

Proposed solution: $y = x - \dfrac{1}{x}$ - Differential equation: $xy' + y = 2x$ --- $y = x - \frac{1}{x} = x - x^{-1}$ $y' = \frac{d}{dx}(x) - \frac{d}{dx}(x^{-1}) = 1 - (-1)x^{-2} = 1 + \frac{1}{x^2}$ --- $$xy' = x\left(1 + \fra...

Proposed solution: $y = x - \dfrac{1}{x}$ - Differential equation: $xy' + y = 2x$ --- $$y = x - \frac{1}{x} = x - x^{-1}$$ $$y' = \frac{d}{dx}(x) - \frac{d}{dx}(x^{-1}) = 1 - (-1)x^{-2} = 1 + \frac{1}{x^2}$$ --- $$xy' = x\left(1 + \fra...

Mathematics I 2080 question paper

Temperature as a Linear Function of Height

STEP 1 - Given Data

STEP 2 - Solution

Building the Linear Model

Part (a): Graph and Slope Meaning

Part (c): Temperature at h=2.5h = 2.5h=2.5 km

Final Results

Temperature as a Linear Function of Height

STEP 1 - Given Data

STEP 2 - Solution

Building the Linear Model

Part (a): Graph and Slope Meaning

Part (c): Temperature at $h = 2.5$ km

Final Results

Convergence/Divergence of ∑n=1∞n25n2+4\sum_{n=1}^{\infty} \dfrac{n^2}{5n^2+4}∑n=1∞​5n2+4n2​

STEP 1 - Given Data

STEP 2 - Solve

Conclusion

Convergence/Divergence of $\sum_{n=1}^{\infty} \dfrac{n^2}{5n^2+4}$

STEP 1 - Given Data

STEP 2 - Solve

Conclusion

Temperature as a Linear Function of Height

STEP 1 - Given Data

STEP 2 - Solution

Building the Linear Model

Part (a): Graph and Slope Meaning

Part (c): Temperature at h=2.5h = 2.5h=2.5 km

Final Results

Convergence/Divergence of ∑n=1∞n25n2+4\sum_{n=1}^{\infty} \dfrac{n^2}{5n^2+4}∑n=1∞​5n2+4n2​

STEP 1 - Given Data

STEP 2 - Solve

Conclusion

Part (c): Temperature at $h = 2.5$ km

Convergence/Divergence of $\sum_{n=1}^{\infty} \dfrac{n^2}{5n^2+4}$

Part (c): Temperature at $h = 2.5$ km

Convergence/Divergence of $\sum_{n=1}^{\infty} \dfrac{n^2}{5n^2+4}$