SL Applications & Interpretation Revision Notes

Worked Example 1: Calculating Percentage Error The exact volume of a sphere is $v_E=1250{\text{ cm}}^3$. A student measures the radius, rounds their calculation, and estimates the volume to be $v_A=1200{\text{ cm}}^3$. Calculate the percentage error.

Solution:
Substitute the values directly into the percentage error formula: $$\begin{array}{rl}\epsilon & =\left|\frac{1200-1250}{1250}\right|\times 100\mathrm{\%}\\ \epsilon & =|-0.04|\times 100\mathrm{\%}=4\mathrm{\%}\end{array}$$

Worked Example 2: Arithmetic, Geometric & Infinite Series

(a) In an arithmetic sequence, $u_1=5$ and $d=3$. Find $u_{20}$ and $S_{20}$.
(b) Find the exact sum of the infinite geometric series: $200+40+8+1.6+\dots$

Solution:
(a) $$\begin{array}{rl}{u}_{20}& =5+(20-1)(3)=62\\ S_{20}& =\frac{20}{2}(2(5)+(19)(3))=10(67)=670\end{array}$$ (b) The first term $u_1=200$.
The common ratio $r=\frac{40}{200}=0.2$.
Because $|0.2|<1$, the infinite sum exists: $$S_{\mathrm{\infty }}=\frac{200}{1-0.2}=\frac{200}{0.8}=250$$

Worked Example 3: Applying Exponents & Logarithms Solve the exponential equation $5\times 3^{2x}=40$.

Solution:
First, isolate the base and exponent by dividing both sides by 5: $$\begin{array}{rl}{3}^{2x}& =8\end{array}$$ Using the logarithmic conversion formula ($a^x=b⟹x={\log }_{a}b$): $$\begin{array}{rl}2x& ={\log }_3(8)\\ 2x& \approx 1.892789\\ x& \approx 0.946\text{ (to 3 s.f.)}\end{array}$$

Worked Example 4: Compound Interest & Depreciation

(a) Compound Interest: $\mathrm{\$}10,000$ is invested at $4\mathrm{\%}$ per annum, compounded quarterly. Find the value after 5 years.
Solution: $PV=10000,r=4,k=4,n=5$. $$FV=10000{(1+\frac{4}{100(4)})}^{4\times 5}=10000(1.01{)}^{20}\approx \mathrm{\$}12,201.90$$

(b) Depreciation Model: A piece of machinery is purchased for $\mathrm{\$}24,000$. It depreciates in value by $18\mathrm{\%}$ each year. Find its value after $5$ years.
Solution: The annual multiplier is $(1-0.18)=0.82$. $$V(5)=24000(0.82{)}^5\approx \mathrm{\$}8898$$

(c) Half-Life Logarithms: Calculate the exact number of years it takes for the machine to depreciate to half of its initial value ($\mathrm{\$}12,000$).
Solution: $$\begin{array}{rl}12000& =24000(0.82{)}^t\\ 0.5& ={0.82}^t\\ t& =\frac{\ln (0.5)}{\ln (0.82)}\approx 3.49\text{ years}\end{array}$$

Worked Example 5: 3x3 Linear Systems A theatre sells Adult ($x$), Child ($y$), and Student ($z$) tickets. A total of $600$ tickets were sold. The revenue was $\mathrm{\$}7816$. Adult tickets cost $\mathrm{\$}15$, Child tickets $\mathrm{\$}10$, and Student tickets $\mathrm{\$}12$. Twice as many Adult tickets were sold as Child tickets. Find the number of each ticket sold.

Solution:
Set up the three linear equations based on the text: $$\begin{array}{rl}x+y+z& =600\text{(Total tickets)}\\ 15x+10y+12z& =7816\text{(Total revenue)}\\ x-2y+0z& =0\text{(Since }x=2y⟹x-2y=0)\end{array}$$ Inputting the matrix into the Casio Equation Solver yields: $$x=308,y=154,z=138$$ Therefore, 308 Adult, 154 Child, and 138 Student tickets were sold.

Worked Example 1: Applying All Linear Formulas

A linear model passes through the points $A(1,4)$ and $B(4,10)$.
(a) Calculate the gradient of the line.
(b) Express the line in point-gradient form.
(c) Convert the equation into gradient-intercept form.
(d) Convert the equation into general form.

Solution:
(a) $m=\frac{y_2-y_1}{x_2-x_1}=\frac{10-4}{4-1}=\frac{6}{3}=2$.
(b) Using point $A(1,4)$ and $m=2$: $y-4=2(x-1)$.
(c) Expand brackets: $y-4=2x-2⟹y=2x+2$.
(d) Move all terms to one side: $2x-y+2=0$.

Worked Example 2: Applying the Axis of Symmetry Formula

A projectile’s height, $h$, in metres, over a horizontal distance $x$, is modelled by the quadratic function: $$h(x)=-2x^2+12x-10$$ (a) Find the horizontal distance required to reach the maximum height.
(b) Calculate the maximum height.
(c) Find the $x$-intercepts (roots) where $h(x)=0$.

Solution:
(a) Apply the axis of symmetry formula: $$\begin{array}{rl}x& =-\frac{b}{2a}=-\frac{12}{2(-2)}=-\frac{12}{-4}=3\text{ m}\end{array}$$ (b) Substitute $x=3$ back into the function: $$\begin{array}{rl}h(3)& =-2(3{)}^2+12(3)-10\\ h(3)& =-18+36-10=8\text{ m}\end{array}$$ (c) Set $h(x)=0$ and factorise (or use GDC ROOT):
$-2(x^2-6x+5)=0⟹-2(x-1)(x-5)=0$.
The roots are $x=1$ and $x=5$.

Worked Example 3: Finding Inverses Algebraically & Graphically

Given the function $f(x)=3x-6$:
(a) Find the inverse function, $f^{-1}(x)$, algebraically.
(b) Demonstrate the composite inverse formula mathematically.

Solution:
(a) Let $y=3x-6$. To find the inverse, swap $x$ and $y$, then isolate the new $y$: $$\begin{array}{rl}x& =3y-6\\ x+6& =3y\\ y& =\frac{x}{3}+2\end{array}$$ Therefore, $f^{-1}(x)=\frac{x}{3}+2$.

(b) We must show that $f(f^{-1}(x))=x$: $$\begin{array}{rl}f(\frac{x}{3}+2)& =3(\frac{x}{3}+2)-6\\ & =x+6-6\\ & =x\end{array}$$ The formula is successfully demonstrated.

Worked Example 4: Interpreting Asymptotes & GDC Intersections

A cooling cup of coffee has its temperature $T$ (in ${}^{\circ }$C) modelled by $T(t)=80(0.8{)}^t+20$, where $t$ is time in minutes.
(a) Find the initial temperature of the coffee.
(b) State the equation of the horizontal asymptote and explain its physical meaning.
(c) Using a GDC, find the time it takes for the coffee to reach ${50}^{\circ }$C.

Solution:
(a) Initial temperature occurs at $t=0$:
$T(0)=80(0.8{)}^0+20=80(1)+20={100}^{\circ }$C.
(b) The horizontal asymptote is $T=20$.
Meaning: As time goes on infinitely, the coffee cools down and approaches ${20}^{\circ }$C, representing the constant room temperature.
(c) Graph Y1 = 80(0.8)^x + 20 and Y2 = 50.
Using G-Solv $\to$ ISCT, the lines cross at $x\approx 4.39$.
It takes $4.39$ minutes.

Worked Example 1: Calculating Arc Length and Sector Area

A sprinkler waters a lawn in the shape of a sector of a circle. The radius of the water spray is $8\text{ m}$ and it rotates through an angle of ${110}^{\circ }$.
(a) Calculate the length of the outer arc of the watered grass.
(b) Calculate the total area of the lawn watered by the sprinkler.

Solution:
(a) Apply the arc length formula with $r=8$ and $\theta =110$: $$\begin{array}{rl}l& =\frac{110}{360}\times 2\pi (8)\\ l& =\frac{11}{36}\times 16\pi \approx 15.4\text{ m}\end{array}$$ (b) Apply the sector area formula: $$\begin{array}{rl}A& =\frac{110}{360}\times \pi (8{)}^2\\ A& =\frac{11}{36}\times 64\pi \approx 61.4{\text{ m}}^2\end{array}$$

Worked Example 2: The Cosine Rule & Area Rule

A triangular plot of land ABC has side lengths $AB=40\text{ m}$ and $BC=30\text{ m}$. The included angle $A\hat{B}C={120}^{\circ }$.
(a) Calculate the length of $AC$.
(b) Find the area of the plot of land.

Solution:
(a) We know SAS, so we use the Cosine Rule: $$\begin{array}{rl}{b}^2& =a^2+c^2-2ac\cos B\\ b^2& ={30}^2+{40}^2-2(30)(40)\cos ({120}^{\circ })\\ b^2& =900+1600-2400(-0.5)\\ b^2& =3700⟹b=\sqrt{3700}\approx 60.8\text{ m}\end{array}$$ (b) Use the Area formula (SAS): $$\text{Area}=\frac{1}{2}ac\sin B=\frac{1}{2}(30)(40)\sin ({120}^{\circ })\approx 520{\text{ m}}^2$$

Worked Example 3: Compound Shapes A solid object consists of a hemisphere (half-sphere) of radius $6\text{ cm}$ placed exactly on top of a cylinder with the same radius and a height of $10\text{ cm}$. Find the total volume and the total exposed surface area.

Solution:
Total Volume = (Volume of Cylinder) + (Volume of Hemisphere) $$\begin{array}{rl}V& =\pi r^{2}h+\frac{1}{2}\left(\frac{4}{3}\pi r^3\right)\\ V& =\pi (6{)}^2(10)+\frac{2}{3}\pi (6{)}^3\\ V& =360\pi +144\pi =504\pi \approx 1580{\text{ cm}}^3\end{array}$$ Total Surface Area = (Curved SA of Cylinder) + (Hemisphere Curved SA) + (Flat Bottom Base) $$\begin{array}{rl}A& =2\pi rh+\frac{1}{2}\left(4\pi r^2\right)+\pi r^2\\ A& =2\pi (6)(10)+2\pi (6{)}^2+\pi (6{)}^2\\ A& =120\pi +72\pi +36\pi =228\pi \approx 716{\text{ cm}}^2\end{array}$$

Worked Example 4: Constructing a Boundary & Nearest Neighbour

(a) Boundary Equation: Site A is at $(2,4)$ and Site B is at $(6,0)$. Find the equation of the Voronoi boundary separating them.

Solution:
1. Find Midpoint $M$: $(\frac{2+6}{2},\frac{4+0}{2})=(4,2)$.
2. Find Gradient of line AB: $m_{AB}=\frac{0-4}{6-2}=-1$.
3. Find Perpendicular Gradient: $m_{\perp }=-\frac{1}{-1}=1$.
4. Substitute $M(4,2)$ and $m_{\perp }=1$ into line equation:
$y-2=1(x-4)⟹y=x-2$.

(b) Toxic Waste: A toxic waste dump is placed at $(3,3)$. Is it closer to Hospital A $(1,1)$ or Hospital B $(6,2)$?

Solution:
Distance to A: $d=\sqrt{(3-1{)}^2+(3-1{)}^2}=\sqrt{8}\approx 2.83$
Distance to B: $d=\sqrt{(3-6{)}^2+(3-2{)}^2}=\sqrt{10}\approx 3.16$
Because $\sqrt{8}<\sqrt{10}$, the dump is closer to Hospital A.

Worked Example 1: Calculating Outliers Graphically

An exam has a lower quartile ($Q_1$) of $32$ and an upper quartile ($Q_3$) of $56$. The lowest score is $10$ and the highest is $98$.
(a) Calculate the Interquartile Range (IQR).
(b) Determine if the maximum and minimum are mathematical outliers.

Solution:
(a) $\text{IQR}=56-32=24$.
(b) Lower boundary: $Q_1-1.5(24)=32-36=-4$.
Since $10>-4$, the minimum is not an outlier.
Upper boundary: $Q_3+1.5(24)=56+36=92$.
Since $98>92$, the maximum is an outlier.

Worked Example 2: Mathematically Exact Scatter Models

Five students logged their wind speed $x$ (km/h) against the time to charge a device $y$ (mins).
$x=\{10,15,21,25,34\}$ $y=\{63,60,55,52,45\}$
Mean $M(\overline{x},\overline{y})$ = $(21,55)$
Regression line: $y=-0.76x+70.96$
Pearson’s $r=-0.995$

(a) Describe the correlation.
(b) Plot the mean point $M$ and draw the line of best fit on the scatter diagram.

Solution:
(a) $r=-0.995$ is a very strong negative linear correlation.
(b) Plot $M(21,55)$. The line must cross the $y$-axis at the intercept ($70.96$) and pass exactly through $M$.

Worked Example 3: Binomial Probabilities & Variance

A biased coin has a probability of $0.72$ of landing on heads. The coin is flipped $5$ times. Let $X$ be the number of heads obtained.
$X\sim B(5,0.72)$.
(a) Find the expected number of heads, $E(X)$.
(b) Find the variance of the distribution.
(c) Find the probability of getting exactly 3 heads.
(d) Find the probability of getting at least 3 heads.

Solution:
(a) $E(X)=np=5\times 0.72=3.6$ heads.
(b) $Var(X)=np(1-p)=5(0.72)(0.28)=1.008$.

(c) We need $P(X=3)$.
Using the Casio Binomial PD function:
$$\begin{gathered} x=3, \\ \text{Numtrial}=5, \\ p=0.72 \end{gathered}$$.
$P(X=3)\approx 0.293$ (to 3 s.f.)

(d) We need $P(X\ge 3)$.
Using the Casio Binomial CD function:
$\text{Lower}=3,\text{Upper}=5,\text{Numtrial}=5,p=0.72$.
$P(X\ge 3)\approx 0.862$ (to 3 s.f.)

Worked Example 4: Normal Distribution Area & InvNorm

The weights of a particular grade of chicken egg are normally distributed with a mean ($\mu$) of $60$ grams and a standard deviation ($\sigma$) of $5$ grams. $X\sim N(60,5^2)$.
(a) Find the probability that an egg weighs between $55\text{g}$ and $68\text{g}$.
(b) The heaviest $10\mathrm{\%}$ of eggs are graded as "Jumbo". Find the minimum weight required for an egg to be graded Jumbo.

Solution:
(a) On the GDC, select Dist \rightarrow NORM \rightarrow Ncd:
Lower = $55$, Upper = $68$, $\sigma =5$, $\mu =60$.
Result: $P(55<X<68)\approx 0.787$ (or $78.7\mathrm{\%}$).
(b) We know the area to the right is $10\mathrm{\%}$ ($0.10$), which means the area to the left is $0.90$.
On the GDC, select Dist \rightarrow NORM \rightarrow InvN:
Area = $0.90$ (Left tail), $\sigma =5$, $\mu =60$.
Result: $x\approx 66.4$.
An egg must weigh at least $66.4\text{g}$.

Worked Example 5: ${\chi }^2$ Test for Independence

A school surveys 110 students to determine if there is an association between a student’s gender and their choice of foreign language. The test is performed at a $5\mathrm{\%}$ significance level.

(a) State the null hypothesis, $H_0$.
(b) The school inputs the observed frequency table into their GDC. The calculator outputs a $p$-value of $0.0130$. State, with a valid reason, whether $H_0$ should be rejected.

Solution:
(a) $H_0$: The choice of language is independent of gender. (Never use the word "correlated" for categorical data).

(b) To make a conclusion, we must compare the $p$-value to the significance level ($0.05$).
Because $0.0130<0.05$, there is sufficient evidence to reject the null hypothesis.
Therefore, we reject ${\mathbf{H}}_{\mathbf{0}}$. (Conclusion: Choice of language is dependent on gender).

Bonus IB Tip: The Two-Sample $t$-test
If a question asks you to test whether the average weight of fish before and after a new diet has changed, your hypotheses would be:
$H_0:{\mu }_{\text{before}}={\mu }_{\text{after}}$
$H_1:{\mu }_{\text{before}}\ne {\mu }_{\text{after}}$
You would use the 2-Sample t-test on your GDC to find the $p$-value, using the exact same rejection rule!

Worked Example 1: Tangents and Normals

Given the function $f(x)=x^2$:
(a) Find the derivative function, $f^{\prime }(x)$.
(b) Find the equation of the tangent line to the curve at the point $P(1,1)$.
(c) Find the equation of the normal line to the curve at $P(1,1)$.

Solution:
(a) Apply the power rule: $f^{\prime }(x)=2x$.
(b) Find tangent gradient at $x=1$:
$m_t=2(1)=2$.
Using $y-y_1=m(x-x_1)$:
$y-1=2(x-1)⟹y=2x-1$
(c) Find normal gradient:
$m_n=-\frac{1}{2}=-0.5$.
$y-1=-0.5(x-1)⟹y=-0.5x+1.5$

Worked Example 2: Algebraic & Graphical Optimisation

A company’s production cost $C$ (in thousands of dollars) for manufacturing $x$ thousand items is modelled by the cubic function: $$C(x)=x^3-6x^2+9x+15$$ (a) Find the derivative $C^{\prime }(x)$.
(b) Find the number of items that should be manufactured to minimize the cost, and state the minimum cost.

Solution:
(a) $C^{\prime }(x)=3x^2-12x+9$.
(b) Algebraic Method: Set the derivative equal to zero to find the turning points: $$\begin{array}{rl}3x^2-12x+9& =0\\ 3(x^2-4x+3)& =0\\ 3(x-1)(x-3)& =0\end{array}$$ The turning points occur at $x=1$ and $x=3$.
Substitute $x=3$ back into the original function: $$C(3)=(3{)}^3-6(3{)}^2+9(3)+15=15$$ They should manufacture $3,000$ items for a minimum cost of $\mathrm{\$}15,000$.
(b) Casio Method: Enter $C(x)$ into your graph menu. Use the G-Solv \rightarrow MIN tool to directly verify the coordinate is $(3,15)$.

Worked Example 3: Finding Area Algebraically & with GDC

Consider the quadratic curve $y=-x^2+4x$.
(a) Find the roots ($x$-intercepts) of the curve.
(b) Calculate the exact area enclosed by the curve and the $x$-axis using integration.

Solution:
(a) Using the GDC’s Root tool, or factoring $-x(x-4)=0$, the roots are $x=0$ and $x=4$.
(b) Algebraic Method: Set up the definite integral from $a=0$ to $b=4$: $$\begin{array}{rl}A& ={\int }_0^4(-x^2+4x)dx\\ A& ={[-\frac{x^3}{3}+\frac{4x^2}{2}]}_0^4\\ A& =(-\frac{4^3}{3}+2(4{)}^2)-(0)\\ A& =-\frac{64}{3}+32=\frac{32}{3}\approx 10.7{\text{ units}}^2\end{array}$$ (b) Casio Method: In the calculator’s run-matrix menu, use the ∫dx tool to input the integral template. This will directly verify the area is $10.666...$

Worked Example 4: Estimating Area Numerically

Use the trapezoidal rule with $n=3$ intervals to approximate the area under $y=x^2+1$ from $x=0$ to $x=3$.

Solution:
1. Calculate width: $h=\frac{3-0}{3}=1$.
2. Calculate the $y$-values:

• $x_0=0⟹y_0=1$

• $x_1=1⟹y_1=2$

• $x_2=2⟹y_2=5$

• $x_3=3⟹y_3=10$

3. Apply the formula: $$\begin{array}{rl}A& \approx \frac{1}{2}(1)[(1+10)+2(2+5)]\\ A& \approx 0.5[11+14]\\ A& \approx 12.5{\text{ units}}^2\end{array}$$