Sec 3 A Math: Linear Law, practice questions & worked solutions

(a)

(b)

$$\begin{aligned} N &= a10^{kt}\\ \lg N &= \lg a10^{kt}\\ \lg N &= \lg a + \lg 10^{kt}\\ \lg N &= \lg a + kt\\ \lg N &= kt + \lg a \end{aligned}$$ $$\begin{aligned} k &= \text{gradient}\\ &= 0.232 \ (\text{3 s.f.})\\ \lg a &= \text{vertical axis} - \text{intercept}\\ \lg a &= 2.5\\ a &= 10^{2.5}\\ &= 316 \ (\text{3 s.f.}) \end{aligned}$$

(a)

Plot $\ln m$ against $t$ and draw a straight line graph.

(b)

$$\begin{aligned} \ln M &= 2.08\\ M &= 8.00 \end{aligned}$$

(c)

$$\begin{aligned} \text{Gradient} &= -0.2\\ k &= 0.2 \end{aligned}$$

(a)

(b)

(i) $$x^a y = b \;\Rightarrow\; a\lg x+\lg y=\lg b \;\Rightarrow\; \lg y=-a\lg x+\lg b$$

Plot $\lg y$ against $\lg x$ and draw a straight line graph.

(ii) $$\begin{aligned} \text{Gradient} &= -5.51\\ a &\approx 5.51 (\pm 0.1)\\ Y\text{-intercept} = \lg b &= 3.8 (\pm 0.1)\\ b &= 6310 (5012-7943) \end{aligned}$$

(a)

(b)

(i) $y = 25.1$ is an abnormal reading of $y$.

correct value of $y$: $\lg y = 1.25$ $$y = 10^{1.25} = 17.8$$

(ii) $$\begin{aligned} y &= \frac{h}{k^x}\\ \lg y &= \lg h - x\lg k\\ \lg y &= (-\lg k)x + \lg h \end{aligned}$$ $$\begin{aligned} \text{gradient} = -\lg k &= \frac{1.75 - 1.50}{1 - 2}\\ \lg k &= 0.25\\ k &= 10^{0.25} = 1.78 \end{aligned}$$ $$\begin{aligned} y\text{-intercept} = \lg h &= 2\\ h &= 100 \end{aligned}$$

(a)

$$\begin{aligned} c &= at + \frac{b}{t}\\ ct &= at^2 + b \end{aligned}$$ Draw $ct$ against $t^2$.

(b)

$$\begin{aligned} ct &= at^2 + b\\ a &= \text{gradient}\\ &= -\frac{20}{4}\\ &= -5\\ b &= \text{vertical intercept}\\ &= 20 \end{aligned}$$

(c)

When $t = 0.5$, $t^2 = 0.25$.

When $t^2 = 0.25$, $$\begin{aligned} ct &= 18.75\\ 0.5c &= 18.75\\ c &= 37.5 > 30 \end{aligned}$$ Since the concentration is greater than $30$ mg/L, the chemical is not safe for plants when $t = 0.5$ h.

(a)

The constant $a$ refers to the initial mass of the radio-active substance.

Or

The constant $a$ refers to the mass of the radio-active substance when $t = 0$.

(b)

$$\begin{aligned} m &= ae^{-bt}\\ \ln m &= \ln\left(ae^{-bt}\right)\\ \ln m &= \ln a - bt\\ \ln m &= -bt + \ln a \end{aligned}$$

$$\begin{aligned} -b &= \frac{-4 - 1}{4.7 - 0.5}\\ -b &\approx -1.1904\\ b &\approx 1.19 \end{aligned}$$ $$\begin{aligned} \ln a &= \text{Y-intercept}\\ \ln a &= 1.6\\ a &\approx e^{1.6}\\ a &\approx 4.9530\\ a &\approx 4.95 \end{aligned}$$

(c)

When $t = 4$, $$\begin{aligned} \ln m &= -3.15\\ m &= e^{-3.15}\\ m &\approx 0.042852\\ m &\approx 0.0429 \end{aligned}$$

(a)

(i) Plot $xy$ against $x^2$ and draw a straight line through the points $(1,\,8.55)$, $(4,\,12.40)$, $(9,\,15.96)$, $(16,\,22.88)$, $(25,\,30.00)$.

(ii)(a) $$\begin{aligned} xy &= px^2+q\\ p &= \frac{26-17.5}{20-10.5} \approx 0.895\\ q &= 8 \end{aligned}$$

(ii)(b) $$\begin{aligned} y &= \frac{20}{x}\\ xy &= 20 \end{aligned}$$ from graph, $x^2=13.25$ $$x = 3.64$$

(b)

$$\begin{aligned} xy &= px^2+q\\ 1 &= \frac{px}{y}+\frac{q}{xy}\\ \frac{px}{y} &= -\frac{q}{xy}+1\\ \frac{x}{y} &= \left(-\frac{q}{p}\right)\frac{1}{xy}+\frac{1}{p} \end{aligned}$$ $$\text{gradient} = -\frac{q}{p},\qquad \text{vertical intercept} = \frac{1}{p}$$

(a)

(b)

Transforming the equation: $$\begin{aligned} y &= 91 - A\mathrm{e}^{kx}\\ 91 - y &= A\mathrm{e}^{kx}\\ \ln(91 - y) &= \ln A + \ln\mathrm{e}^{kx}\\ \ln(91 - y) &= \ln A + kx \end{aligned}$$ Using 2 points on the line $(1, 3.8)$ and $(3.5, 3)$, $$\text{gradient} = \frac{3.8 - 3}{1 - 3.5} = -0.32$$ Reading off the $Y$-intercept, we have $4.15$

The equation of line is $$\ln(91 - y) = -0.32x + 4.15$$ By comparing, $$\begin{aligned} \ln A &= 4.15 & k &= -0.32\\ A &= 63.4 \ (\text{3 s.f.}) \end{aligned}$$

(c)

• (-) The lower mark on week 5 was likely to be due to a harder paper.

• (-) Andy might have been ill.

(d)

The datapoint is $(5, 71)$

Based on the straight line graph, the reading should be $2.5$.

The expected marks is $$\begin{aligned} \ln(91 - y) &= 2.5\\ y &= 79 \end{aligned}$$