题型分析

Question Type 1: Standard Linearization ($y = mx + c$)

如何识别

Equation can be rearranged directly to $y = mx + c$ without logarithms. Variables are linearly related.

标准解题方法

1. Rearrange given equation to $y = mx + c$ form
2. Identify what goes on $y$-axis and $x$-axis
3. Write expression for gradient and y-intercept
4. Check that the expression is dimensionally correct

评分标准（MS 模式）

• M1: Correct rearrangement to $y = mx + c$
• A1: Correct identification of $y$ and $x$ variables
• B1: Correct expression for gradient
• B1: Correct expression for y-intercept

Example 1 — 9702/s20/qp/51 Q2(a)

$V = \frac{Q_0}{C} e^{-t/RC}$ A graph is plotted of $\ln V$ on the $y$-axis against $t$ on the $x$-axis. Determine expressions for the gradient and $y$-intercept.

MS 展开查看

MS:

• B1: $\text{gradient} = -\frac{1}{CR}$
• B1: $y\text{-intercept} = \ln\left(\frac{Q_0}{C}\right)$

Example 2 — 9702/w20/qp/51 Q2(a)

$\frac{h_i}{h_o} = \frac{d}{f} - \frac{t}{2f} - 1$ A graph is plotted of $\frac{h_i}{h_o}$ on the $y$-axis against $d$ on the $x$-axis. Determine expressions for gradient and $y$-intercept.

MS 展开查看

MS:

• B1: $\text{gradient} = \frac{1}{f}$
• B1: $y\text{-intercept} = -\frac{t}{2f} - 1$

Example 3 — 9702/s21/qp/52 Q2(a)

$E = I(R_1 + R_2 + r)$ A graph is plotted of $\frac{1}{I}$ on the $y$-axis against $(R_1 + R_2)$ on the $x$-axis. Determine expressions for gradient and $y$-intercept.

MS 展开查看

MS:

• B1: $\text{gradient} = \frac{1}{E}$
• B1: $y\text{-intercept} = \frac{r}{E}$

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Question Type 2: Log-Log Linearization ($\lg y = \lg a + n \lg x$)

如何识别

Equation is a power law: $y = ax^n$

标准解题方法

1. Take $\lg$ of both sides: $\lg y = \lg a + n \lg x$
2. Plot $\lg y$ against $\lg x$
3. Gradient $= n$, y-intercept $= \lg a$
4. Hence $a = 10^{\text{intercept}}$

Example 1 — 9702/s22/qp/51 Q2(a)

$L = SZM^n$ A graph is plotted of $\lg L$ on the $y$-axis against $\lg M$ on the $x$-axis. Determine expressions for the gradient and $y$-intercept.

MS 展开查看

MS:

• B1: $\text{gradient} = n$
• B1: $y\text{-intercept} = \lg(SZ)$

Example 2 — 9702/s22/qp/52 Q2(a)

$L = SKT^a$ A graph is plotted of $\lg L$ against $\lg T$. Determine expressions for gradient and $y$-intercept.

MS 展开查看

MS:

• B1: $\text{gradient} = a$
• B1: $y\text{-intercept} = \lg(SK)$

Example 3 — 9702/s24/qp/52 Q2(a)

$T = \frac{2L^n}{C}$ A graph is plotted of $\lg T$ against $\lg L$. Determine gradient and $y$-intercept.

MS 展开查看

MS:

• B1: $\text{gradient} = n$
• B1: $y\text{-intercept} = \lg(2/C)$

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Question Type 3: Log-Linear Linearization ($\ln y = \ln a + kx$)

如何识别

Equation is exponential: $y = ae^{kx}$

标准解题方法

1. Take $\ln$ of both sides: $\ln y = \ln a + kx$
2. Plot $\ln y$ against $x$
3. Gradient $= k$, y-intercept $= \ln a$
4. Hence $a = e^{\text{intercept}}$

Example 1 — 9702/s20/qp/52 Q2(a)

$\eta = He^{E/kT}$ A graph is plotted of $\ln\eta$ against $1/T$. Determine expressions for gradient and $y$-intercept.

MS 展开查看

MS:

• B1: $\text{gradient} = \frac{E}{k}$
• B1: $y\text{-intercept} = \ln H$

Example 2 — 9702/w21/qp/52 Q2(a)

$R = R_0 e^{-\mu t}$ A graph is plotted of $\ln R$ against $t$. Determine gradient and $y$-intercept.

MS 展开查看

MS:

• B1: $\text{gradient} = -\mu$
• B1: $y\text{-intercept} = \ln R_0$

常见陷阱

• $\ln$ of a quantity with units: write $\ln(V/\text{V})$, NOT $\ln V$
• For $\eta = He^{E/kT}$, the x-axis is $1/T$, NOT $T$
• Sign errors: check if the exponent is negative

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Question Type 4: Reciprocal Linearization ($y = a + b/x$)

如何识别

Equation contains an inverse term: $y = a + \frac{b}{x}$

标准解题方法

1. Rearrange to $y = a + b \cdot \frac{1}{x}$
2. Plot $y$ against $\frac{1}{x}$
3. Gradient $= b$, y-intercept $= a$

Example 1 — 9702/s23/qp/51 Q2(a)

$V = E - \frac{I}{k}$ A graph is plotted of $V$ on the $y$-axis against $I$ on the $x$-axis. Determine expressions for gradient and $y$-intercept.

MS 展开查看

MS:

• B1: $\text{gradient} = -\frac{1}{k}$
• B1: $y\text{-intercept} = E$

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题型对比总结

题型	原始关系	作图方式	Gradient	Y-intercept
Standard	$y = mx + c$	$y$ vs $x$	$m$	$c$
Log-log	$y = ax^n$	$\lg y$ vs $\lg x$	$n$	$\lg a$
Log-linear	$y = ae^{kx}$	$\ln y$ vs $x$	$k$	$\ln a$
Reciprocal	$y = a + b/x$	$y$ vs $1/x$	$b$	$a$