| Exam Board | OCR |
|---|---|
| Module | H240/01 (Pure Mathematics) |
| Year | 2021 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Topic | Parametric differentiation |
| Type | Tangent parallel to axis condition |
| Difficulty | Standard +0.3 This is a standard parametric equations question requiring routine differentiation and algebraic manipulation. Parts (a)-(c) involve straightforward observations and chain rule application, while part (d) requires setting dx/dt = 0 and solving a simple exponential equation. All techniques are textbook exercises with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.06a Exponential function: a^x and e^x graphs and properties1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| \(e^{kt} > 0\) for all \(t\), so \(y > 0\) for \(t \geq 0\), hence never crosses \(x\)-axis | B1 | Or show that \(2e^{-3t} = 0\) has no solutions. Need to see \(y \neq 0\) or \(y > 0\) and reason relating to exponential (or logarithmic) function. Must clearly be referring to \(y\) or \(2e^{-3t}\). |
| Answer | Marks | Guidance |
|---|---|---|
| \(e^{2t} - 4e^t + 3 = 0\); \((e^t-1)(e^t-3)=0\); \(e^t=1,\ e^t=3\) | M1 | Equate to 0 and attempt to solve disguised quadratic. 'determine' so some evidence of method needed. |
| \(t=0,\ t=\ln 3\) | A1 | Obtain both correct values. A0 for \(\ln 1\) and not 0. |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{\mathrm{d}x}{\mathrm{d}t} = 2e^{2t} - 4e^t\) | B1 | Correct \(\frac{\mathrm{d}x}{\mathrm{d}t}\). Mark derivative and condone no/wrong label. |
| \(\frac{\mathrm{d}y}{\mathrm{d}t} = -6e^{-3t}\) | B1 | Correct \(\frac{\mathrm{d}y}{\mathrm{d}t}\). Mark derivative and condone no/wrong label. |
| \(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{-6e^{-3t}}{2e^{2t}-4e^t}\) | M1 | Attempt correct method to combine derivatives. Combine their derivatives correctly. |
| \(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{-3e^{-3t}}{e^{2t}-2e^t} = \frac{3}{e^{3t}(2e^t - e^{2t})} = \frac{3}{2e^{4t}-e^{5t}}\) A.G. | A1 | Show manipulation to given answer. Need to see some evidence of how \(e^{-3t}\) is dealt with. AG so method must be fully correct. |
| Answer | Marks | Guidance |
|---|---|---|
| \(2e^{4t} - e^{5t} = 0\) | M1 | Equate denominator to 0. Or \(\frac{\mathrm{d}x}{\mathrm{d}y}=0\) or \(\frac{\mathrm{d}x}{\mathrm{d}t}=0\). |
| \(e^{4t}(2-e^t)=0\); \(t=\ln 2\) | A1 | Solve for \(t\) to obtain \(t=\ln 2\). No need to see \(e^{4t}=0\) discounted. |
| \(\left(-1,\ \frac{1}{4}\right)\) | A1 | Obtain correct coordinate. Or \(x=-1, y=\frac{1}{4}\). |
# Question 9:
## Part (a)
$e^{kt} > 0$ for all $t$, so $y > 0$ for $t \geq 0$, hence never crosses $x$-axis | B1 | Or show that $2e^{-3t} = 0$ has no solutions. Need to see $y \neq 0$ or $y > 0$ and reason relating to exponential (or logarithmic) function. Must clearly be referring to $y$ or $2e^{-3t}$.
## Part (b)
$e^{2t} - 4e^t + 3 = 0$; $(e^t-1)(e^t-3)=0$; $e^t=1,\ e^t=3$ | M1 | Equate to 0 and attempt to solve disguised quadratic. 'determine' so some evidence of method needed.
$t=0,\ t=\ln 3$ | A1 | Obtain both correct values. A0 for $\ln 1$ and not 0.
## Part (c)
$\frac{\mathrm{d}x}{\mathrm{d}t} = 2e^{2t} - 4e^t$ | B1 | Correct $\frac{\mathrm{d}x}{\mathrm{d}t}$. Mark derivative and condone no/wrong label.
$\frac{\mathrm{d}y}{\mathrm{d}t} = -6e^{-3t}$ | B1 | Correct $\frac{\mathrm{d}y}{\mathrm{d}t}$. Mark derivative and condone no/wrong label.
$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{-6e^{-3t}}{2e^{2t}-4e^t}$ | M1 | Attempt correct method to combine derivatives. Combine their derivatives correctly.
$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{-3e^{-3t}}{e^{2t}-2e^t} = \frac{3}{e^{3t}(2e^t - e^{2t})} = \frac{3}{2e^{4t}-e^{5t}}$ **A.G.** | A1 | Show manipulation to given answer. Need to see some evidence of how $e^{-3t}$ is dealt with. AG so method must be fully correct.
## Part (d)
$2e^{4t} - e^{5t} = 0$ | M1 | Equate denominator to 0. Or $\frac{\mathrm{d}x}{\mathrm{d}y}=0$ or $\frac{\mathrm{d}x}{\mathrm{d}t}=0$.
$e^{4t}(2-e^t)=0$; $t=\ln 2$ | A1 | Solve for $t$ to obtain $t=\ln 2$. No need to see $e^{4t}=0$ discounted.
$\left(-1,\ \frac{1}{4}\right)$ | A1 | Obtain correct coordinate. Or $x=-1, y=\frac{1}{4}$.
---9 A particle moves in the $x - y$ plane so that at time $t$ seconds, where $t \geqslant 0$, its coordinates are given by $x = \mathrm { e } ^ { 2 t } - 4 \mathrm { e } ^ { t } + 3 , y = 2 \mathrm { e } ^ { - 3 t }$.
\begin{enumerate}[label=(\alph*)]
\item Explain why the path of the particle never crosses the $x$-axis.
\item Determine the exact values of $t$ when the path of the particle intersects the $y$-axis.
\item Show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 3 } { 2 \mathrm { e } ^ { 4 t } - \mathrm { e } ^ { 5 t } }$.
\item Hence find the coordinates of the particle when its path is parallel to the $y$-axis.
\end{enumerate}
\hfill \mbox{\textit{OCR H240/01 2021 Q9 [10]}}