| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Year | 2018 |
| Session | October |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differentiating Transcendental Functions |
| Type | Applied rate of change |
| Difficulty | Standard +0.3 This is a straightforward applied calculus question requiring: (a) substitution of t=0, (b) algebraic manipulation and logarithms to solve for t, and (c) quotient rule differentiation followed by numerical evaluation. All techniques are standard C3/C4 material with no novel insight required, making it slightly easier than average. |
| Spec | 1.06a Exponential function: a^x and e^x graphs and properties1.06d Natural logarithm: ln(x) function and properties1.06g Equations with exponentials: solve a^x = b1.07n Stationary points: find maxima, minima using derivatives1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates1.08k Separable differential equations: dy/dx = f(x)g(y) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P = 450\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(315 = \frac{900e^{\frac{1}{4}t}}{3e^{\frac{1}{4}t}-1} \Rightarrow 45e^{\frac{1}{4}t} = 315\) | M1A1 | M1: substitutes \(P=315\), cross-multiplies to form \(Ae^{\frac{t}{4}}=B\); A1: \(45e^{\frac{t}{4}}=315\) or equivalent |
| \(\Rightarrow e^{\frac{1}{4}t} = 7 \Rightarrow t = 4\ln 7\) | M1A1 | M1: \(e^{\frac{t}{4}}=D\,(D>0)\Rightarrow t=\ldots\) using ln's; A1: \(t=4\ln 7\) or \(t=2\ln 49\) or \(\ln 2401\) (not \(4\ln\frac{315}{45}\)) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{dP}{dt} = \frac{\left(3e^{\frac{1}{4}t}-1\right)\times 225e^{\frac{1}{4}t} - 900e^{\frac{1}{4}t}\times\frac{3}{4}e^{\frac{1}{4}t}}{\left(3e^{\frac{1}{4}t}-1\right)^2} = \frac{-225e^{\frac{1}{4}t}}{\left(3e^{\frac{1}{4}t}-1\right)^2}\) | M1A1 | Quotient rule applied correctly |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\left.\frac{dP}{dt}\right | _{t=8} = \frac{(3e^2-1)\times 225e^2 - 900e^2\times\frac{3}{4}e^2}{(3e^2-1)^2} = \frac{-225e^2}{(3e^2-1)^2} = -3.71\) | M1A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Attempts quotient rule on \(P = \frac{900e^{\frac{1}{4}t}}{3e^{\frac{1}{4}t}-1}\) with \(u = 900e^{\frac{1}{4}t}\), \(v = 3e^{\frac{1}{4}t}-1\) | M1 | Condone slips on coefficients |
| \(\frac{dP}{dt} = \frac{\left(3e^{\frac{1}{4}t}-1\right)\times Ae^{\frac{1}{4}t} - 900e^{\frac{1}{4}t}\times Be^{\frac{1}{4}t}}{\left(3e^{\frac{1}{4}t}-1\right)^2}\), \(A,B>0\) | Score | Award if candidate incorrectly multiplies out before writing \(\frac{dP}{dt}\) |
| \(\frac{dP}{dt} = \frac{\left(3e^{\frac{1}{4}t}-1\right)\times 225e^{\frac{1}{4}t} - 900e^{\frac{1}{4}t}\times\frac{3}{4}e^{\frac{1}{4}t}}{\left(3e^{\frac{1}{4}t}-1\right)^2}\) | A1 | May be left unsimplified |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dP}{dt} = 225e^{\frac{1}{4}t}\times\left(3e^{\frac{1}{4}t}-1\right)^{-1} - 675e^{\frac{1}{4}t}\times e^{\frac{1}{4}t}\left(3e^{\frac{1}{4}t}-1\right)^{-2}\) | A1 | May be left unsimplified |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Substitutes \(t=8\) into \(\frac{dP}{dt}\) and calculates a value | M1 | |
| \(-3.71\) only | A1 | Not awrt; if candidate writes \(3.71\) this is A0 |
## Question 9:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P = 450$ | B1 | |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $315 = \frac{900e^{\frac{1}{4}t}}{3e^{\frac{1}{4}t}-1} \Rightarrow 45e^{\frac{1}{4}t} = 315$ | M1A1 | M1: substitutes $P=315$, cross-multiplies to form $Ae^{\frac{t}{4}}=B$; A1: $45e^{\frac{t}{4}}=315$ or equivalent |
| $\Rightarrow e^{\frac{1}{4}t} = 7 \Rightarrow t = 4\ln 7$ | M1A1 | M1: $e^{\frac{t}{4}}=D\,(D>0)\Rightarrow t=\ldots$ using ln's; A1: $t=4\ln 7$ or $t=2\ln 49$ or $\ln 2401$ (not $4\ln\frac{315}{45}$) |
### Part (c)(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dP}{dt} = \frac{\left(3e^{\frac{1}{4}t}-1\right)\times 225e^{\frac{1}{4}t} - 900e^{\frac{1}{4}t}\times\frac{3}{4}e^{\frac{1}{4}t}}{\left(3e^{\frac{1}{4}t}-1\right)^2} = \frac{-225e^{\frac{1}{4}t}}{\left(3e^{\frac{1}{4}t}-1\right)^2}$ | M1A1 | Quotient rule applied correctly |
### Part (c)(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\left.\frac{dP}{dt}\right|_{t=8} = \frac{(3e^2-1)\times 225e^2 - 900e^2\times\frac{3}{4}e^2}{(3e^2-1)^2} = \frac{-225e^2}{(3e^2-1)^2} = -3.71$ | M1A1 | Substitutes $t=8$ into their $\frac{dP}{dt}$ expression |
# Question c(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempts quotient rule on $P = \frac{900e^{\frac{1}{4}t}}{3e^{\frac{1}{4}t}-1}$ with $u = 900e^{\frac{1}{4}t}$, $v = 3e^{\frac{1}{4}t}-1$ | M1 | Condone slips on coefficients |
| $\frac{dP}{dt} = \frac{\left(3e^{\frac{1}{4}t}-1\right)\times Ae^{\frac{1}{4}t} - 900e^{\frac{1}{4}t}\times Be^{\frac{1}{4}t}}{\left(3e^{\frac{1}{4}t}-1\right)^2}$, $A,B>0$ | Score | Award if candidate incorrectly multiplies out before writing $\frac{dP}{dt}$ |
| $\frac{dP}{dt} = \frac{\left(3e^{\frac{1}{4}t}-1\right)\times 225e^{\frac{1}{4}t} - 900e^{\frac{1}{4}t}\times\frac{3}{4}e^{\frac{1}{4}t}}{\left(3e^{\frac{1}{4}t}-1\right)^2}$ | A1 | May be left unsimplified |
**Alternatively (product rule):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dP}{dt} = 225e^{\frac{1}{4}t}\times\left(3e^{\frac{1}{4}t}-1\right)^{-1} - 675e^{\frac{1}{4}t}\times e^{\frac{1}{4}t}\left(3e^{\frac{1}{4}t}-1\right)^{-2}$ | A1 | May be left unsimplified |
---
# Question c(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Substitutes $t=8$ into $\frac{dP}{dt}$ and calculates a value | M1 | |
| $-3.71$ only | A1 | **Not** awrt; if candidate writes $3.71$ this is A0 |
---9. A rare species of mammal is being studied. The population $P$, $t$ years after the study started, is modelled by the formula
$$P = \frac { 900 \mathrm { e } ^ { \frac { 1 } { 4 } t } } { 3 \mathrm { e } ^ { \frac { 1 } { 4 } t } - 1 } , \quad t \in \mathbb { R } , \quad t \geqslant 0$$
Using the model,
\begin{enumerate}[label=(\alph*)]
\item calculate the number of mammals at the start of the study,
\item calculate the exact value of $t$ when $P = 315$
Give your answer in the form $a \ln k$, where $a$ and $k$ are integers to be determined.
\item \begin{enumerate}[label=(\roman*)]
\item Find $\frac { \mathrm { d } P } { \mathrm {~d} t }$
\item Hence find the value of $\frac { \mathrm { d } P } { \mathrm {~d} t }$ when $t = 8$, giving your answer to 2 decimal places.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{Edexcel C34 2018 Q9 [9]}}