Edexcel C1 — Question 8 11 marks

Exam BoardEdexcel
ModuleC1 (Core Mathematics 1)
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicApplied differentiation
TypeSpreading stain or growing patch area
DifficultyModerate -0.8 This is a straightforward related rates problem requiring substitution to find constants, basic differentiation using the chain rule, and evaluation at a point. All steps are routine C1 techniques with no problem-solving insight needed, making it easier than average but not trivial due to the multi-part structure.
Spec1.02e Complete the square: quadratic polynomials and turning points1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates

8. Some ink is poured onto a piece of cloth forming a stain that then spreads. The area of the stain, \(A \mathrm {~cm} ^ { 2 }\), after \(t\) seconds is given by $$A = ( p + q t ) ^ { 2 } ,$$ where \(p\) and \(q\) are positive constants.
Given that when \(t = 0 , A = 4\) and that when \(t = 5 , A = 9\),
  1. find the value of \(p\) and show that \(q = \frac { 1 } { 5 }\),
  2. find \(\frac { \mathrm { d } A } { \mathrm {~d} t }\) in terms of \(t\),
  3. find the rate at which the area of the stain is increasing when \(t = 15\).

AnswerMarks Guidance
(a) \(t = 0, A = 4 \Rightarrow 4 = p^2\)M1
\(p > 0 \therefore p = 2\)A1
\(t = 5, A = 9 \Rightarrow 9 = (2 + 5q)^2\)M1
\(2 + 5q = \pm 3\)
\(q = \frac{1}{5}(-2 \pm 3)\)M1
\(q > 0 \therefore q = \frac{1}{5}\)A1
(b) \(A = (2 + \frac{1}{5}t)^2 = 4 + \frac{4}{5}t + \frac{1}{25}t^2\)M1 A1
\(\frac{dA}{dt} = \frac{4}{5} + \frac{2}{25}t\)M1 A1
(c) \(t = 15 \therefore \frac{dA}{dt} = \frac{4}{5} + \frac{2}{25}(15) = 2 \text{ cm}^2\text{s}^{-1}\)M1 A1 (11)
(a) $t = 0, A = 4 \Rightarrow 4 = p^2$ | M1 |
$p > 0 \therefore p = 2$ | A1 |
$t = 5, A = 9 \Rightarrow 9 = (2 + 5q)^2$ | M1 |
$2 + 5q = \pm 3$ | |
$q = \frac{1}{5}(-2 \pm 3)$ | M1 |
$q > 0 \therefore q = \frac{1}{5}$ | A1 |
(b) $A = (2 + \frac{1}{5}t)^2 = 4 + \frac{4}{5}t + \frac{1}{25}t^2$ | M1 A1 |
$\frac{dA}{dt} = \frac{4}{5} + \frac{2}{25}t$ | M1 A1 |
(c) $t = 15 \therefore \frac{dA}{dt} = \frac{4}{5} + \frac{2}{25}(15) = 2 \text{ cm}^2\text{s}^{-1}$ | M1 A1 | (11)
8. Some ink is poured onto a piece of cloth forming a stain that then spreads.

The area of the stain, $A \mathrm {~cm} ^ { 2 }$, after $t$ seconds is given by

$$A = ( p + q t ) ^ { 2 } ,$$

where $p$ and $q$ are positive constants.\\
Given that when $t = 0 , A = 4$ and that when $t = 5 , A = 9$,
\begin{enumerate}[label=(\alph*)]
\item find the value of $p$ and show that $q = \frac { 1 } { 5 }$,
\item find $\frac { \mathrm { d } A } { \mathrm {~d} t }$ in terms of $t$,
\item find the rate at which the area of the stain is increasing when $t = 15$.
\end{enumerate}

\hfill \mbox{\textit{Edexcel C1  Q8 [11]}}