OCR MEI C4 — Question 3 18 marks

Exam BoardOCR MEI
ModuleC4 (Core Mathematics 4)
Marks18
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicDifferential equations
TypeTank/container - variable cross-section (cone/hemisphere/other)
DifficultyStandard +0.3 This is a standard C4 related rates and differential equations question with clear scaffolding. Parts (i) and (iii) involve routine differentiation and chain rule application, while parts (ii) and (iv) require separating variables and integrating—all standard techniques. The question guides students through each step with 'show that' targets, making it slightly easier than average for C4.
Spec1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates1.08k Separable differential equations: dy/dx = f(x)g(y)
\includegraphics{figure_3} Fig. 9 shows a hemispherical bowl, of radius 10 cm, filled with water to a depth of \(x\) cm. It can be shown that the volume of water, \(V\) cm\(^3\), is given by $$V = \pi(10x^2 - \frac{1}{3}x^3).$$ Water is poured into a leaking hemispherical bowl of radius 10 cm. Initially, the bowl is empty. After \(t\) seconds, the volume of water is changing at a rate, in cm\(^3\) s\(^{-1}\), given by the equation $$\frac{dV}{dt} = k(20 - x),$$ where \(k\) is a constant.
  1. Find \(\frac{dV}{dx}\), and hence show that \(\pi x \frac{dx}{dt} = k\). [4]
  2. Solve this differential equation, and hence show that the bowl fills completely after \(T\) seconds, where \(T = \frac{50\pi}{k}\). [5]
Once the bowl is full, the supply of water to the bowl is switched off, and water then leaks out at a rate of \(kx\) cm\(^3\) s\(^{-1}\).
  1. Show that, \(t\) seconds later, \(\pi(20 - x) \frac{dx}{dt} = -k\). [3]
  2. Solve this differential equation. Hence show that the bowl empties in 37 seconds. [6]
Question 3:
AnswerMarks Guidance
3(i) dV/dx = (20x  x2)
dV dV dx
  .
dt dx dt
dx
πx(20x). k(20x)
dt
dx
 πx k*
AnswerMarks
dtB1
M1
AA11
A1
AnswerMarks
[4]o
ag
AnswerMarks Guidance
.3(ii) πxdxkdt
 ½ x2 = kt + c
When t = 0, x = 0  c = 0
 ½ x2 = kt
Full when x = 10, t = T
 50 = kT
AnswerMarks
 T = 50/k *M1
A1
B1
MM11
A1
AnswerMarks
[5]separate variables and attempt integration of both sides
condone absence of c
c=0 www
substitut t or T=50 π/k or x=10 and rearranging for the other
(dependent on first M1) oe
ag, need to have c=0
AnswerMarks Guidance
3(iii) dV/dt = kx
dx
 πx(20x). kx
dt
dx
 π(20x) k*
AnswerMarks
dtB1
M1
A1
AnswerMarks
[3]correct
dV/dx.dx/dt= ±kx ft
ag
AnswerMarks Guidance
QuueessttiionAnswer Marks
3(iv) π(20x)dxkdt
(20x  ½ x2) = kt + c
When t = 0, x = 10
 (200  50) = c
 c = 150
 (20x  ½ x2) = 150  kt
x = 0 when 150  kt = 0
AnswerMarks
 t = 150/k = 3T*M1
B1
A1
AA11
MM11
A1
AnswerMarks
[6]separate variables and intend to integrate both sides
LHS (not dependent on M1)
RHS ie –kt +c (condone absence of c)
evaluatio of c cao oe (x=10, t=0)
substitut x=0 and rearrange for t -dependent on first M1 and non-
zero c ,oe
ag
4(i) When t = 0, v = 5(1 – e0) = 0
As t  , e–2t  0,  v  5
AnswerMarks
When t = 0.5, v = 3.16 m s–1E1
E1
B1
[3]
dv
(ii) 5(2)e2t 10e2t
dt
10 – 2v = 10 – 10(1 – e–2t) = 10e–2t
dv
 102v
AnswerMarks
dtB1
M1
E1
[3]
dv
(iii) 100.4v2
dt
10 dv
 1
1004v2 dt
10 dv
 4
25v2 dt
10 dv
 4*
(5v)(5v) dt
10 A B
 
(5v)(5v) 5v 5v
 10 = A(5 + v) + B(5 – v)
v = 5  10 = 10A  A = 1
v = –5  10 = 10B  B = 1
10 1 1
  
(5v)(5v) 5v 5v
1 1
 (  )dv4dt
5v 5v
 ln(5v)ln(5v)4tc
when t = 0, v = 0,  0 = 4×0 + c  c = 0
5v
 ln 4t
5v
1 5v
 t  ln  *
AnswerMarks
4 5vM1
E1
M1
A1
M1
A1
A1
E1
AnswerMarks
[8]for both A=1,B=1
separating variables correctly and
indicating integration
ft their A,B, condone absence of c
ft finding c from an expression of correct
form
(iv) When t  , e–4t  0,  v  5/1 = 5
5(1e2)
when t = 0.5, t  3.8ms1
AnswerMarks
1e2E1
M1A1
[3]
AnswerMarks
(v) The first modelE1
[1]www 
Question 3:
3 | (i) | dV/dx = (20x  x2)
dV dV dx
  .
dt dx dt
dx
πx(20x). k(20x)
dt
dx
 πx k*
dt | B1
M1
AA11
A1
[4] | o
ag
.3 | (ii) | πxdxkdt
 ½ x2 = kt + c
When t = 0, x = 0  c = 0
 ½ x2 = kt
Full when x = 10, t = T
 50 = kT
 T = 50/k * | M1
A1
B1
MM11
A1
[5] | separate variables and attempt integration of both sides
condone absence of c
c=0 www
substitut t or T=50 π/k or x=10 and rearranging for the other
(dependent on first M1) oe
ag, need to have c=0
3 | (iii) | dV/dt = kx
dx
 πx(20x). kx
dt
dx
 π(20x) k*
dt | B1
M1
A1
[3] | correct
dV/dx.dx/dt= ±kx ft
ag
Quueessttiion | Answer | Marks | Guidance
3 | (iv) | π(20x)dxkdt
(20x  ½ x2) = kt + c
When t = 0, x = 10
 (200  50) = c
 c = 150
 (20x  ½ x2) = 150  kt
x = 0 when 150  kt = 0
 t = 150/k = 3T* | M1
B1
A1
AA11
MM11
A1
[6] | separate variables and intend to integrate both sides
LHS (not dependent on M1)
RHS ie –kt +c (condone absence of c)
evaluatio of c cao oe (x=10, t=0)
substitut x=0 and rearrange for t -dependent on first M1 and non-
zero c ,oe
ag
4(i) When t = 0, v = 5(1 – e0) = 0
As t  , e–2t  0,  v  5
When t = 0.5, v = 3.16 m s–1 | E1
E1
B1
[3]
dv
(ii) 5(2)e2t 10e2t
dt
10 – 2v = 10 – 10(1 – e–2t) = 10e–2t
dv
 102v
dt | B1
M1
E1
[3]
dv
(iii) 100.4v2
dt
10 dv
 1
1004v2 dt
10 dv
 4
25v2 dt
10 dv
 4*
(5v)(5v) dt
10 A B
 
(5v)(5v) 5v 5v
 10 = A(5 + v) + B(5 – v)
v = 5  10 = 10A  A = 1
v = –5  10 = 10B  B = 1
10 1 1
  
(5v)(5v) 5v 5v
1 1
 (  )dv4dt
5v 5v
 ln(5v)ln(5v)4tc
when t = 0, v = 0,  0 = 4×0 + c  c = 0
5v
 ln 4t
5v
1 5v
 t  ln  *
4 5v | M1
E1
M1
A1
M1
A1
A1
E1
[8] | for both A=1,B=1
separating variables correctly and
indicating integration
ft their A,B, condone absence of c
ft finding c from an expression of correct
form
(iv) When t  , e–4t  0,  v  5/1 = 5
5(1e2)
when t = 0.5, t  3.8ms1
1e2 | E1
M1A1
[3]
(v) The first model | E1
[1] | www
\includegraphics{figure_3}

Fig. 9 shows a hemispherical bowl, of radius 10 cm, filled with water to a depth of $x$ cm. It can be shown that the volume of water, $V$ cm$^3$, is given by
$$V = \pi(10x^2 - \frac{1}{3}x^3).$$

Water is poured into a leaking hemispherical bowl of radius 10 cm. Initially, the bowl is empty. After $t$ seconds, the volume of water is changing at a rate, in cm$^3$ s$^{-1}$, given by the equation
$$\frac{dV}{dt} = k(20 - x),$$
where $k$ is a constant.

\begin{enumerate}[label=(\roman*)]
\item Find $\frac{dV}{dx}$, and hence show that $\pi x \frac{dx}{dt} = k$. [4]

\item Solve this differential equation, and hence show that the bowl fills completely after $T$ seconds, where $T = \frac{50\pi}{k}$. [5]
\end{enumerate}

Once the bowl is full, the supply of water to the bowl is switched off, and water then leaks out at a rate of $kx$ cm$^3$ s$^{-1}$.

\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{2}
\item Show that, $t$ seconds later, $\pi(20 - x) \frac{dx}{dt} = -k$. [3]

\item Solve this differential equation.

Hence show that the bowl empties in 37 seconds. [6]
\end{enumerate}

\hfill \mbox{\textit{OCR MEI C4  Q3 [18]}}
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