C4 June 2014 (R) Q7

EdexcelOld spec10 marksIntegration Modelling

7. The rate of increase of the number, \(N\), of fish in a lake is modelled by the differential equation \[\frac{\mathrm{d}N}{\mathrm{d}t} = \frac{(kt - 1)(5000 - N)}{t} \qquad t > 0,\quad 0 < N < 5000\]

In the given equation, the time \(t\) is measured in years from the start of January 2000 and \(k\) is a positive constant.

(a) By solving the differential equation, show that \[N = 5000 - At\mathrm{e}^{-kt}\] where \(A\) is a positive constant. (5)

After one year, at the start of January 2001, there are 1200 fish in the lake.

After two years, at the start of January 2002, there are 1800 fish in the lake.

(b) Find the exact value of the constant \(A\) and the exact value of the constant \(k\). (4)

Mark scheme (a)

Scheme

Marks

\(\dfrac{\mathrm{d}N}{\mathrm{d}t} = \dfrac{(kt - 1)(5000 - N)}{t},\quad t > 0,\ 0 < N < 5000\)

\(\displaystyle\int \dfrac{1}{5000 - N}\,\mathrm{d}N = \displaystyle\int \dfrac{(kt - 1)}{t}\,\mathrm{d}t\quad \left\{\text{or } = \displaystyle\int\left(k - \dfrac{1}{t}\right)\mathrm{d}t\right\}\)
See notes

\(-\ln\left(5000 - N\right) = kt - \ln t;\ + c\)
See notes

M1 A1; A1

then eg either...	or...	or...
\(-kt + c = \ln\left(5000 - N\right) - \ln t\) \(-kt + c = \ln\left(\dfrac{5000 - N}{t}\right)\) \(\mathrm{e}^{-kt + c} = \dfrac{5000 - N}{t}\)	\(kt + c = \ln t - \ln\left(5000 - N\right)\) \(kt + c = \ln\left(\dfrac{t}{5000 - N}\right)\) \(\mathrm{e}^{kt + c} = \dfrac{t}{5000 - N}\)	\(\ln\left(5000 - N\right) = -kt + \ln t + c\) \(5000 - N = \mathrm{e}^{-kt + \ln t + c}\) \(5000 - N = t\mathrm{e}^{-kt + c}\)

leading to \(N = 5000 - At\mathrm{e}^{-kt}\) with no incorrect working/statements. See notes

A1 * cso

(5)

Notes

B1: Separates variables as shown. \(\mathrm{d}N\) and \(\mathrm{d}t\) should be in the correct positions, though this mark can be implied by later working. Ignore the integral signs.

M1: Either \(\pm\lambda\ln\left(5000 - N\right)\) or \(\pm\lambda\ln\left(N - 5000\right)\) or \(kt - \ln t\) where \(\lambda \neq 0\) is a constant.

A1: For \(-\ln\left(5000 - N\right) = kt - \ln t\) or \(\ln\left(5000 - N\right) = -kt + \ln t\) or \(-\dfrac{1}{k}\ln\left(5000 - N\right) = t - \dfrac{1}{k}\ln t\) oe

A1: which is dependent on the 1^st M1 mark being awarded.
For applying a constant of integration, eg. \(+ c\) or \(+ \ln\mathrm{e}^c\) or \(+ \ln c\) or \(A\) to their integrated equation

Note: \(+ c\) can be on either side of their equation for the 2^nd A1 mark.

A1: Uses a constant of integration eg. “\(c\)” or “\(\ln\mathrm{e}^c\)” “\(\ln c\)” or applies a fully correct method to prove the result \(N = 5000 - At\mathrm{e}^{-kt}\) with no incorrect working seen. (Correct solution only.)

NOTE: IMPORTANT
There needs to be an intermediate stage of justifying the \(A\) and the \(\mathrm{e}^{-kt}\) in \(At\mathrm{e}^{-kt}\) by for example

either \(5000 - N = \mathrm{e}^{\ln t - kt + c}\)
or \(5000 - N = t\mathrm{e}^{-kt + c}\)
or \(5000 - N = t\mathrm{e}^{-kt}\mathrm{e}^c\)

or equivalent needs to be stated before achieving \(N = 5000 - At\mathrm{e}^{-kt}\)

Mark scheme (b)

Scheme	Marks
\(\{t = 1,\ N = 1200 \Rightarrow\}\quad 1200 = 5000 - A\mathrm{e}^{-k}\) \(\{t = 2,\ N = 1800 \Rightarrow\}\quad 1800 = 5000 - 2A\mathrm{e}^{-2k}\) At least one correct statement written down using the boundary conditions	B1
So \(\quad A\mathrm{e}^{-k} = 3800\) and \(\quad 2A\mathrm{e}^{-2k} = 3200\) or \(A\mathrm{e}^{-2k} = 1600\)
Eg. \(\dfrac{\mathrm{e}^{-k}}{2\mathrm{e}^{-2k}} = \dfrac{3800}{3200}\) or \(\dfrac{2\mathrm{e}^{-2k}}{\mathrm{e}^{-k}} = \dfrac{3200}{3800}\) So \(\dfrac{1}{2}\mathrm{e}^{k} = \dfrac{3800}{3200}\) or \(2\mathrm{e}^{-k} = \dfrac{3200}{3800}\) An attempt to eliminate \(A\) by producing an equation in only \(k\).	M1
\(k = \ln\left(\dfrac{7600}{3200}\right)\) or equivalent \(\left\{\text{eg } k = \ln\left(\dfrac{19}{8}\right)\right\}\) At least one of \(A = 9025\) cao or \(k = \ln\left(\dfrac{7600}{3200}\right)\) or exact equivalent	A1
\(\left\{A = 3800\left(\mathrm{e}^{k}\right) = 3800\left(\dfrac{19}{8}\right) \Rightarrow\right\}\ A = 9025\) Both \(A = 9025\) cao and \(k = \ln\left(\dfrac{7600}{3200}\right)\) or exact equivalent (corrected from the printed mark scheme: “Both … or …”)	A1
	(4)

Alternative Method for the M1 mark in (b)

Scheme	Marks
\(\mathrm{e}^{-k} = \dfrac{3800}{A}\) \(2A\left(\dfrac{3800}{A}\right)^2 = 3200\) An attempt to eliminate \(k\) by producing an equation in only \(A\)	M1

Notes

B1: At least one of either \(1200 = 5000 - A\mathrm{e}^{-k}\) (or equivalent) or \(1800 = 5000 - 2A\mathrm{e}^{-2k}\) (or equivalent)

M1: Either an attempt to eliminate \(A\) by producing an equation in only \(k\).
or an attempt to eliminate \(k\) by producing an equation in only \(A\)

A1: At least one of \(A = 9025\) cao or \(k = \ln\left(\dfrac{7600}{3200}\right)\) or equivalent

A1: Both \(A = 9025\) cao and \(k = \ln\left(\dfrac{7600}{3200}\right)\) or equivalent (corrected from the printed mark scheme: “Both … or …”)

Note: Alternative correct values for \(k\) are \(k = \ln\left(\dfrac{19}{8}\right)\) or \(k = -\ln\left(\dfrac{8}{19}\right)\) or \(k = \ln 7600 - \ln 3200\)
or \(k = -\ln\left(\dfrac{3800}{9025}\right)\) or equivalent.

Note: \(k = 0.8649\ldots\) without a correct exact equivalent is A0.

Mark scheme (c)

Scheme	Marks
\(\left\{t = 5,\ N = 5000 - 9025(5)\mathrm{e}^{-5\ln\left(\frac{19}{8}\right)}\right\}\)
\(N = 4402.828401\ldots = 4400\) (fish) (nearest 100) anything that rounds to 4400	B1
	(1)
	(10 marks)

Notes

B1: anything that rounds to 4400