OCR Further Pure Core 1 2018 March — Question 9 8 marks

Exam BoardOCR
ModuleFurther Pure Core 1 (Further Pure Core 1)
Year2018
SessionMarch
Marks8
TopicFirst order differential equations (integrating factor)
TypeModelling with rate proportional to amount
DifficultyStandard +0.8 This is a non-standard differential equation requiring separation of variables with partial fractions (arctan integration), parameter determination from two conditions, and careful algebraic manipulation. While the setup is straightforward, the integration technique and solving for the constant with the given conditions elevates this above typical first-order DE questions.
Spec1.08k Separable differential equations: dy/dx = f(x)g(y)1.08l Interpret differential equation solutions: in context
9 In an experiment, at time \(t\) minutes there is \(Q\) grams of substance present.
It is known that the substance decays at a rate that is proportional to \(1 + Q ^ { 2 }\). Initially there are 100 grams of the substance present and after 100 minutes there are 50 grams present. Find the amount of the substance present after 400 minutes.
AnswerMarks Guidance
\(\frac{dQ}{dt} = -k\left(1 + Q^2\right) \Rightarrow \int \frac{dQ}{1+Q^2} = \int -k\,dt\)B1, M1 3.3, 3.1b
\(\Rightarrow \tan^{-1}Q = c - kt\)A1 1.1
\(t = 0, Q = 100 \Rightarrow c = \tan^{-1}100\)A1 3.4
\(t = 100, Q = 50 \Rightarrow \tan^{-1}50 = \tan^{-1}100 - 100k\)A1 3.4
\(\Rightarrow k = \frac{1}{100}\left(\tan^{-1}100 - \tan^{-1}50\right) \approx 0.0001\)M1 1.1a
When \(t = 400\), \(Q = 20\)A1 3.4
\(20\,\text{g}\)A1 3.2b
[8]
Question 10 (i) (a):
AnswerMarks Guidance
\(\sec\theta = \frac{r}{x}\)B1 1.1
\(\Rightarrow r = 2 - \frac{r}{x}\)B1 3.1a
\(\Rightarrow r = \frac{2x}{x+1}\)B1 3.1a
\(y^2 = r^2 - x^2 \Rightarrow y^2 = \left(\frac{2x}{x+1}\right)^2 - x^2\)A1 2.1
[4]
Question 10 (i) (b):
AnswerMarks Guidance
Because the curve is \(y^2 = ...\) and so \(x = p\) gives \(y = \pm q\)B1 2.4
[1]
Question 10 (i) (c):
AnswerMarks Guidance
\(a = -1\)B1 2.2a
[1]
Question 10 (ii):
AnswerMarks Guidance
\(V = \pi\int y^2 dx\)M1 1.1a
\(= \pi\int\left[\left(\frac{2x}{x+1}\right)^2 - x^2\right]dx = \pi\left[4 - x^2 - 4\left(\frac{2x+2}{x^2+2x+1}\right) + \frac{4}{(x+1)^2}\right]dx\)M1 3.1a
\(\pi\left[4x - \frac{x^3}{3} - 4\ln(x+1)^2 - \frac{4}{(x+1)}\right]\)M1, A1 3.1a, 1.1
Limits are 0 and 1B1, M1 3.1a, 1.1
\(\Rightarrow V = \pi\left[\left(4 - \frac{1}{3} - 4\ln 4 - 2\right) - (0 - 0 - 0 - 4)\right]\)A1 1.1
\(= \pi\left(\frac{17}{3} - 4\ln 4\right)\)
[8] 
$\frac{dQ}{dt} = -k\left(1 + Q^2\right) \Rightarrow \int \frac{dQ}{1+Q^2} = \int -k\,dt$ | B1, M1 | 3.3, 3.1b |

$\Rightarrow \tan^{-1}Q = c - kt$ | A1 | 1.1 |

$t = 0, Q = 100 \Rightarrow c = \tan^{-1}100$ | A1 | 3.4 |

$t = 100, Q = 50 \Rightarrow \tan^{-1}50 = \tan^{-1}100 - 100k$ | A1 | 3.4 |

$\Rightarrow k = \frac{1}{100}\left(\tan^{-1}100 - \tan^{-1}50\right) \approx 0.0001$ | M1 | 1.1a |

When $t = 400$, $Q = 20$ | A1 | 3.4 |

$20\,\text{g}$ | A1 | 3.2b | Must have unit |

| [8] | |

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# Question 10 (i) (a):

$\sec\theta = \frac{r}{x}$ | B1 | 1.1 |

$\Rightarrow r = 2 - \frac{r}{x}$ | B1 | 3.1a |

$\Rightarrow r = \frac{2x}{x+1}$ | B1 | 3.1a |

$y^2 = r^2 - x^2 \Rightarrow y^2 = \left(\frac{2x}{x+1}\right)^2 - x^2$ | A1 | 2.1 | AG |

| [4] | |

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# Question 10 (i) (b):

Because the curve is $y^2 = ...$ and so $x = p$ gives $y = \pm q$ | B1 | 2.4 |

| [1] | |

---

# Question 10 (i) (c):

$a = -1$ | B1 | 2.2a | Accept $x = -1$ |

| [1] | |

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# Question 10 (ii):

$V = \pi\int y^2 dx$ | M1 | 1.1a | Use of formula |

$= \pi\int\left[\left(\frac{2x}{x+1}\right)^2 - x^2\right]dx = \pi\left[4 - x^2 - 4\left(\frac{2x+2}{x^2+2x+1}\right) + \frac{4}{(x+1)^2}\right]dx$ | M1 | 3.1a | any equivalent form that can be integrated |

$\pi\left[4x - \frac{x^3}{3} - 4\ln(x+1)^2 - \frac{4}{(x+1)}\right]$ | M1, A1 | 3.1a, 1.1 | Attempt to integrate |

Limits are 0 and 1 | B1, M1 | 3.1a, 1.1 |

$\Rightarrow V = \pi\left[\left(4 - \frac{1}{3} - 4\ln 4 - 2\right) - (0 - 0 - 0 - 4)\right]$ | A1 | 1.1 | $4\ln 4$ and $8\ln 2$ are equivalent |

$= \pi\left(\frac{17}{3} - 4\ln 4\right)$ | |

| [8] | |
9 In an experiment, at time $t$ minutes there is $Q$ grams of substance present.\\
It is known that the substance decays at a rate that is proportional to $1 + Q ^ { 2 }$. Initially there are 100 grams of the substance present and after 100 minutes there are 50 grams present.

Find the amount of the substance present after 400 minutes.

\hfill \mbox{\textit{OCR Further Pure Core 1 2018 Q9 [8]}}
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