C3 June 2011 Q5
5. The mass, \(m\) grams, of a leaf \(t\) days after it has been picked from a tree is given by
\[m = p\mathrm{e}^{-kt}\]
where \(k\) and \(p\) are positive constants.
When the leaf is picked from the tree, its mass is 7.5 grams and 4 days later its mass is 2.5 grams.
(a) Write down the value of \(p\). (1)
(b) Show that \(k = \dfrac{1}{4}\ln 3\). (4)
(c) Find the value of \(t\) when \(\dfrac{\mathrm{d}m}{\mathrm{d}t} = -0.6\ln 3\). (6)
| Scheme | Marks |
|---|---|
| \(p = 7.5\) | B1 |
| (1) |
| Scheme | Marks |
|---|---|
| \(2.5 = 7.5e^{-4k}\) | M1 |
| \(e^{-4k} = \dfrac{1}{3}\) | M1 |
| \(-4k = \ln\left(\dfrac{1}{3}\right)\) | dM1 |
| \(-4k = -\ln(3)\) | |
| \(k = \dfrac{1}{4}\ln(3)\) | A1* |
| See notes for additional correct solutions and the last A1 | |
| (4) |
| Scheme | Marks |
|---|---|
| \(\dfrac{dm}{dt} = -kpe^{-kt}\) ft on their \(p\) and \(k\) | M1A1ft |
| \(-\dfrac{1}{4}\ln 3 \times 7.5e^{-\frac{1}{4}(\ln 3)t} = -0.6\ln 3\) | |
| \(e^{-\frac{1}{4}(\ln 3)t} = \dfrac{2.4}{7.5} = (0.32)\) | M1A1 |
| \(-\frac{1}{4}(\ln 3)t = \ln(0.32)\) | dM1 |
| \(t = 4.1486\ldots\) 4.15 or awrt 4.1 | A1 |
| (6) | |
| (11 marks) |