| Exam Board | OCR |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2011 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differential equations |
| Type | Separable variables - standard (applied/contextual) |
| Difficulty | Standard +0.3 This is a straightforward separable differential equations question requiring standard integration technique (substitution with u = x - 8) and evaluation at given limits. The method is routine for C4, though the fractional power and two-part structure adds minor complexity beyond the most basic examples. |
| Spec | 1.08k Separable differential equations: dy/dx = f(x)g(y) |
| Answer | Marks | Guidance |
|---|---|---|
| Attempt to sep variables in the form \(\int \frac{p}{(x-8)^q}dx = \int q\,dt\) | M1 | Or invert as \(\frac{dr}{dx} = \frac{p}{(x-8)^q}\) ; \(p,q,r\) costs |
| \(\int \frac{1}{(x-8)^q}dx = k(x-8)^{p}\) | A1 | \(k\) cost |
| All correct \((+ c)\) | A1 | |
| For equation containing 'c'; substitute \(t = 0, x = 72\) | M1 | M2 for \(\int = \) or \(\int = \) |
| Correct corresponding value of \(c\) from correct eqn | A1 | |
| Subst their \(c\) & \(x = 35\) back into eqn | M1 | |
| \(t = \frac{21}{8}\) or 2.63 / 2.625 [C.A.O] | A1 7 | A2: \(t = \frac{21}{8}\) or 2.63 / 2.625 WWW |
| Answer | Marks |
|---|---|
| State/imply in some way that \(x = 8\) when flow stops | B1 |
| Substitute \(x = 8\) back into equation containing numeric 'c' | M1 |
| \(t = 6\) | A1 3 |
**(i)**
Attempt to sep variables in the form $\int \frac{p}{(x-8)^q}dx = \int q\,dt$ | M1 | Or invert as $\frac{dr}{dx} = \frac{p}{(x-8)^q}$ ; $p,q,r$ costs
$\int \frac{1}{(x-8)^q}dx = k(x-8)^{p}$ | A1 | $k$ cost
All correct $(+ c)$ | A1 |
For equation containing 'c'; substitute $t = 0, x = 72$ | M1 | M2 for $\int = $ or $\int = $
Correct corresponding value of $c$ from correct eqn | A1 |
Subst their $c$ & $x = 35$ back into eqn | M1 |
$t = \frac{21}{8}$ or 2.63 / 2.625 [C.A.O] | A1 7 | A2: $t = \frac{21}{8}$ or 2.63 / 2.625 WWW
**(ii)**
State/imply in some way that $x = 8$ when flow stops | B1 |
Substitute $x = 8$ back into equation containing numeric 'c' | M1 |
$t = 6$ | A1 3 |9 Paraffin is stored in a tank with a horizontal base. At time $t$ minutes, the depth of paraffin in the tank is $x \mathrm {~cm}$. When $t = 0 , x = 72$. There is a tap in the side of the tank through which the paraffin can flow. When the tap is opened, the flow of the paraffin is modelled by the differential equation
$$\frac { \mathrm { d } x } { \mathrm {~d} t } = - 4 ( x - 8 ) ^ { \frac { 1 } { 3 } }$$
(i) How long does it take for the level of paraffin to fall from a depth of 72 cm to a depth of 35 cm ?\\
(ii) The tank is filled again to its original depth of 72 cm of paraffin and the tap is then opened. The paraffin flows out until it stops. How long does this take?
\hfill \mbox{\textit{OCR C4 2011 Q9 [10]}}