# Question 13(a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $V = \frac{1}{3}\pi h^2(30-h) = 10\pi h^2 - \frac{1}{3}\pi h^3 \Rightarrow \frac{dV}{dh} = 20\pi h - \pi h^2$ or $\frac{dV}{dh} = \frac{2}{3}\pi h(30-h) - \frac{1}{3}\pi h^2$ | M1A1 | M1: Attempts $\frac{dV}{dh}$ by multiplying out and differentiating, or by product rule giving form $\alpha h(30-h)\pm\beta h^2$. A1: Any correct (possibly un-simplified) form for $\frac{dV}{dh}$ |
| Uses $\frac{dV}{dt} = \frac{dV}{dh}\times\frac{dh}{dt} \Rightarrow -\frac{1}{10}V = (20\pi h - \pi h^2)\times\frac{dh}{dt}$ | M1 | Uses correct form of chain rule $\frac{dV}{dt}=\frac{dV}{dh}\times\frac{dh}{dt}$ with their $\frac{dV}{dh}$ and $\frac{dV}{dt}=-\frac{1}{10}V$ |
| $\Rightarrow -\frac{1}{10}\times\frac{1}{3}\pi h^2(30-h) = \pi h(20-h)\times\frac{dh}{dt}$ | M1 | Substitutes $V=\frac{1}{3}\pi h^2(30-h)$ and rearranges to obtain $\frac{dh}{dt}$ in terms of $h$ |
| $\frac{dh}{dt} = -\frac{h(30-h)}{30(20-h)}$* | A1* | Given answer. Must have intermediate lines, correct factorisation, no errors, and "$\frac{dh}{dt} =$" seen. **(5)** |

# Question 13(b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{30(20-h)}{h(30-h)} \equiv \frac{A}{h} + \frac{B}{30-h}$ | B1 | Correct form for partial fractions |
| $30(20-h) \equiv A(30-h) + Bh$; $h=30 \Rightarrow 30B=-300 \Rightarrow B=-10$; $h=0 \Rightarrow 30A=600 \Rightarrow A=20$ | M1 | Attempts to get both constants by correct method e.g. substituting, comparing coefficients, cover-up rule |
| $\frac{30(20-h)}{h(30-h)} \equiv \frac{20}{h} - \frac{10}{30-h}$ | A1 | Correct partial fractions (or states "$A$" $= 20$, "$B$" $= -10$) **(3)** |

# Question 13(c) Way 1:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dh}{dt} = -\frac{h(30-h)}{30(20-h)} \Rightarrow \int\frac{30(20-h)}{h(30-h)}\,dh = -1\int\,dt$ | B1 | Correct statement (minus sign must be present on one side or other). Omission of "$dh$" and "$dt$" condoned if intention clear. |
| $20\ln h + 10\ln(30-h)$ | M1A1ft | M1: Integrates partial fractions to obtain $\pm P\ln h \pm Q\ln(30-h)$. A1ft: Correct integration following through their "$A$" and "$B$" |
| $t=0, h=10 \Rightarrow c = 20\ln 10 + 10\ln 20$ | M1 | Substitutes $h=10$ and $t=0$ to find value for $c$ |
| $h=5 \Rightarrow t = 20\ln 10 + 10\ln 20 - 10\ln 25 - 20\ln 5$ | ddM1 | Substitutes $h=5$ and uses their value of $c$ to find $t$ |
| $t = 11.63$ (secs) | A1cso | Awrt 11.63 only **(6) (14 marks)** |

# Question 13(c) Way 2:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dh}{dt} = -\frac{h(30-h)}{30(20-h)} \Rightarrow \int\frac{30(20-h)}{h(30-h)}\,dh = -1\int\,dt$ | B1 | Correct statement with minus sign present. |
| $20\ln h + 10\ln(30-h)$ | M1A1ft | M1: Integrates partial fractions to obtain $\pm P\ln h \pm Q\ln(30-h)$. A1ft: Correct integration following through their values. |
| $(t=)\left[20\ln h + 10\ln(30-h)\right]_{5}^{10}$ or $\left[20\ln h + 10\ln(30-h)\right]_{10}^{5}$ | M1 | Attempts limits 5 and 10 for $h$ |
| $(t=)[20\ln 10 + 10\ln 20]-[20\ln 5 + 10\ln 25]$ | ddM1 | Substitutes $h=5$ and $h=10$ to find value for $t$ |
| $t = 11.63$ | A1cso | Awrt 11.63 only **(6)** |

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