# Question 7:

## Part (a):

$\frac{dx}{dt} = 81\sec^2 t \sec t \tan t$, $\frac{dy}{dt} = 3\sec^2 t$ | B1 | At least one of $\frac{dx}{dt}$ or $\frac{dy}{dt}$ correct

| B1 | Both $\frac{dx}{dt}$ and $\frac{dy}{dt}$ are correct

$\frac{dy}{dx} = \frac{3\sec^2 t}{81\sec^3 t \tan t} = \frac{1}{27\sec t \tan t} = \frac{\cos t}{27\tan t} = \frac{\cos^2 t}{27\sin t}$ | M1 | Applies their $\frac{dy}{dt}$ divided by their $\frac{dx}{dt}$

At $t = \frac{\pi}{6}$: $\frac{dy}{dx} = \frac{3\sec^2\!\left(\frac{\pi}{6}\right)}{81\sec^3\!\left(\frac{\pi}{6}\right)\tan\!\left(\frac{\pi}{6}\right)} = \frac{4}{72} = \frac{3}{54} = \frac{1}{18}$ | A1 cao cso | $\frac{4}{72}$

**[4]**

### Notes (a):
- **B1:** At least one of $\frac{dx}{dt}$ or $\frac{dy}{dt}$ correct. Note: can be implied from working.
- **B1:** Both $\frac{dx}{dt}$ and $\frac{dy}{dt}$ correct. Note: can be implied from working.
- **M1:** Applies their $\frac{dy}{dt}$ divided by their $\frac{dx}{dt}$, where both are trigonometric functions of $t$.
- **A1:** $\frac{4}{72}$ or any equivalent correct rational answer not involving surds. Allow $0.0\dot{5}$ with recurring symbol.
- Note: check that $\frac{dx}{dt}$ is differentiated correctly. E.g. $x = 27\sec^3 t = 27(\cos t)^{-3} \Rightarrow \frac{dx}{dt} = -81(\cos t)^{-2}(-\sin t)$ is correct.

### Way 2 (Alternative using Cartesian equation):
$y = \left(x^{\frac{2}{3}}-9\right)^{\frac{1}{2}} \Rightarrow \frac{dy}{dx} = \frac{1}{2}\left(x^{\frac{2}{3}}-9\right)^{-\frac{1}{2}}\left(\frac{2}{3}x^{-\frac{1}{3}}\right)$ | M1 | $\frac{dy}{dx} = \pm Kx^{-\frac{1}{3}}\left(x^{\frac{2}{3}}-9\right)^{-\frac{1}{2}}$

| A1 | $\frac{dy}{dx} = \frac{1}{2}\left(x^{\frac{2}{3}}-9\right)^{-\frac{1}{2}}\left(\frac{2}{3}x^{-\frac{1}{3}}\right)$ oe

At $t = \frac{\pi}{6}$, $x = 27\sec^3\!\left(\frac{\pi}{6}\right) = 24\sqrt{3}$ | dM1 | Uses $t = \frac{\pi}{6}$ to find $x$ and substitutes into $\frac{dy}{dx}$

$\frac{dy}{dx} = \frac{1}{2}\!\left(\frac{1}{\sqrt{3}}\right)\!\left(\frac{1}{3\sqrt{3}}\right) = \frac{1}{18}$ | A1 cao cso | $\frac{1}{18}$

Note: Way 2 marked as M1 A1 dM1 A1. Second M1 dependent on first M1.

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## Part (b):

$\{1+\tan^2 t = \sec^2 t\} \Rightarrow 1 + \left(\frac{y}{3}\right)^2 = \left(\sqrt[3]{\frac{x}{27}}\right)^2 = \left(\frac{x}{27}\right)^{\frac{2}{3}}$ | M1 | Applying a correct trig identity (usually $1+\tan^2 t = \sec^2 t$) to give a Cartesian equation in $x$ and $y$ only

$\Rightarrow 1 + \frac{y^2}{9} = \frac{x^{\frac{2}{3}}}{9} \Rightarrow 9 + y^2 = x^{\frac{2}{3}} \Rightarrow y = \left(x^{\frac{2}{3}}-9\right)^{\frac{1}{2}}$ | A1* cso |

$a = 27$ and $b = 216$ or $27 \leqslant x \leqslant 216$ | B1 | Both $a=27$ and $b=216$

**[3]**

### Notes (b):
- **M1:** Either: applying correct trig identity; or starting from RHS to achieve $\sqrt{9\tan^2 t}$; or starting from LHS to achieve $\sqrt{9\sec^2 t - 9}$.
- **A1\*:** For a correct proof of $y = \left(x^{\frac{2}{3}}-9\right)^{\frac{1}{2}}$. Note this result is printed on Question Paper, so no incorrect working allowed.
- **B1:** Both $a=27$ and $b=216$. Note $27 \leqslant x \leqslant 216$ is also fine.

### Way 2 (Starting from RHS):
$\left(x^{\frac{2}{3}}-9\right)^{\frac{1}{2}} = \sqrt{(27\sec^3 t)^{\frac{2}{3}}-9} = \sqrt{9\sec^2 t - 9} = \sqrt{9\tan^2 t}$ | M1 | Applying $1+\tan^2 t = \sec^2 t$ to achieve $\sqrt{9\tan^2 t}$

$= 3\tan t = y$ {= LHS} cso | A1* | Correct proof from $\left(x^{\frac{2}{3}}-9\right)^{\frac{1}{2}}$ to $y$

### Way 3 (Starting from LHS):
$y = 3\tan t = \sqrt{9\tan^2 t} = \sqrt{9\sec^2 t - 9}$ | M1 | Applying $1+\tan^2 t = \sec^2 t$ to achieve $\sqrt{9\sec^2 t-9}$

$= \sqrt{9\!\left(\frac{x}{27}\right)^{\frac{2}{3}}-9} = \sqrt{9\!\left(\frac{x^{\frac{2}{3}}}{9}\right)-9} = \left(x^{\frac{2}{3}}-9\right)^{\frac{1}{2}}$ cso | A1* | Correct proof from $y$ to $\left(x^{\frac{2}{3}}-9\right)^{\frac{1}{2}}$

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## Part (c):

$V = \pi\displaystyle\int_{27}^{125}\!\left(\left(x^{\frac{2}{3}}-9\right)^{\frac{1}{2}}\right)^2 dx$ or $\pi\displaystyle\int_{27}^{125}\!\left(x^{\frac{2}{3}}-9\right)dx$ | B1 | For $\pi\int\!\left(\left(x^{\frac{2}{3}}-9\right)^{\frac{1}{2}}\right)^2$ or $\pi\int\!\left(x^{\frac{2}{3}}-9\right)$. Ignore limits and $dx$. Can be implied.

$= \{\pi\}\!\left[\frac{3}{5}x^{\frac{5}{3}} - 9x\right]_{27}^{125}$ | M1 | Either $\pm Ax^{\frac{5}{3}} \pm Bx$, $A\neq 0$, $B\neq 0$ or integrates $x^{\frac{2}{3}}$ correctly to give $\frac{3}{5}x^{\frac{5}{3}}$ oe

| A1 | $\frac{3}{5}x^{\frac{5}{3}} - 9x$ oe

$= \{\pi\}\!\left[\left(\frac{3}{5}(125)^{\frac{5}{3}}-9(125)\right) - \left(\frac{3}{5}(27)^{\frac{5}{3}}-9(27)\right)\right]$ | dM1 | Substitutes limits of 125 and 27 into integrated function and subtracts correct way round

$= \{\pi\}\!\left[(1875-1125)-(145.8-243)\right]$

$= \dfrac{4236\pi}{5}$ or $847.2\pi$ | A1 | $\dfrac{4236\pi}{5}$ or $847.2\pi$

**[5]**

### Notes (c):
- **B1:** For a correct statement of $\pi\int\!\left(\left(x^{\frac{2}{3}}-9\right)^{\frac{1}{2}}\right)^2$ or $\pi\int\!\left(x^{\frac{2}{3}}-9\right)$. Ignore limits and $dx$. Can be implied.
- **M1:** Either integrates to give $\pm Ax^{\frac{5}{3}} \pm Bx$, $A\neq 0$, $B\neq 0$ or integrates $x^{\frac{2}{3}}$ correctly to give $\frac{3}{5}x^{\frac{5}{3}}$ oe.
- **A1:** $\frac{3}{5}x^{\frac{5}{3}}-9x$ or $\dfrac{x^{\frac{5}{3}}}{\left(\frac{5}{3}\right)}-9x$ oe.
- **dM1:** Substitutes limits of 125 and 27 into an integrated function and subtracts the correct way round. Dependent on previous M1.
- **A1:** A correct exact answer of $\dfrac{4236\pi}{5}$ or $847.2\pi$.
- Note: $\pi$ in volume formula only required for B1 and final A1.
- Note: decimal answer of 2661.557... without correct exact answer is A0.
- Note: if candidate gains B1M1A1 then writes 2661 or awrt 2662 with no method for substituting limits, award final M1A0.

### Way 2 (Parametric integration):
$V = \pi\displaystyle\int 9\tan^2 t\,(81\sec^2 t\sec t\tan t)\,dt$ | B1 | $\pi\int 3\tan t\,(81\sec^2 t\sec t\tan t)\,dt$. Ignore limits and $dx$. Can be implied.

$= \{\pi\}\displaystyle\int 729\sec^2 t(\sec^2 t-1)\sec t\tan t\,dt$

$= \{\pi\}\displaystyle\int 729(\sec^4 t - \sec^2 t)\sec t\tan t\,dt$

$= \{\pi\}\!\left[729\!\left(\frac{1}{5}\sec^5 t - \frac{1}{3}\sec^3 t\right)\right]$ | M1 | $\pm A\sec^5 t \pm B\sec^3 t$

| A1 | $729\!\left(\frac{1}{5}\sec^5 t - \frac{1}{3}\sec^3 t\right)$

$V = \{\pi\}\!\left[729\!\left(\frac{1}{5}\!\left(\frac{5}{3}\right)^5 - \frac{1}{3}\!\left(\frac{5}{3}\right)^3\right) - 729\!\left(\frac{1}{5}1^5 - \frac{1}{3}1^3\right)\right]$ | dM1 | Substitutes $\sec t = \frac{5}{3}$ and $\sec t = 1$ into integrated function and subtracts correct way round

$= 729\pi\!\left[\left(\frac{250}{243}\right)-\left(-\frac{2}{15}\right)\right] = \dfrac{4236\pi}{5}$ or $847.2\pi$ | A1 | $\dfrac{4236\pi}{5}$ or $847.2\pi$

**[5]**

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