| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Year | 2019 |
| Session | January |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric differentiation |
| Type | Tangent/normal meets curve again |
| Difficulty | Challenging +1.2 This is a multi-part parametric question requiring chain rule differentiation, finding a tangent equation, and solving a simultaneous system. Parts (a) and (b) are standard C3/C4 techniques. Part (c) requires substituting the tangent equation back into parametric equations and solving algebraically, which is more demanding but still a familiar exam pattern. The algebra in part (c) elevates this above routine (0.0) but doesn't require novel insight. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dx}{dt}=2t-1\) | B1 | Correct derivative |
| Quotient rule: \(\frac{dy}{dt}=\frac{(1-t)\times4-4t\times(-1)}{(1-t)^2}\); obtains \(\frac{dy}{dt}=\frac{\alpha(1-t)\pm\beta t}{(1-t)^2}\), \(\alpha>0,\beta>0\); or product rule: \(\frac{dy}{dt}=4t(1-t)^{-2}+4(1-t)^{-1}\) | M1 | If incorrect formula quoted scores M0 |
| \(\frac{dy}{dx}=\frac{dy}{dt}\times\frac{dt}{dx}=\frac{(1-t)\times4-4t\times(-1)}{(1-t)^2}\times\frac{1}{2t-1}\) | M1 | Correct application of chain rule; their \(\frac{dy}{dt}\) divided by their \(\frac{dx}{dt}\) |
| \(\frac{dy}{dx}=\frac{4}{(2t-1)(1-t)^2}\) | A1 | Allow e.g. \(\frac{4}{(2t-1)(1-2t+t^2)}\); but not \(\frac{1}{(2t-1)}\times\frac{4}{(1-t)^2}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(t=-1\rightarrow(2,-2)\) or \(x=2,y=-2\) | B1 | Correct coordinates for \(P\) |
| \(t=-1\Rightarrow\frac{dy}{dx}=\frac{4}{(2(-1)-1)(1-(-1))^2}\left(=-\frac{1}{3}\right)\) | M1 | Attempts gradient; may be implied by their value for gradient |
| \(y+2=-\frac{1}{3}(x-2)\) | M1 | Correct straight line method for tangent not normal; if using \(y=mx+c\) must reach as far as finding \(c\) |
| \(x+3y+4=0\) | A1 | Any integer multiple |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(t^2-t+3\left(\frac{4t}{1-t}\right)+4=0\) | M1 | Substitutes to obtain equation in \(t\) only |
| \(t^3-2t^2-7t-4=0\) | A1 | Correct cubic |
| \((t+1)(t^2-3t-4)=0\) or \((t+1)^2(t-4)=0\); attempt to factorise using \((t\pm1)\) or \((t\pm1)^2\) as factor | M1 | Dependent on correct cubic with constant term |
| \(t=4\) | A1 | Correct value of \(t\) |
| \(\left(12,-\frac{16}{3}\right)\) | A1 | Correct coordinates |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y=\frac{4t}{1-t}\Rightarrow t=\frac{y}{4+y}\Rightarrow x=\left(\frac{y}{4+y}\right)^2-\frac{y}{4+y}\); \(\Rightarrow\left(\frac{y}{4+y}\right)^2-\frac{y}{4+y}+3y+4=0\) | M1 | Finds \(x\) in terms of \(y\) eliminating \(t\); algebra must be correct (allow sign errors only for making \(t\) the subject from \(y\)) |
| \(3y^3+28y^2+76y+64=0\) | A1 | Correct cubic |
| \((y+2)(3y^2+22y+32)=0\) or \((y+2)^2(3y+16)=0\); attempt to factorise using \((y\pm2)\) or \((y\pm2)^2\) as factor | M1 | Dependent on correct cubic with constant term |
| \(y=-\frac{16}{3}\) | A1 | Correct value of \(y\) |
| \(\left(12,-\frac{16}{3}\right)\) | A1 | Correct coordinates |
# Question 8:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dx}{dt}=2t-1$ | B1 | Correct derivative |
| Quotient rule: $\frac{dy}{dt}=\frac{(1-t)\times4-4t\times(-1)}{(1-t)^2}$; obtains $\frac{dy}{dt}=\frac{\alpha(1-t)\pm\beta t}{(1-t)^2}$, $\alpha>0,\beta>0$; or product rule: $\frac{dy}{dt}=4t(1-t)^{-2}+4(1-t)^{-1}$ | M1 | If incorrect formula quoted scores M0 |
| $\frac{dy}{dx}=\frac{dy}{dt}\times\frac{dt}{dx}=\frac{(1-t)\times4-4t\times(-1)}{(1-t)^2}\times\frac{1}{2t-1}$ | M1 | Correct application of chain rule; their $\frac{dy}{dt}$ divided by their $\frac{dx}{dt}$ |
| $\frac{dy}{dx}=\frac{4}{(2t-1)(1-t)^2}$ | A1 | Allow e.g. $\frac{4}{(2t-1)(1-2t+t^2)}$; but **not** $\frac{1}{(2t-1)}\times\frac{4}{(1-t)^2}$ |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $t=-1\rightarrow(2,-2)$ or $x=2,y=-2$ | B1 | Correct coordinates for $P$ |
| $t=-1\Rightarrow\frac{dy}{dx}=\frac{4}{(2(-1)-1)(1-(-1))^2}\left(=-\frac{1}{3}\right)$ | M1 | Attempts gradient; may be implied by their value for gradient |
| $y+2=-\frac{1}{3}(x-2)$ | M1 | Correct straight line method for tangent **not normal**; if using $y=mx+c$ must reach as far as finding $c$ |
| $x+3y+4=0$ | A1 | Any integer multiple |
## Part (c) Way 1:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $t^2-t+3\left(\frac{4t}{1-t}\right)+4=0$ | M1 | Substitutes to obtain equation in $t$ only |
| $t^3-2t^2-7t-4=0$ | A1 | Correct cubic |
| $(t+1)(t^2-3t-4)=0$ or $(t+1)^2(t-4)=0$; attempt to factorise using $(t\pm1)$ or $(t\pm1)^2$ as factor | M1 | Dependent on correct cubic with constant term |
| $t=4$ | A1 | Correct value of $t$ |
| $\left(12,-\frac{16}{3}\right)$ | A1 | Correct coordinates |
## Part (c) Way 2:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y=\frac{4t}{1-t}\Rightarrow t=\frac{y}{4+y}\Rightarrow x=\left(\frac{y}{4+y}\right)^2-\frac{y}{4+y}$; $\Rightarrow\left(\frac{y}{4+y}\right)^2-\frac{y}{4+y}+3y+4=0$ | M1 | Finds $x$ in terms of $y$ eliminating $t$; algebra must be correct (allow sign errors only for making $t$ the subject from $y$) |
| $3y^3+28y^2+76y+64=0$ | A1 | Correct cubic |
| $(y+2)(3y^2+22y+32)=0$ or $(y+2)^2(3y+16)=0$; attempt to factorise using $(y\pm2)$ or $(y\pm2)^2$ as factor | M1 | Dependent on correct cubic with constant term |
| $y=-\frac{16}{3}$ | A1 | Correct value of $y$ |
| $\left(12,-\frac{16}{3}\right)$ | A1 | Correct coordinates |8. A curve has parametric equations
$$x = t ^ { 2 } - t , \quad y = \frac { 4 t } { 1 - t } \quad t \neq 1$$
\begin{enumerate}[label=(\alph*)]
\item Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in terms of $t$, giving your answer as a simplified fraction.
\item Find an equation for the tangent to the curve at the point $P$ where $t = - 1$, giving your answer in the form $a x + b y + c = 0$ where $a , b$ and $c$ are integers.
The tangent to the curve at $P$ cuts the curve at the point $Q$.
\item Use algebra to find the coordinates of $Q$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C34 2019 Q8 [13]}}