Standard +0.3 This is a straightforward parametric differentiation question requiring the product rule and chain rule. Students must find dx/dt and dy/dt using standard techniques, then divide to get dy/dx and simplify using trigonometric identities (specifically the tan addition formula). While it requires careful algebraic manipulation, it follows a completely standard method with no novel insight required, making it slightly easier than average.
4 The parametric equations of a curve are
$$x = \mathrm { e } ^ { - t } \cos t , \quad y = \mathrm { e } ^ { - t } \sin t$$
Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \tan \left( t - \frac { 1 } { 4 } \pi \right)\).
Use correct product or quotient rule at least once
M1*
Obtain \(\frac{dy}{dt}=e^{-t}\sin t-e^{-t}\cos t\) or \(\frac{dy}{dt}=e^{-t}\cos t-e^{-t}\sin t\), or equivalent
A1
Use \(\frac{dy}{dx}=\frac{dy}{dt} \div \frac{dx}{dt}\)
M1
Obtain \(\frac{dy}{dx}=\frac{\sin t-\cos t}{\sin t+\cos t}\), or equivalent
A1
EITHER: Express \(\frac{dy}{dx}\) in terms of \(\tan t\) only
M1(dep*)
Show expression is identical to \(\tan\left(t-\frac{1}{4}\pi\right)\)
A1
OR: Express \(\tan\left(t-\frac{1}{4}\pi\right)\) in terms of \(\tan t\)
M1
Show expression is identical to \(\frac{dy}{dx}\)
A1
[6]
Use correct product or quotient rule at least once | M1* |
Obtain $\frac{dy}{dt}=e^{-t}\sin t-e^{-t}\cos t$ or $\frac{dy}{dt}=e^{-t}\cos t-e^{-t}\sin t$, or equivalent | A1 |
Use $\frac{dy}{dx}=\frac{dy}{dt} \div \frac{dx}{dt}$ | M1 |
Obtain $\frac{dy}{dx}=\frac{\sin t-\cos t}{\sin t+\cos t}$, or equivalent | A1 |
**EITHER:** Express $\frac{dy}{dx}$ in terms of $\tan t$ only | M1(dep*) |
Show expression is identical to $\tan\left(t-\frac{1}{4}\pi\right)$ | A1 |
**OR:** Express $\tan\left(t-\frac{1}{4}\pi\right)$ in terms of $\tan t$ | M1 |
Show expression is identical to $\frac{dy}{dx}$ | A1 | [6]
4 The parametric equations of a curve are
$$x = \mathrm { e } ^ { - t } \cos t , \quad y = \mathrm { e } ^ { - t } \sin t$$
Show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = \tan \left( t - \frac { 1 } { 4 } \pi \right)$.
\hfill \mbox{\textit{CAIE P3 2013 Q4 [6]}}