Standard +0.3 This is a straightforward parametric differentiation question requiring the chain rule (dy/dx = dy/dt ÷ dx/dt), product rule for both derivatives, and trigonometric simplification. While it has 6 marks indicating multiple steps, the techniques are standard and the algebraic manipulation, though requiring care with the tan(t - π/4) identity, follows a predictable path without requiring novel insight.
Use correct product or quotient rule at least once
M1*
Obtain \(\frac{dy}{dt} = e^{-t}\sin t - e^{-t}\cos t\) or \(\frac{dy}{dr} = e^{-t}\cos t - e^{-t}\sin t\), 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 $\frac{dy}{dr} = e^{-t}\cos t - e^{-t}\sin t$, 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]
The parametric equations of a curve are
$$x = e^{-t}\cos t, \quad y = e^{-t}\sin t.$$
Show that $\frac{dy}{dx} = \tan(t - \frac{1}{4}\pi)$. [6]
\hfill \mbox{\textit{CAIE P3 2013 Q4 [6]}}