Standard +0.3 This is a straightforward parametric differentiation question requiring the chain rule formula dy/dx = (dy/dt)/(dx/dt) and substitution at a specific point. The differentiation steps are routine (logarithm and exponential with chain rule), and simplification at t=e is mechanical. Slightly above average difficulty due to the algebraic manipulation required, but still a standard textbook exercise with no novel insight needed.
2 The parametric equations of a curve are
$$x = ( \ln t ) ^ { 2 } , \quad y = \mathrm { e } ^ { 2 - t ^ { 2 } }$$
for \(t > 0\).
Find the gradient of the curve at the point where \(t = \mathrm { e }\), simplifying your answer.
\(\frac{dy}{dx} = \frac{dy}{dt} \div \frac{dx}{dt}\) and substitute \(t = e\)
M1
Correct use of chain rule for \(\frac{dy}{dx}\left(\frac{-2e^2e^{2-e^2}}{2\ln e}\right)\). Condone an error between correct combination of the derivatives and attempt to substitute \(e\)
Obtain \(-e^{4-e^2}\)
A1
ISW. Accept \(-0.0337(405..)\). Accept \(-e^4e^{-e^2}\), \(\frac{-e^4}{e^{e^2}}\) and \(-e^2e^{2-e^2}\). Allow M1A1 for a correct decimal answer following B1B1 seen
## Question 2:
| Answer | Mark | Guidance |
|--------|------|----------|
| Obtain $\frac{dx}{dt} = \frac{2}{t}\ln t$ | B1 | Any equivalent form |
| Obtain $\frac{dy}{dt} = -2te^{2-t^2}$ | B1 | Any equivalent form |
| $\frac{dy}{dx} = \frac{dy}{dt} \div \frac{dx}{dt}$ and substitute $t = e$ | M1 | Correct use of chain rule for $\frac{dy}{dx}\left(\frac{-2e^2e^{2-e^2}}{2\ln e}\right)$. Condone an error between correct combination of the derivatives and attempt to substitute $e$ |
| Obtain $-e^{4-e^2}$ | A1 | ISW. Accept $-0.0337(405..)$. Accept $-e^4e^{-e^2}$, $\frac{-e^4}{e^{e^2}}$ and $-e^2e^{2-e^2}$. Allow M1A1 for a correct decimal answer following B1B1 seen |
---
2 The parametric equations of a curve are
$$x = ( \ln t ) ^ { 2 } , \quad y = \mathrm { e } ^ { 2 - t ^ { 2 } }$$
for $t > 0$.\\
Find the gradient of the curve at the point where $t = \mathrm { e }$, simplifying your answer.\\
\hfill \mbox{\textit{CAIE P3 2023 Q2 [4]}}