Standard +0.3 This is a straightforward parametric differentiation question requiring students to find dy/dx using the chain rule, then determine the parameter value where x=0 (y-axis crossing), and substitute. It involves standard techniques (differentiating ln and quotient rule) with no conceptual challenges, making it slightly easier than average.
3 The parametric equations of a curve are
$$x = \ln ( 2 t + 3 ) , \quad y = \frac { 3 t + 2 } { 2 t + 3 }$$
Find the gradient of the curve at the point where it crosses the \(y\)-axis.
Use quotient of product rule, or equivalent, for derivative of \(y\)
M1
Obtain \(\frac{5}{(2t+3)^2}\) or unsimplified equivalent
A1
Obtain \(t = -1\)
B1
Use \(\frac{dy}{dx} = \frac{dy/dt}{dx/dt}\) in algebraic or numerical form
M1
Obtain gradient \(\frac{5}{2}\)
A1
[6]
Obtain $\frac{2}{2t+3}$ for derivative of $x$ | B1 |
Use quotient of product rule, or equivalent, for derivative of $y$ | M1 |
Obtain $\frac{5}{(2t+3)^2}$ or unsimplified equivalent | A1 |
Obtain $t = -1$ | B1 |
Use $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$ in algebraic or numerical form | M1 |
Obtain gradient $\frac{5}{2}$ | A1 | [6]
3 The parametric equations of a curve are
$$x = \ln ( 2 t + 3 ) , \quad y = \frac { 3 t + 2 } { 2 t + 3 }$$
Find the gradient of the curve at the point where it crosses the $y$-axis.
\hfill \mbox{\textit{CAIE P3 2014 Q3 [6]}}