2023 H2 Mathematics Paper 1 Question 1

Differentiation I: Tangents and Normals, Parametric Curves

Answers

y=−10ex+21e.y = - 10 \mathrm{e} x + 21 \mathrm{e}.

Full solutions

ln⁡y=(11−5x)21yd⁡ ⁣yd⁡ ⁣x=2(11−5x)(−5)\begin{align*} \ln y &= \left( 11 - 5 x \right)^2 \\ \frac{1}{y} \frac{\operatorname{d}\!y}{\operatorname{d}\!x} &= 2 \left( 11 - 5 x \right) \left( - 5 \right) \\ \end{align*}

When x=2,x=2,

y=e1ed⁡ ⁣yd⁡ ⁣x=−10d⁡ ⁣yd⁡ ⁣x=−10e\begin{align*} y &= \mathrm{e} \\ \frac{1}{\mathrm{e}} \frac{\operatorname{d}\!y}{\operatorname{d}\!x} &= - 10 \\ \frac{\operatorname{d}\!y}{\operatorname{d}\!x} &= - 10 \mathrm{e} \end{align*}
y−e=−10e(x−2)y=−10ex+21e  ■\begin{gather*} y - \mathrm{e} = - 10 \mathrm{e} \left( x - 2 \right) \\ y = - 10 \mathrm{e} x + 21 \mathrm{e} \; \blacksquare \end{gather*}

Question Commentary

A rather typical application of differentiation (equation of tangent) question. Implicit differentiation is the most straightforward way to approach, and we will have to make sure we leave our answers in exact values.