2021 H2 Mathematics Paper 1 Question 3

Differentiation I: Tangents and Normals, Parametric Curves

Answers

(b)

y=12x+72{y = \frac{1}{2}x + \frac{7}{2}}

Full solutions

(a)

Differentiating implicitly w.r.t. x,{x,}
12x−12+12y−12dydx=0y−12dydx=−x−12dydx=−(yx)12  ■\begin{gather*} \frac{1}{2}x^{-\frac{1}{2}} + \frac{1}{2}y^{-\frac{1}{2}} \frac{\mathrm{d}y}{\mathrm{d}x} &= 0 \\ y^{-\frac{1}{2}} \frac{\mathrm{d}y}{\mathrm{d}x} = - x^{-\frac{1}{2}} \\ \frac{\mathrm{d}y}{\mathrm{d}x} = - \left( \frac{y}{x} \right)^{\frac{1}{2}} \; \blacksquare \end{gather*}

(b)

When x=1,{x=1,}
1+y12=3y=4dydx=−(41)12=−2\begin{align*} 1 + y^{\frac{1}{2}} &= 3 \\ y &= 4 \\ \frac{\mathrm{d}y}{\mathrm{d}x} &= -\left(\frac{4}{1}\right)^{\frac{1}{2}} \\ &= -2 \end{align*}
Hence the gradient of the normal at x=1{x=1} is 12{\frac{1}{2}}

Equation of normal at x=1:{x=1:}

y−4=12(x−1)y=12x+72  ■\begin{gather*} y - 4 = \frac{1}{2} (x-1) \\ y = \frac{1}{2}x + \frac{7}{2} \; \blacksquare \end{gather*}