Pearson Edexcel IAL Mathematics P3.4.3 dy 1 dy

Practise using dy/dx = 1/(dx/dy) for parametric-style curve equations, then rewrite gradients in terms of x or y.

Exam points

find dx/dy first, then invert it to obtain dy/dx in simplest form
use trig identities or algebraic substitution to express dy/dx purely in x
solve a gradient condition to find a point on the curve or a tangent or normal

[Maximum number: 6]

A curve C has equation y=f(x), where

f(x)=\arcsin\left(\frac{x}{2}\right),\qquad -2\leqslant x\leqslant2,\qquad -\frac{\pi}{2}\leqslant y\leqslant\frac{\pi}{2}

(a)

Given $x=2\sin y$ , show that

\frac{dy}{dx}=\frac{1}{\sqrt{A-x^2}}

where A is a constant to be found.

[ 3 ]

x=2\sin y\Rightarrow \frac{dx}{dy}=2\cos y

\frac{dy}{dx}=\frac{1}{2\cos y}

Since $\cos y=\sqrt{1-\sin^2y}=\sqrt{1-\frac{x^2}{4}}$ ,

\frac{dy}{dx}=\frac{1}{2\sqrt{1-\frac{x^2}{4}}}=\frac{1}{\sqrt{4-x^2}}

Thus A=4.

M1 A1 A1cso

(b)

The point P lies on C and has y coordinate $\frac{\pi}{4}$ .

Find the equation of the tangent to C at P. Write your answer in the form y=mx+c, where m and c are constants to be found.

[ 3 ]

At P, $y=\frac{\pi}{4}$ , so

x=2\sin\frac{\pi}{4}=\sqrt2

\left.\frac{dy}{dx}\right|_{x=\sqrt2}=\frac{1}{\sqrt{4-2}}=\frac{\sqrt2}{2}

y-\frac{\pi}{4}=\frac{\sqrt2}{2}(x-\sqrt2)

y=\frac{\sqrt2}{2}x-1+\frac{\pi}{4}

M1 dM1 A1

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P3.4.3 - Reciprocal derivative relationship question 1

Question (a)

Question (b)