Parametric Special Tangents and Direction | CAIE 9709 Pure Mathematics 2 Notes

Pure Mathematics 2

Mathematics9709

Use the uncancelled rates

Use the uncancelled rates $dx/dt$ and $dy/dt$ when deciding whether the tangent is horizontal or vertical.

Horizontal tangent: $dy/dt=0$ and $dx/dt\ne0$ .
Vertical tangent: $dx/dt=0$ and $dy/dt\ne0$ .
If both rates are zero, the ratio is $0/0$ . The tangent needs separate local analysis.

Worked example

Cancellation can hide a zero-over-zero point

x=t^2,\qquad y=t^3

\frac{dy}{dx}=\frac{3t^2}{2t}

\frac{dy}{dx}=\frac32t\qquad(t\ne0)

At $t=0$ , both original derivatives are zero. The simplified expression cannot be substituted there as if the ratio had been valid at that point.

Interpret the sign

At a point with a finite gradient, $dy/dx>0$ means $y$ increases as $x$ increases; $dy/dx<0$ means it decreases. A zero gradient is neither positive nor negative: it identifies a horizontal tangent, but does not by itself decide whether the whole function is increasing or decreasing over an interval.

Examiner note

If the question asks “increasing, decreasing or neither”, state the sign of the gradient as the reason. A value without the interpretation does not answer the question.

Find the parameter and point

Your turnIndependent check [7 marks]

A curve is given by $x=1+2\sin^2\theta$ and $y=4\tan\theta$ , where $\pi/2<\theta<3\pi/2$ . Find the point where the tangent is vertical, and justify the condition you use.

Show worked answer

\frac{dx}{d\theta}=4\sin\theta\cos\theta

\frac{dy}{d\theta}=4\sec^2\theta

A vertical tangent needs $dx/d\theta=0$ while $dy/d\theta\ne0$ . In the stated interval, $\theta=\pi$ ; there $dy/d\theta=4$ .

Answer

(x,y)=(1,0)