Horizontal and vertical parametric tangents

A curve CC is defined by the parametric equations x=θ−cos⁡θx=\theta-\cos\theta, y=θ+cos⁡θy=\theta+\cos\theta, 0≤θ≤2π0\le\theta\le2\pi.

(i) Find the coordinate of the point(s) where the tangent is parallel to xx-axis. marks

(ii) Find the equation of tangent to the curve CC that is parallel to the yy-axis. marks

(iii) Find the exact equation of the tangent to the curve at the point PP where θ=π/6\theta=\pi/6. marks

(iv) Hence, determine whether the tangent at point PP will meet the curve CC again. marks

[modified 2011/SRJC/Prelim/II/3]

Total marks: 11

Solution

(i)

Parametric differentiation

x=θ−cos⁡θx=\theta-\cos\theta, y=θ+cos⁡θy=\theta+\cos\theta, 0≤θ≤2π0\le\theta\le2\pi. Find the coordinates of the points where the tangent is parallel to the xx-axis.

Explore this idea

Hint

  1. Differentiate both coordinates with respect to θ\theta.

  2. A horizontal tangent has dydθ=0\frac{dy}{d\theta}=0 with dxdθ≠0\frac{dx}{d\theta}\ne0.

Answer · (i)

(π/2,π/2)(\pi/2,\pi/2).

Explore the idea

A curve with one horizontal and one vertical tangent

The point moves with θ. Jump to the special points.

Position along the curve(1.5708, 1.5708)θ = 1.5708
Try an example
Tangent gradient
0
Normal gradient
vertical

θ=π2\theta=\frac\pi2: dydθ=0\frac{dy}{d\theta}=0, so the tangent is horizontal.

The tangent is horizontal and the normal is vertical. Both axes use equal scales.

  • Parametric curve
  • Tangent
  • Normal
  • A right angle between the tangent and normal when the guide is enabled
Key idea

For parametric curves, horizontal tangents come from dydθ=0\frac{dy}{d\theta}=0 and vertical tangents from dxdθ=0\frac{dx}{d\theta}=0.

Ideas to try
  1. Find the two places where the tangent is horizontal or vertical.
  2. What is the gradient at the start θ=0\theta=0?
(ii)

Vertical tangents

Find the equation of the tangent to CC that is parallel to the yy-axis.

Hint

  1. Vertical: dxdθ=0\frac{dx}{d\theta}=0.

  2. The equation is x=x= the xx-coordinate there.

Answer · (ii)

x=3π/2x=3\pi/2.

(iii)

Tangent at a parameter

Find the exact equation of the tangent to CC at the point PP where θ=π6\theta=\frac\pi6.

Hint

  1. Find the coordinates of PP.

  2. Find the gradient: 1−sin⁡π61+sin⁡π6\frac{1-\sin\frac\pi6}{1+\sin\frac\pi6}.

  3. Write the line and simplify.

Answer · (iii)

y=x/3+π/9+2/3y=x/3+\pi/9+2/\sqrt3.

(iv)

Tangent meeting the curve again

Hence determine whether the tangent at PP meets CC again.

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Hint

  1. Substitute x(θ)x(\theta) and y(θ)y(\theta) into the tangent equation.

  2. Solve for θ\theta in [0,2π][0,2\pi] and look for solutions other than π6\frac\pi6.

Answer · (iv)

Yes: one further intersection at θ≈3.816801916\theta\approx3.816801916.

0≤θ≤2π0\le\theta\le2\pi; distinct from the tangent point θ=π/6.

Explore the idea

The tangent at P crosses the curve again

The curve is the dark teal path; the tangent is the straight line.

Switch on a feature to add it to the sketch and see how it is found. A full-mark sketch shows all of them.

  • Parametric curve
  • Second intersection
Ideas to try
Switch on both points and compare their xx and yy values with the tangent line.