Two tangents to an exponential-logarithmic curve

A curve CC is given by the equation ex2y=ln⁡x2e^{x^2}y=\ln x^2, where x≠0x\ne0.

(i) Show that 2xln⁡x2+ex2dydx=2x2x\ln x^2+e^{x^2}\frac{dy}{dx}=\frac2x. marks

(ii) Find the equation of the tangent to CC at the point PP where x=1x=1. Given that the tangent at PP meets CC again at the point QQ, find the coordinates of QQ, leaving your answer correct to 5 significant figures. marks

(iii) Determine the acute angle, in degrees, between the tangents to CC at the points PP and QQ. marks

[2022/HCI/Prelim/I/9]

Total marks: 9

Solution

If this part uses an earlier result, revisit (ii) second intersection Intersections by calculator or open all working together.

(iii)

Angle between tangents

Determine the acute angle, in degrees, between the tangents to CC at PP and QQ.

Explore this idea

Hint

  1. Find the gradient at QQ using part (i).

  2. Convert gradients into angles with the xx-axis.

  3. Use the angle sum in the triangle formed.

Answer · (iii)

66.9∘66.9^\circ (1 d.p.).

Explore the idea

Open interactive exploration

Two tangents, one triangle

The tangents at P (purple) and Q (amber) meet the x-axis at the angles α and β.

Position along the curve(-0.49516, -1.10008)
Try an example
Tangent gradient
-4.25026

At QQ: β=tan⁡−14.250=76.8∘\beta=\tan^{-1}4.250=76.8^\circ (a falling tangent).

For finite nonzero gradients, tangent gradient × normal gradient = −1. Both axes use equal scales.

  • Curve
  • Tangent at P
  • Tangent at Q
Key idea

Angle between lines of opposite slope sign: 180∘−α−β180^\circ-\alpha-\beta.

Ideas to try
Estimate the acute angle between the purple and blue lines, then compare with 66.9∘66.9^\circ.