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

(i)

Implicit differentiation with a product

A curve CC is given by ex2y=ln⁡x2e^{x^2}y=\ln x^2, x≠0x\ne0. Show that 2xln⁡x2+ex2dydx=2x2x\ln x^2+e^{x^2}\dfrac{dy}{dx}=\dfrac2x.

Hint

  1. Differentiate both sides, using the product rule on ex2ye^{x^2}y.

  2. Replace ex2ye^{x^2}y by ln⁡x2\ln x^2 using the original equation.

Answer · (i)

2xln⁡(x2)+ex2y′=2/x2x\ln(x^2)+e^{x^2}y'=2/x.

x≠0x\ne0.

(ii) tangent

Tangent at a point

Find the equation of the tangent to CC at the point PP where x=1x=1.

Hint

  1. Find yy at x=1x=1.

  2. Use part (i) to find the gradient there.

Answer · (ii) tangent

y=(2/e)(x−1)y=(2/e)(x-1).

(ii) second intersection

Intersections by calculator

The tangent at PP meets CC again at QQ. Find the coordinates of QQ, correct to 5 significant figures.

Explore this idea

Hint

  1. Set ln⁡x2ex2=2e(x−1)\frac{\ln x^2}{e^{x^2}}=\frac2e(x-1).

  2. Solve with the graphing calculator and exclude x=1x=1.

Answer · (ii) second intersection

Q=(−0.49516,−1.1001)Q=(-0.49516,-1.1001) (5 s.f.).

Explore the idea

The tangent at P cuts the curve again at Q

Jump to P and Q. The fixed purple line is the tangent at P.

Position along the curve(1, 0)
Try an example
Tangent gradient
0.73576

At P: gradient 2e≈0.736\frac2e\approx0.736. The tangent here is the purple line.

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

  • E^(x²) y = ln x²
  • Tangent at the moving point
Key idea

The tangent at PP is also a chord of the curve, meeting it again at QQ.

Ideas to try
Slide to QQ and compare the two tangent lines.
(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

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.