Trigonometric and Inverse Trigonometric Functions

Lessons

Senior 5 · Mathematics

Trigonometric and Inverse Trigonometric Functions

5 lessons

Lesson 5 of 5

Applications: refraction in a prism, simple harmonic motion and optimisation

25 min

Calculus at work

This lesson uses the derivatives of Lessons 3 and 4 on three problems from science and daily life. In a glass prism, which angle of incidence bends a ray the least? For a vibrating body, how fast is it moving and how hard is it being pushed at each instant? And where should you stand to see a screen best, or how should a metal sheet be folded so that a gutter carries the most rainwater? Each answer comes from the same steps: write the quantity as a function of an angle or a time, differentiate, and look where the derivative is zero.

Definition

Prism: a block of glass with two plane faces meeting at the angle of the prism $A$ . A ray enters one face with angle of incidence $i$ and leaves the other with angle of emergence $e$ .

Angle of deviation $D$ : the angle between the incoming and the outgoing ray. Since the angles of refraction inside satisfy $r_{1} + r_{2} = A$ , geometry gives $D = i + e - A$ .

Minimum deviation $D_{\min}$ : the smallest value of $D$ as $i$ varies.

Example: for $A = 60^\circ$ , $i = 50^\circ$ and $e = 47^\circ$ , the deviation is $50^\circ + 47^\circ - 60^\circ = 37^\circ$ .

Refraction in a prism and minimum deviation

At the two faces Snell's law (Unit 1) gives $\sin i = n\sin r_1$ and $\sin e = n\sin r_2$ , with $r_{2} = A - r_{1}$ . So $D$ is a function of $i$ alone (Figure 2). Differentiating each equation with respect to $i$ :

\cos i = n\cos r_1\,\frac{dr_1}{di}, \qquad \cos e\,\frac{de}{di} = n\cos r_2\,\frac{dr_2}{di} = -n\cos r_2\,\frac{dr_1}{di},

\text{so} \qquad \frac{dD}{di} = 1 + \frac{de}{di} = 1 - \frac{\cos i\cos r_2}{\cos e\cos r_1}.

This is zero when $\cos i\cos r_2 = \cos e\cos r_1$ , which holds for the symmetric path $i = e$ , $r_1 = r_2 = \frac{A}{2}$ (and one can show that this is the only solution). Then $D_{\min} = 2i - A$ , so $i = \frac{A + D_{\min}}{2}$ , and Snell's law becomes

n = \frac{\sin\frac{A + D_{\min}}{2}}{\sin\frac{A}{2}}.

Measuring $A$ and $D_{\min}$ with a spectrometer is the standard way to find the refractive index of glass.

Figure 1. Minimum deviation in a prism of angle

A = 60^\circ

and

n = 1.5

, drawn to scale: the path is symmetric,

r_1 = r_2 = \frac{A}{2} = 30^\circ

, the ray inside is parallel to the base, and

i = e \approx 48.6^\circ

Figure 2.

D (i) = i + e - A

, computed from Snell's law at both faces. Its minimum, where

D^{'} (i) = 0

, is at the symmetric path of Figure 1.

Worked example 1: refractive index from a measurement

A glass prism with $A = 60^\circ$ gives a minimum deviation of $38.9^\circ$ . Find $n$ , and the angle of incidence at minimum deviation.

Solution

n = \frac{\sin\frac{60^\circ + 38.9^\circ}{2}}{\sin 30^\circ} = \frac{\sin 49.45^\circ}{0.5} = \frac{0.7598}{0.5} = 1.52.

The angle of incidence is $i = \frac{A + D_{\min}}{2} = 49.45^\circ \approx 49.5^\circ$ . For a thin prism (small $A$ ), $\sin\theta \approx \theta$ gives $D_{\min} \approx (n - 1)A$ .

Simple harmonic motion: velocity and acceleration

In Unit 1 a body in SHM had displacement $x = A\sin(\omega t + \varphi)$ . Differentiating with respect to time:

v = \frac{dx}{dt} = A\omega\cos(\omega t + \varphi), \qquad a = \frac{dv}{dt} = -A\omega^2\sin(\omega t + \varphi) = -\omega^2 x.

The acceleration always points back to the centre and is proportional to the distance from it: $a = -\omega^2 x$ is the defining equation of SHM in physics. Since $\cos^2 + \sin^2 = 1$ , the speed at displacement $x$ is

v^2 = \omega^2\left(A^2 - x^2\right):

greatest, $A\omega$ , at the centre; zero at the ends $x = \pm A$ , where the acceleration is greatest, $A\omega^2$ .

Worked example 2: a mass on a spring and a guitar string

(a) A mass moves with $x = 0.1\cos 5t$ (metres, seconds). Find its greatest speed and acceleration, and its speed when $x = 0.06\,\mathrm{m}$ . (b) A point on a guitar string playing the note A (440 Hz) moves with $x = 0.002\sin(880\pi t)$ . Find its greatest speed and acceleration.

Solution

$v = -0.5\sin 5t$ and $a = -2.5\cos 5t = -25x$ . Greatest speed $A\omega = 0.5\,\mathrm{m\,s^{-1}}$ ; greatest acceleration $A\omega^2 = 2.5\,\mathrm{m\,s^{-2}}$ . At $x = 0.06$ : $v^{2} = 25 (0.01 - 0.0036) = 0.16$ , so the speed is $0.4\,\mathrm{m\,s^{-1}}$ .
$\omega = 880\pi$ : greatest speed $0.002 \times 880\pi \approx 5.53\,\mathrm{m\,s^{-1}}$ ; greatest acceleration $0.002 \times (880\pi)^2 \approx 15\,300\,\mathrm{m\,s^{-2}}$ , more than 1500 times the acceleration due to gravity, even though the point moves only 2 mm.

Definition

Optimisation: finding the greatest or least value of a quantity that depends on a variable. On an interval, the extreme values occur where the derivative is zero (a stationary point) or at an end of the interval; the sign of the derivative on each side tells a maximum from a minimum.

Example: $f(\theta) = \sin 2\theta$ on $\left[0, \frac{\pi}{2}\right]$ : $f'(\theta) = 2\cos 2\theta = 0$ at $\theta = \frac{\pi}{4}$ , where $f$ takes its greatest value 1.

Worked example 3: the best place to stand

The bottom of a screen is 1 m and its top 3 m above a viewer's eye level. At what horizontal distance $x$ from the wall is the viewing angle $\theta$ greatest?

Solution The top and bottom of the screen make angles $\arctan\frac{3}{x}$ and $\arctan\frac{1}{x}$ with the horizontal, so $\theta(x) = \arctan\frac{3}{x} - \arctan\frac{1}{x}$ for $x \gt 0$ . With $\left(\arctan\frac{k}{x}\right)' = \frac{-\frac{k}{x^2}}{1 + \frac{k^2}{x^2}} = -\frac{k}{x^2 + k^2}$ :

\theta'(x) = -\frac{3}{x^2 + 9} + \frac{1}{x^2 + 1} = 0 \iff x^2 + 9 = 3x^2 + 3 \iff x = \sqrt{3}.

$\theta' \gt 0$ before $\sqrt{3}$ and $\theta' \lt 0$ after, so this is a maximum: $\theta = \arctan\sqrt{3} - \arctan\frac{1}{\sqrt{3}} = \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6} = 30^\circ$ at $x \approx 1.73$ m (Figure 3).

Figure 3.

\theta(x) = \arctan\frac{3}{x} - \arctan\frac{1}{x}

: standing too close or too far both shrink the angle; the best place is

x = \sqrt{3}

Worked example 4: a rain gutter

A metal sheet 30 cm wide is bent into a gutter: a 10 cm base with two 10 cm sides, each tilted outwards at an angle $\theta$ to the vertical. Which $\theta$ gives the largest cross-section, so that the gutter carries the most rainwater from a roof?

Solution The cross-section is a trapezium of height $10\cos\theta$ and parallel sides 10 and $10 + 20\sin\theta$ :

S(\theta) = \frac{1}{2}(20 + 20\sin\theta)(10\cos\theta) = 100\cos\theta\,(1 + \sin\theta), \qquad 0 \le \theta \lt \frac{\pi}{2}.

S'(\theta) = 100\left[-\sin\theta(1 + \sin\theta) + \cos^2\theta\right] = 100(1 - \sin\theta - 2\sin^2\theta) = 100(1 - 2\sin\theta)(1 + \sin\theta).

$S^{'} = 0$ when $\sin\theta = \frac{1}{2}$ , $\theta = 30^\circ$ ; $S' \gt 0$ before and $S' \lt 0$ after, so it is a maximum: $S = 100 \times \frac{\sqrt{3}}{2} \times \frac{3}{2} = 75\sqrt{3} \approx 130\,\mathrm{cm^2}$ , compared with $100\,\mathrm{cm^2}$ for upright sides.

Warning

Check the ends and the units. A zero of the derivative may be a minimum, or the greatest value may sit at an end of the interval, so always test the sign of the derivative or compare with the end values. Calculus formulas need angles in radians; convert to degrees only for the final answer.

Tip

A pattern worth remembering: for a screen from height $a$ to height $b$ above the eye, the best distance is $x = \sqrt{ab}$ , the geometric mean of Unit 2. Check it with Worked example 3: $\sqrt{1 \times 3} = \sqrt{3}$ .

Khan Academy: Introduction to harmonic motion. The first part graphs the motion of a mass on a spring; in the last part Sal differentiates x = A cos(ωt) twice and uses F = ma to find the angular frequency, exactly the calculus of this lesson.

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Practice

Give angles to one decimal place. The last two are harder.

A prism with $A = 60^\circ$ has a minimum deviation of $40^\circ$ . Find its refractive index.
Find the minimum deviation of a $60^\circ$ prism with $n = 1.6$ .
Estimate the deviation produced by a thin prism with $A = 6^\circ$ and $n = 1.5$ .
A body moves with $x = 0.04\sin 10t$ (metres, seconds). Find its greatest speed, its greatest acceleration, and its speed when $x = 0.02$ m.
Show that $x = A\sin(\omega t + \varphi)$ satisfies $\frac{d^2x}{dt^2} = -\omega^2 x$ .
A rectangle is drawn in a semicircle of radius 5, with one side on the diameter. Use the angle $\theta$ of a top corner to show that its area is $25\sin 2\theta$ , and find the largest area.
A 5 m ladder leans against a wall, making angle $\theta$ with the ground. Its foot slides away from the wall at $0.3\,\mathrm{m\,s^{-1}}$ . How fast is $\theta$ changing when the foot is 3 m from the wall?
A board on a wall runs from 2 m to 8 m above a pupil's eye level. Find the distance from the wall that gives the largest viewing angle, and that angle.

Show answersHide answers

$n = \frac{\sin 50^\circ}{\sin 30^\circ} = 1.53$ .
$\sin\frac{60^\circ + D}{2} = 1.6 \times 0.5 = 0.8$ , so $\frac{60^\circ + D}{2} = 53.13^\circ$ and $D_{\min} = 46.3^\circ$ .
$D \approx (n - 1)A = 0.5 \times 6^\circ = 3^\circ$ .
$A\omega = 0.4\,\mathrm{m\,s^{-1}}$ ; $A\omega^2 = 4\,\mathrm{m\,s^{-2}}$ ; $v = 10\sqrt{0.0016 - 0.0004} \approx 0.346\,\mathrm{m\,s^{-1}}$ .
$x' = A\omega\cos(\omega t + \varphi)$ , $x'' = -A\omega^2\sin(\omega t + \varphi) = -\omega^2 x$ .
The corner is $(5\cos\theta, 5\sin\theta)$ : width $10\cos\theta$ , height $5\sin\theta$ , area $50\sin\theta\cos\theta = 25\sin 2\theta$ . Largest 25 when $\theta = \frac{\pi}{4}$ (a $5\sqrt{2}$ by $\frac{5\sqrt{2}}{2}$ rectangle).
$x = 5\cos\theta$ , so $\frac{dx}{dt} = -5\sin\theta\,\frac{d\theta}{dt}$ . At $x = 3$ , $\sin\theta = \frac{4}{5}$ : $0.3 = -4\,\frac{d\theta}{dt}$ , so $\frac{d\theta}{dt} = -0.075\,\mathrm{rad\,s^{-1}}$ (the angle decreases).
$x = \sqrt{2 \times 8} = 4$ m; $\theta = \arctan 2 - \arctan 0.5 \approx 36.9^\circ$ .

1Calculus at work

2Refraction in a prism and minimum deviation

3Simple harmonic motion: velocity and acceleration

Calculus at work

Refraction in a prism and minimum deviation

Simple harmonic motion: velocity and acceleration