Question

Allocated function: y = A cos kt

Values assigned: A = 4, k = 5

Graph of displacement y (metres) against time t (seconds) for time interval from t = 0s to t = 2π/k seconds.

Q: Calculate the turning points of the function using differential calculus and show for each turning point whether it is a maximum, a minimum or a point of inflexion, by using the second derivative of the function.

Answer 1

$y=A\cos(kt)$

but A=4 and k=5

$\implies y=4 \cos(5t)$

at turning points,first derivative is zero

$\implies {dy\over dx}=4(-\sin(5t)*5)=-20 \sin(5t)=0$

$\implies \sin(5t)=0$ ......................(i)

let $5t=\theta \implies \sin(\theta)=0$

from the unit circle,we have two angles whose sine is $\theta$ ,that is 0 and $\pi$

$\implies \theta=0+2\pi c=2\pi c$

$\theta=\pi +2 \pi c$ , where c in an interger

but $\theta=5t$

$\theta=0+2 \pi c=2 \pi c \implies 2 \pi c=5t \implies t={2 \pi c \over 5}..............(ii)$

$\theta=\pi + 2 \pi c \implies \pi+2 \pi c=5t \implies t={\pi\over 5}+{2 \pi c\over 5}.................(iii)$

we then use equation (ii) and (iii) avove to find all of the infinitely many solutions for equation (i)

we are asked to find those solutions which are in the interval $[0 , {2 \pi\over 5}]$

this means that we shall stop at $t={2 \pi\over 5}$

we then substitute v arious values of c to equation (ii) and (iii) as follows

when $c=0$

$t={2 \pi (0)\over 5}=0$ from equation (ii)

$t={\pi\over 5}+{2 \pi (0)\over 5}={\pi\over 5}$ from equation (iii)

when $c=1$

$t={2 \pi (1)\over 5}={2 \pi\over 5}$ from equation (ii)

$t={\pi\over 5}+{2 \pi (1)\over 5}={3\pi\over 5}$ from equation (iii)

from interval $[0 , {2 \pi\over 5}]$

t=0 , $t={\pi\over 5}$ , $t={2 \pi\over 5}$

we then find y values using the following equation

$y=4 \cos(5t)$

when $t=0$

$y=4 \cos(5*0)=4$ $\implies$ turning point $(0 , 4)$

when $t={\pi\over 5}$

$y=4 \cos(5*{\pi\over 5})=-4$ $\implies$ turning point $({\pi\over 5} ,- 4)$

when $t={2 \pi\over 5}$

$y=4 \cos(5*{2 \pi\over 5})=4$ $\implies$ turning point $({2\pi\over 5} , 4)$

we use the second derivative to determine the nature of the turning points

${d^2y\over dt^2}=-20(\cos(5t)*5)$

$=-100\cos(5t)$

when $t=0$

${d^2y\over dt^2}=-100\cos(5*0)=-100\lt0$ $\implies (0 , 4)$ is a local maximum

when $t={\pi\over 5}$

${d^2y\over dt^2}=-100\cos(5*{\pi\over 5}=100\gt0$ $\implies ({\pi\over 5} , -4)$ is a local minimum

when $t={2 \pi\over 5}$

${d^2y\over dt^2}=-100\cos(5*{2\pi\over 5})=-100\lt0$ $\implies (0 , 4)$ is a local maximum