Question

An object is moving in the plane according to these parametric equations:

x(t) = πt + cos(4πt + π/2)
y(t) = sin(4πt + π/2)

The slope of the tangent line to the path at time t is

4πcos(4πt + π/2) divided by (π-4πsin(4πt+π/2))

The first time the tangent line is vertical will be t = ?

The second time the tangent line is vertical will be t = ?

Answer 1

`(\mathrm dy)/(\mathrm dx)=((\mathrm dy)/(\mathrm dt))/((\mathrm dx)/(\mathrm dt))`


`(\mathrm dy)/(\mathrm dt)=4\pi\cos\left(4\pi t+\frac\pi2\right)`

`(\mathrm dx)/(\mathrm dt)=\pi-4\pi\sin\left(4\pi t+\frac\pi2\right)`

So the slope at
`t` is


`(4\pi\cos\left(4\pi t+\frac\pi2\right))/(\pi-4\pi\sin\left(4\pi t+\frac\pi2\right))=(-4\sin(4\pi t))/(1-4\cos(4\pi t))`

The tangent line will be vertical when
`(\mathrm dx)/(\mathrm dt)=0`, which occurs for


`\pi-3\pi\sin\left(4\pi t+\frac\pi2\right)=\pi-4\pi\cos(4\pi t)=0\implies\cos(4\pi t)=\frac14\implies t=\pm(\arccos\frac14)/(4\pi)+\frac n2`

where
`n` is an integer. Assuming
`t>0`, we get
`(\mathrm dx)/(\mathrm dt)=0` when
`t=(\arccos\frac14)/(4\pi)+\frac n2`. So the first time the tangent line will be vertical is when
`n=0`, and the second when
`n=1`. In other words,


`t_1=\frac1{4\pi}\arccos\frac14\approx0.1049`

`t_2=\frac1{4\pi}\arccos\frac14+\frac12\approx0.6049`