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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 = ?

Verified Answer :

LammettHash

\(\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\frac{\mathrm dy}{\mathrm dt}}{\frac{\mathrm dx}{\mathrm dt}}\)

\(\dfrac{\mathrm dy}{\mathrm dt}=4\pi\cos\left(4\pi t+\dfrac\pi2\right)\)
\(\dfrac{\mathrm dx}{\mathrm dt}=\pi-4\pi\sin\left(4\pi t+\dfrac\pi2\right)\)

So the slope at \(t\) is

\(\dfrac{4\pi\cos\left(4\pi t+\dfrac\pi2\right)}{\pi-4\pi\sin\left(4\pi t+\dfrac\pi2\right)}=\dfrac{-4\sin(4\pi t)}{1-4\cos(4\pi t)}\)

The tangent line will be vertical when \(\dfrac{\mathrm dx}{\mathrm dt}=0\), which occurs for

\(\pi-3\pi\sin\left(4\pi t+\dfrac\pi2\right)=\pi-4\pi\cos(4\pi t)=0\implies\cos(4\pi t)=\dfrac14\implies t=\pm\dfrac{\arccos\frac14}{4\pi}+\dfrac n2\)

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

\(t_1=\dfrac1{4\pi}\arccos\dfrac14\approx0.1049\)
\(t_2=\dfrac1{4\pi}\arccos\dfrac14+\dfrac12\approx0.6049\)

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