Course:MATH102/Question Challenge/2002 December Q7

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Question

A robotic arm rotates at the rate of 1 revolution per minute around the point O. As it rotates, the length $r$ of the arm changes. The relationship between $r$ (in meters) and the angle $\theta$ is given by

$r=4-2\cos(\theta ),\quad {\mbox{ for }}0\leq \theta \leq 2\pi$

The robotic arm rotates about point O.

(a) At what angle is the arm longest?

(b) Find the rate of change of the arm with respect to time.

(c) At what angle is the rate of change largest?

Hints

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Solutions

[show]Solution

$r=4-2\cos \theta$ ${\frac {d\theta }{dt}}=$ 1 revolution per minute$=2\pi$ radians per minute To find the only angle for which the robotic arm is longest, we find the angle for which $cos\theta =-1$. This angle is $\pi$: $r=4-2cos(\pi )$ $r=4-(-2)$ $r=6$ To find the rate of change of the length of the robotic arm with respect to time we take the first derivative: ${\frac {dr}{dt}}=-2(-\sin \theta )\cdot {\frac {d\theta }{dt}}$ ${\frac {dr}{dt}}=2\sin \theta \cdot {\frac {d\theta }{dt}}$ This is the angle for $\theta$ in which the arm is at maximum length. This is because in the domain $\theta$, $cos\theta$ only has one minimum of $-1$ To find when the rate of change of the robotic arm is largest, we take the second derivative of the function for the length of the robotic arm: ${\frac {dr}{dt}}=4\pi \cdot \sin \theta$ $d^{2}r/dt=4\pi \cdot \cos \theta$ $0=4\pi \cdot \cos \theta$ $0=\cos \theta$ When $\theta =\pi /2,3\pi /2$, $cos=0$ so we have found our critical points. $sin$ is maximum when $\theta =pi/2$, so we substitute that value of $\theta$ into the first derivative for the robotic arm to find when it is growing fastest: ${\frac {dr}{dt}}=4\pi \cdot \sin \theta$ ${\frac {dr}{dt}}=4/pi\cdot \sin(\pi /2)$ ${\frac {dr}{dt}}=4\pi (1)$ ${\frac {dr}{dt}}=4\pi$ Conversely, the robotic arm is shrinking fastest when $\theta =3\pi /2$: ${\frac {dr}{dt}}=4\pi \cdot \sin(3\pi /2)$ ${\frac {dr}{dt}}=4/\pi (-1)$ ${\frac {dr}{dt}}=-4\pi$ Part A) The arm is longest when $\theta =\pi$ Part B) ${\frac {dr}{dt}}=4\pi \sin \theta$ Part C) The angle at which the arm is growing the fastest when $\theta =\pi /2$. The arm is shrinking the fastest when $\theta =3\pi /2$.