Question

Particle P moves along the y-axis so that its position at time t is given by y(t)=4t−23 for all times t. A second particle, particle Q, moves along the x-axis so that its position at time t is given by x(t)=sin(πt)2−t for all times t≠2.
As time t approaches 2, what is the limit of the position of particle Q ? Show the work that leads to your answer.
Show that the velocity of particle Q is given by vQ(t)=2πcos(πt)−πtcos(πt)+sin(πt)(2−t)2 for all times t≠2.
Find the rate of change of the distance between particle P and particle Q at time t=12. Show the work that leads to your answer.

Answer 1

a) The limit of the position of particle $Q$ when time approaches 2 is $-\pi$.

b) The velocity of particle $Q$ is $v_{Q}(t) = \frac{2\pi\cdot \cos \pi t-\pi\cdot t \cdot \cos \pi t -\sin \pi t}{(2-t)^{2}}$ for all $t \ne 2$.

c) The rate of change of the distance between particle $P$ and particle $Q$ at time $t = \frac{1}{2}$ is $\frac{4\sqrt{82}}{9}$.

How to apply limits and derivatives to the study of particle motion

a) To determine the limit for $t = 2$, we need to apply the following two algebraic substitutions:

$u = \pi t$ (1)

$k = 2\pi - u$ (2)

Then, the limit is written as follows:

$x(t) = \lim_{t \to 2} \frac{\sin \pi t}{2-t}$

$x(t) = \lim_{t \to 2} \frac{\pi\cdot \sin \pi t}{2\pi - \pi t}$

$x(u) = \lim_{u \to 2\pi} \frac{\pi\cdot \sin u}{2\pi - u}$

$x(k) = \lim_{k \to 0} \frac{\pi\cdot \sin (2\pi-k)}{k}$

$x(k) = -\pi\cdot \lim_{k \to 0} \frac{\sin k}{k}$

$x(k) = -\pi$

The limit of the position of particle $Q$ when time approaches 2 is $-\pi$. $\blacksquare$

b) The function velocity of particle $Q$ is determined by the derivative formula for the division between two functions, that is:

$v_{Q}(t) = \frac{f'(t)\cdot g(t)-f(t)\cdot g'(t)}{g(t)^{2}}$ (3)

Where:

• $f(t)$ - Function numerator.
• $g(t)$ - Function denominator.
• $f'(t)$ - First derivative of the function numerator.
• $g'(x)$ - First derivative of the function denominator.

If we know that $f(t) = \sin \pi t$, $g(t) = 2 - t$, $f'(t) = \pi \cdot \cos \pi t$ and $g'(x) = -1$, then the function velocity of the particle is:

$v_{Q}(t) = \frac{\pi \cdot \cos \pi t \cdot (2-t)-\sin \pi t}{(2-t)^{2}}$

$v_{Q}(t) = \frac{2\pi\cdot \cos \pi t-\pi\cdot t \cdot \cos \pi t -\sin \pi t}{(2-t)^{2}}$

The velocity of particle $Q$ is $v_{Q}(t) = \frac{2\pi\cdot \cos \pi t-\pi\cdot t \cdot \cos \pi t -\sin \pi t}{(2-t)^{2}}$ for all $t \ne 2$. $\blacksquare$

c) The vector rate of change of the distance between particle P and particle Q ($\dot r_{Q/P} (t)$) is equal to the vectorial difference between respective vectors velocity:

$\dot r_{Q/P}(t) = \vec v_{Q}(t) - \vec v_{P}(t)$ (4)

Where $\vec v_{P}(t)$ is the vector velocity of particle P.

If we know that $\vec v_{P}(t) = (0, 4)$, $\vec v_{Q}(t) = \left(\frac{2\pi\cdot \cos \pi t - \pi\cdot t \cdot \cos \pi t + \sin \pi t}{(2-t)^{2}}, 0 \right)$ and $t = \frac{1}{2}$, then the vector rate of change of the distance between the two particles:

$\dot r_{P/Q}(t) = \left(\frac{2\pi \cdot \cos \pi t - \pi\cdot t \cdot \cos \pi t + \sin \pi t}{(2-t)^{2}}, -4 \right)$

$\dot r_{Q/P}\left(\frac{1}{2} \right) = \left(\frac{2\pi\cdot \cos \frac{\pi}{2}-\frac{\pi}{2}\cdot \cos \frac{\pi}{2} +\sin \frac{\pi}{2}}{\frac{3}{2} ^{2}}, -4 \right)$

$\dot r_{Q/P} \left(\frac{1}{2} \right) = \left(\frac{4}{9}, -4 \right)$

The magnitude of the vector rate of change is determined by Pythagorean theorem:

$|\dot r_{Q/P}| = \sqrt{\left(\frac{4}{9} \right)^{2}+(-4)^{2}}$

$|\dot r_{Q/P}| = \frac{4\sqrt{82}}{9}$

The rate of change of the distance between particle $P$ and particle $Q$ at time $t = \frac{1}{2}$ is $\frac{4\sqrt{82}}{9}$. $\blacksquare$

Remark

The statement is incomplete and poorly formatted. Correct form is shown below:

Particle $P$ moves along the y-axis so that its position at time $t$ is given by $y(t) = 4\cdot t - 23$ for all times $t$. A second particle, $Q$, moves along the x-axis so that its position at time $t$ is given by $x(t) = \frac{\sin \pi t}{2-t}$ for all times $t \ne 2$.

a) As times approaches 2, what is the limit of the position of particle $Q?$ Show the work that leads to your answer.

b) Show that the velocity of particle $Q$ is given by $v_{Q}(t) = \frac{2\pi\cdot \cos \pi t-\pi\cdot t \cdot \cos \pi t +\sin \pi t}{(2-t)^{2}}$.

c) Find the rate of change of the distance between particle $P$ and particle $Q$ at time $t = \frac{1}{2}$. Show the work that leads to your answer.

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