# At time t minutes, robot A is at (t, 2t + 1) and robot B is at (t^2, 2t^2 + 1). (a) Where is each...

## Question:

At time {eq}t {/eq} minutes, robot {eq}A {/eq} is at {eq}(t, 2t + 1) {/eq} and robot {eq}B {/eq} is at {eq}(t^2, 2t^2 + 1) {/eq}.

(a) Where is each robot when {eq}t = 0 {/eq} and {eq}t = 1? {/eq}

(b) Find the slope of the tangent line, {eq}\frac {dy}{dx} {/eq}, to the path of each robot at {eq}t = 1 {/eq} minute.

(c) Find the speed of each robot at {eq}t = 1 {/eq} minute.

(d) Discuss the motion of a robot that follows the path {eq}(sin(t), 2sin(t) + 1) {/eq} for {eq}20 {/eq} minutes.

## Parametric Equations in the Plane:

A parametric equation in the plane is a function {eq}r(t)=(x(t),y(t)) {/eq} from an interval in the real number line to the plane. Usually, the functions {eq}x(t) {/eq} and {eq}y(t) {/eq} are taken to be differentiable. The tangent vector to this equation at time {eq}t {/eq} is then the vector {eq}r'(t)=\left<x'(t),y'(t)\right> {/eq}.

## Answer and Explanation:

**a)** Let {eq}r_a(t)=(t,2t+1) {/eq} give the position of robot A at time {eq}t {/eq}, and let {eq}r_b(t)=(t^2,2t^2+1) {/eq} give the position of robot B at time {eq}t {/eq}. Then:

{eq}\begin{align*} r_a(0)&=(0,2(0)+1)\\ &=(0,1)\\ r_b(0)&=(0^2,2(0^2)+1)\\ &=(0,1) \, . \end{align*} {/eq}

So at time {eq}t=0 {/eq}, robot {eq}A {/eq} and robot {eq}B {/eq} are both at the point {eq}(0,1) {/eq}.

Similarly,

{eq}\begin{align*} r_a(1)&=(1,2(1)+1)\\ &=(1,3)\\ r_b(0)&=(1^2,2(1^2)+1)\\ &=(1,3) \, . \end{align*} {/eq}

So at time {eq}t=0 {/eq}, robot {eq}A {/eq} and robot {eq}B {/eq} are both at the point {eq}(1,3) {/eq}.

**(b)** Differentiating {eq}r_a(t)=(t,2t+1) {/eq} with respect to time gives {eq}r_a'(t)=\left<1,2\right> {/eq}. So {eq}\frac{dy}{dx}=\frac{2}{1}=2 {/eq}, at time {eq}t=1 {/eq} or at any other time.

Similarly, differentiating {eq}r_b(t)=(t^2,2t^2+1) {/eq} with respect to time gives {eq}r_b'(t)=\left<2t,4t\right> {/eq} so {eq}\frac{dy}{dx}=\frac{4t}{2t}=2 {/eq}, at time {eq}t=1 {/eq} or at any other time when {eq}t \ne 0 {/eq}.

**(c)** As previously computed, we have {eq}r_a'(t)=\left<1,2\right> {/eq}. So the speed of robot A is

{eq}\begin{align*} \|r_a'(t)\|&=\sqrt{1^2+2^2}\\ &=\sqrt{5} \, , \end{align*} {/eq}

at time {eq}t=1 {/eq} or at any other time.

Similarly, we have {eq}r_b'(t)=\left<2t,4t\right> {/eq}. So the speed of robot B is

{eq}\begin{align*} \|r_b'(t)\|&=\sqrt{(2t)^2+(4t)^2}\\ &=\sqrt{4t^2+16t^2}\\ &=\sqrt{20t^2}\\ &=2|t|\sqrt{5} \, . \end{align*} {/eq}

In particular, at time {eq}t=1 {/eq} we have {eq}|t|=1 {/eq} and so the speed of robot B is {eq}2\sqrt{5} {/eq}.

**(d)** Suppose a robot follows the path {eq}(\sin t, 2 \sin t + 1) {/eq} for 20 minutes. At any {eq}t {/eq}-value, we have {eq}y=2x+1 {/eq}, and so the robot will remain on the line {eq}y=2x+1 {/eq} for all time. The {eq}x {/eq}-coordinate of the robot will oscillate every {eq}2\pi {/eq} minutes, because {eq}x=\sin t {/eq}, so the {eq}y {/eq}-coordinate will also oscillate every {eq}2\pi {/eq} minutes.

In summary, the robot will move back and forth along the line {eq}y=2x+1 {/eq}, returning to its starting point every {eq}2\pi {/eq} minutes.