Question

Consider an object moving in the plane whose location at time t seconds is given by the parametric equations: x(t)=5cos(πt) y(t)=3sin(πt). Assume the distance units in the plane are meters. (a) The object is moving around an ellipse (as in the previous problem) with equation: x2 a2 + y2 b2 =1 where a= Correct: Your answer is correct. and b= Correct: Your answer is correct. . (b) The location of the object at time t=1/3 seconds is ( Correct: Your answer is correct. , Correct: Your answer is correct. ). (c) The horizontal velocity of the object at time t is x ' (t)= Correct: Your answer is correct. m/s. (d) The horizontal velocity of the object at time t=1/3 seconds is

Answer 1

Answer:

a)

a = 1/5

b = 1/3

b)

x(1/3) = 2.500

y(1/3) = 2.598

c)

x'(t) = -5π sin(π t)

d)

x'(1/3) = -13.603

Explanation:

Hi!

a)

We can notice that

x(t)/5 = cos(πt)

y(t)/3 = sin(πt)

Therefore:

( x(t) / 5 )^2 + ( y(t) / 3 )^2 = cos^2(πt) + sin^2(πt) = 1

That is:

a = 1/5

b = 1/3

b)

At t=1/3

x(1/3) = 5 cos(π/3)

y(1/3) = 3 sin(π/3)

But

cos(π/3) = 1/2 = 0.5

sin(π/3) = √3 / 2 = 0.866

That is:

x(1/3) = 2.5

y(1/3) = 2.598

c)

The horizontal velocity is:

x'(t) = -5π sin(π t)

d)

at time t =1/3

x'(1/3) = -5π sin(π/3) = -13.603

Answer 2

a) When the parametric equations is: $x(t)=5cos(πt) y(t)=3sin(πt)$

Then a = 1/5

After that b = 1/3

b) x(1/3) = 2.500

Then y(1/3) = 2.598

c)$x'(t) = -5π sin(π t)$

d) x'(1/3) = -13.603

$(A) x(t)/5 = cos(πt)$

Now y(t)/3 = sin(πt)

Horizontal velocity

Therefore:

The parametric equations is:

$(x(t) / 5 )^2 + ( y(t) / 3 )^2 = cos^2(πt) + sin^2(πt) = 1$

That is:

So, a = 1/5

Then b = 1/3

b)At t=1/3

$x(1/3) = 5 cos(π/3)$

$y(1/3) = 3 sin(π/3)$

But

Then cos(π/3) = 1/2 = 0.5

After that $sin(π/3) = √3 / 2 = 0.866$

That is:

x(1/3) = 2.5

Thus, y(1/3) = 2.598

c) The horizontal velocity is:

$x'(t) = -5π sin(π t)$

d) at time t =1/3

Thus, x'(1/3) = -5π sin(π/3) = -13.603

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