Question

Check my answers?

Answer 1

The vector function r(t) represents the position of a particle at time t. Find the velocity, speed, and acceleration at the given value of t.
r(t) = <4cost, 3sint>, t = pi/3

Answer 2

My answers:
velocity: <-2, 3sqrt3/2>
speed: sqrt43
acceleration: <-2, -3sqrt3/2>

If someone could verify these, I would really appreciate it! I'm not sure if they're correct.

Answer 3

Did you start by doing this
r(t)=<4cost, 3sint>
r'(t)=<-4sin(t),3cos(t)>, t=pi/3
r'(pi/3)=<-4sin(pi/3),3cos(pi/3)> = r'(pi/3)=<-2sqrt(3),3/2>

Answer 4

Yes, looks right so far. Thanks :)

Answer 5

Still there @sweetburger ?

Answer 6

Oh sorry didn't know if you wanted me to continue.

Answer 7

Did my answers look right? For the speed and acceleration too?

Answer 8

Yea so
v(t)=r'(t)=<-4sin(t),3cos(t)>
r''(t)=<-4cos(t),-3sin(t)>, t=pi/3
a(t)=r''(pi/3)=<-4cos(pi/3),-3sin(pi/3)>,t=pi/3 = r''(pi/3)=<-2,(-3sqrt(3))/(2))>
speed = \[\sqrt{(-2\sqrt(3))^2+(3/2)^2}=\frac{ \sqrt(57) }{ 2 }=speed\]

Answer 9

I'm not sure whether are results align. Its possible i could have an error in my work.

Answer 10

I'm really confused by how you got your speed... I have the same velocity as you, so let's start there.
v(t) = <-4sint, 3cost>
Evaluate at pi/3
-4(1/2), 3(sqrt3/2)
-2, 3sqrt3/2
And then square and add both, then take the square root...
4 + (9+3)/4
16/4 + 27/4
43/4
Now take the square root...
sqrt(43)/2

Shouldn't that be the speed?

Answer 11

@sweetburger

Answer 12

let me check could have made a calculation error

Answer 13

I think we came to different velocity values.
I came to <-2sqrt(3),3/2>
and you came to <-2, 3sqrt3/2>

Answer 14

which is why we get different speeds.

Answer 15

Ayy, there it is. Mistake on my part. Thanks sweetburger!

Answer 16

Np glad we made some progress :)