Question

Some more calc questions. How do I do 1606c, 1607, and 1609bc?

Answer 1

@justjm have ye done questions like this

Answer 2

yeah I'm in that unit too, let me take a look

Answer 3

For 1606c you have to convert the vector to cartesian (scalar) if you remember from precalc

from R(t)=(sin t)i + (cos 2t)j
x= sin t and y = cos 2t
So you have to eliminate the parameter t by solving for t in one of them, I'll do sin t because that's a bit easier
x = sin t
arcsin(x)=t
then substitute t into y=cos 2t

y= cos(2arcsin(x))

Answer 4

I did that for 1606c, but that wasn't the answer at all

Answer 5

Oh wait nvm I looked at the answer key wrong oof

Answer 6

oof lol it happens

Answer 7

For 1607 you would need to first integrate the a=3i-j to get the velocity, and for (c1, c2), you would need to find the unit vector from the finding the direction of the particle using the two points. Those are c1 and c2, and you would put them within the base vectors. Then you would integrate again and use the initial point (1,2).

@imqwerty

Answer 8

That's right. You can integrate the acceleration over time to get velocity, and the velocity vector will include the constants of integration, integrate the velocity again to find the displacement, which will also contain some constants of integration.
To find the constants of integration use the given conditions of position and velocity

Answer 9

1609
The R(t) vector at any time t is essentially the direction in which the particle is headed.

Find the time of collision of the particles by equating their i and j coordinates and then plug this time in their respective R(t) to get their direction of motion at collision

To find the speed at collision just differentiate R(t) wrt t and get their velocity vectors, plug in the t which you calculated above. Don't forget to take the mod tho bcz speed is always positive

Answer 10

Thank you both so much! That really helped clear my confusion.