Vector Calculus: Find Position & Velocity Vectors

[danny_manny]

Homework Statement

Use the given information to find the position and velocity
vectors of the particle.

a(t) = −cos t i − sin t j; v(0) = i; r(0) = j

Homework Equations

The Attempt at a Solution

Ok first step integrate a(t).

which i get to be
-sin(t)i +cos(t)j + c

now using the initial condition v(0) = i
i get v(t) = 1-sin(t)i +cos(t)j

now integrate v(t)

= (cos(t)+t)i +sin(t)j
using initial condition r(0) = j
r(t) = (cos(t) + t)i + (sin(t)+1)j

they are my two answers in bold but the back of my book has different answers and i don't know what's going wrong.

Thanks for the assistance

 

[SammyS]

Homework Statement

Use the given information to find the position and velocity
vectors of the particle.

a(t) = −cos t i − sin t j; v(0) = i; r(0) = j

Homework Equations

The Attempt at a Solution

Ok first step integrate a(t).

which i get to be
-sin(t)i +cos(t)j + c

now using the initial condition v(0) = i
i get v(t) = 1-sin(t)i +cos(t)j

now integrate v(t)

= (cos(t)+t)i +sin(t)j
using initial condition r(0) = j
r(t) = (cos(t) + t)i + (sin(t)+1)j

they are my two answers in bold but the back of my book has different answers and i don't know what's going wrong.

Thanks for the assistance

Your v(t) is incorrect. It gives you that v(0) = 1 + j. That's neither a vector nor a scalar. You can't add a vector & a scalar.

The constant, c, you have after integrating a(t) is a vector constant, c.

 

[danny_manny]

Your v(t) is incorrect. It gives you that v(0) = 1 + j. That's neither a vector nor a scalar. You can't add a vector & a scalar.

The constant, c, you have after integrating a(t) is a vector constant, c.

so the vector constant should be equal to i no?

 

[SammyS]

so the vector constant should be equal to i no?

No, although I see that you may have meant to have v(t) = (1-sin(t))i + cos(t)j. But that still ignores the fact that cos(0) = 1, not zero.

 

[danny_manny]

so the answer must be
(1-sin(t))i+(cos(t)+1)j ?

in the book however the answer is given
(1-sin(t))i+(cos(t)-1)j

I don't know where I'm going wrong.
Thanks again.

 

[SammyS]

so the answer must be
(1-sin(t))i+(cos(t)+1)j ?

in the book however the answer is given
(1-sin(t))i+(cos(t)-1)j

I don't know where I'm going wrong.
Thanks again.

cos(0) - 1 = 1 - 1 =0

 

[danny_manny]

Sorry Sammy I'm still lost :(

 

[vela]

so the answer must be
(1-sin(t))i+(cos(t)+1)j ?

in the book however the answer is given
(1-sin(t))i+(cos(t)-1)j

How did you come up with your answer? What do you get when you set t=0 in your answer?

 

[danny_manny]

v(0) = -sin(0)i +cos(0)t + c
= 0i+1j+c

 

[vela]

I don't see how you solved for c.

 

[danny_manny]

That's the part I'm stuck at do i set the lhs equal to zero?

 

[vela]

What do the initial conditions given in the problem statement say?

 

[danny_manny]

oh I see, thanks.