# Thread: a particle is moving.. w/ equation.

1. ## a particle is moving.. w/ equation.

A particle is moving so that the position at time t is given by:
r(t) = e^t(cost i - sint j)

Find the velocity vector at time t:

find the acceleration vector at time t:

find the speed at time t:

i think i need to find the first and second derivatives for velocity and acceleration, not sure about speed?

i wasn't sure on the derivatives of a natural log.

thanks for any help.

2. Originally Posted by rcmango
A particle is moving so that the position at time t is given by:
r(t) = e^t(cost i - sint j)

Find the velocity vector at time t:

find the acceleration vector at time t:

find the speed at time t:

i think i need to find the first and second derivatives for velocity and acceleration, not sure about speed?

i wasn't sure on the derivatives of a natural log.

thanks for any help.
you are right about first and second derivatives for velocity and acceleration. speed is just velocity without direction, so i believe you just remove the i and j. derivative of lnx is 1/x, of course if x is a function you have to multiply by it's derivative

3. Originally Posted by Jhevon
speed is just velocity without direction, so i believe you just remove the i and j.
Yes, and no.

Speed in this case is the magnitude of the velocity vector. So if you have a velocity:
v = ai + bj
then the speed would be:
|v| = sqrt(a^2 + b^2)

-Dan

4. Originally Posted by topsquark
Yes, and no.

Speed in this case is the magnitude of the velocity vector. So if you have a velocity:
v = ai + bj
then the speed would be:
|v| = sqrt(a^2 + b^2)

-Dan
A yes, now i remember. speed = |v|. There i go running my mouth off again without knowing the facts

5. might need some help fixing my differentiation skills..

for the first derivative my guess would be to use the chain rule to differentiate?

so the first derivative is: -1/x(cost i - sin j)(sint i - cost j)

and for the 2nd derivative i guessed to use the chain and product rule in combination: so i got something like:- 1/x^2(cost i - sin j)(sint i - cost j)[(sint i - cost j)(sint i - cost j) + (cost i - sint j)(cost i + sint i)]

how am i doing.

thanks.

6. Originally Posted by rcmango
A particle is moving so that the position at time t is given by:
r(t) = e^t(cost i - sint j)

Find the velocity vector at time t:

find the acceleration vector at time t:

find the speed at time t:
Originally Posted by rcmango
might need some help fixing my differentiation skills..

for the first derivative my guess would be to use the chain rule to differentiate?

so the first derivative is: -1/x(cost i - sin j)(sint i - cost j)
What is x??

I think you are making this too hard. Given h(x) = f(x)*g(x)
h'(x) = f'(x)*g(x) + f(x)*g'(x)

So:
r(t) = e^t*(cos(t)i - sin(t)j)
r'(t) = e^t*(cos(t)i - sin(t)j) + e^t*(-sin(t)i - cos(t)j)
r'(t) = e^t*([cos(t) - sin(t)]i + [-sin(t) - cos(t)]j)

r''(t) = e^t*([cos(t) - sin(t)]i + [-sin(t) - cos(t)]j) + e^t*([-sin(t) - cos(t)]i + [-cos(t) + sin(t)]j)
r''(t) = e^t*(-2sin(t)i + -2cos(t)j)
r''(t) = -2e^t*(sin(t)i + cos(t)j)

For the speed:
r'(t) = e^t*([cos(t) - sin(t)]i + [-sin(t) - cos(t)]j)

So
|r'(t)| = sqrt{e^(2t)[cos(t) - sin(t)]^2 + e^(2t)[-sin(t) - cos(t)]^2}

|r'(t)| = e^t*sqrt{cos^2(t) + sin^2(t) - 2sin(t)cos(t) + sin^2(t) + cos^2(t) + 2sin(t)cos(t)}

|r'(t)| = e^t*sqrt{1 + 1}

|r'(t)| = e^t*sqrt{2}

-Dan