# a particle is moving.. w/ equation.

• Feb 27th 2007, 05:22 PM
rcmango
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.
• Feb 27th 2007, 06:07 PM
Jhevon
Quote:

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
• Feb 27th 2007, 07:49 PM
topsquark
Quote:

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
• Feb 27th 2007, 07:51 PM
Jhevon
Quote:

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
• Feb 27th 2007, 09:39 PM
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)

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.
• Feb 28th 2007, 09:00 AM
topsquark
Quote:

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:

Quote:

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
The content structure of the text passage is a discussion thread from a math help forum, where a user named rcmango posts a question about a particle's motion described by a vector function, and other users, including Jhevon and topsquark, provide guidance and solutions.
single_question_with_answer
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
# a particle is moving.. w/ equation.

• Feb 27th 2007, 05:22 PM
rcmango
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.
• Feb 27th 2007, 06:07 PM
Jhevon
Quote:

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
• Feb 27th 2007, 07:49 PM
topsquark
Quote:

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
• Feb 27th 2007, 07:51 PM
Jhevon
Quote:

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
• Feb 27th 2007, 09:39 PM
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)

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.
• Feb 28th 2007, 09:00 AM
topsquark
Quote:

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:

Quote:

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
user
The content structure of the text passage is a discussion thread from a math help forum, where a user named rcmango posts a question about a particle's motion described by a vector function, and other users, including Jhevon and topsquark, provide guidance and solutions.

TYPE: SINGLE_QUESTION_WITH_ANSWER