Position of a particle

1. Mar 6, 2009

Emethyst

1. The problem statement, all variables and given/known data
Suppose the following function describes the position of a particle moving along the x-axis, where s in metres and t is greater than or equal to 0 is in seconds: s(t)=t/t^2+3. (a) Where is the particle when it is stationary? (b) What is the velocity when it has no net force acting on it (ie. when the acceleration is zero)?

2. Relevant equations
All rules for the simplification of derivatives.

3. The attempt at a solution
My problems are from the simplification of the derivatives. What I got for (a) was -1/t^2+3, but this does not get the answer of sqrt3 for t as the answer key says when I set the position derivative to 0 to find the time when the particle is stationary. For (b) again the simplification did not go as it should, because I should have ended up with t=0 and t=3 when setting the velocity derivative to 0, but I only got t=0. I am using the quotient rule to try and get the derivatives. If anyone could be of assistance it would be great. (the answers are sqrt3/6 for (a) and 1/3, -1/24 for (b)).

2. Mar 6, 2009

lanedance

Hi Emethyst

I think the issue is in your derivative, for s(t)=t/t^2+3

the dervatitive is not -1/t^2+3... not sur how you are doind it with the quotient rule maybe you can show your working?

deosn;t s(t) just simplify to
s(t)=t/t^2+3 = 1/t + 3
or do you actually mean
s(t)=t/(t^2+3)

the quotient rule comes from the product rule, and i find it easier to use the product rule sometimes

ie
re-write

s(t) = u(x)/v(x) = u(x)*(1/v(x)) = u(x)*w(x) with w(x) = 1/v(x)
then do product rule with u & w

3. Mar 6, 2009

Emethyst

the equation for s(t) is t/(t^2+3) (it is shown without brackets on the sheet). Thanks for showing me that, that may be where I went wrong. What I did was use the formula dy/dx=q(x)p'(x)-p(x)q'(x)/(q(x))^2, where q(x) is t^2+3 and p(x) is t. I will have to try cancelling the t on top, but I thought you could not do that unless it's multiplied? And if you use the product rule, will it not in the end come out the same as using the quotient rule? I am planning to try it, but I thought I would ask either way just to make sure :tongue:

4. Mar 6, 2009

lanedance

can't cancel it the way it is written

the quotient rule you have shown is the way to go

have a try & see how you go

i was just making the point the quotient can be easily derived from the product rule, so i find it easier to just remember the product rule

Last edited: Mar 6, 2009
5. Mar 7, 2009

Emethyst

Well I tried it again and came up with the same answer. What i get is -t^2+3/(t^2+3)^2, or simplified -1/(t^2+3). Only the first can be right, because it would give a sqrt3 for an answer, which is what I need, but then it gets messy for b because I end up with a quintic function that I have no idea how to solve :tongue:. I might just have to keep fooling around with it to see if I get anywhere, any suggestions to help with the simplification?

6. Mar 7, 2009

lanedance

check you brcakets looking ok if you actually have
(-t^2+3)/(t^2+3)^2

can't make that simplification you mentioned as (-t^2+3)/(t^2+3) does not equal 1
ie
(-t^2+3) doed not equal (t^2+3)

some BIG hints, if you ever get really stuck, try & be really explicit about what you are doing

s(t)=t/(t^2+3) = u(t).v(t)

with
u(t) = t
v(t) = 1/(t^2+3)

by product rule
s'(t)= u'(t).v(t) + u(t).v'(t)

what are u' & v' ?
u'(t) = 1, easy v'(t) looks a little trickier

how about we write v(t) in form easy for the chain rule?

v(t) = x(y(t))
with
x(y(t)) = 1/y(t)
y(t) = t^2+3

so then what is v'(t) by the chain rule?

v'(t) = x'(y(t)).y'(t)
or its easier to tink of as

v'(t) = (dx(y)/dy).(dy(t)/(t))