Question

The displacement (in meters) of a particle moving in a straight line is given by the equation of motion s = 1/(t^2), where t is measured in seconds. Find the velocity of the particle at times t=a, t=1, t=2, and t=3.

Obviously I just need help getting v(a). Thanks much

Answer 1

they want you to find the derivative ds/dt which will be the velocity

Answer 2

okay. i know. sorry. v(a) = lim(h->0) [(1/(a+h)^2) - 1/a^2]/h.... is what the textbook leads me to believe i should use... that not it?

Answer 3

or i mean ds/dt (1/(t^2)) = ds/dt (t^-2) = (2(1/t)) = 2t. so I don't use that formula above or what?

Answer 4

if not what's the difference.....

Answer 5

oh wait that's the average rate of change, i'm supposed to go for the instantaneous rate of change, right..

Answer 6

WHAT AM I NOT GETTING

Answer 7

You are trying to do the derivative using first principles, but hopefully you will learn very soon that
\[\frac{d}{dt}\frac{1}{t^2}= \frac{d}{dt}t^{-2}= -2t^{-3}\]

Answer 8

Find the velocity of the particle at times t=a, t=1, t=2, and t=3.
so V(a) = -2/a^3
V(1)= -2
V(2)= -1/4
V(3)= -2/27

Answer 9

ohhhh whoops stupid mistake up there sorry thanks i see

Answer 10

If you use
lim(h->0) [(1/(t+h)^2) - 1/t^2]/h.
It gets quite involved. It eventually gets to the answer... But quite quickly you learn the power rule
\[\frac{d}{dx}x^n= nx^{n-1}\]
which works for negative and fractional n, It even works for n=0

\[ \frac{d}{dx}x^0= 0\cdot x^{-1}= 0\]
which makes sense, because d/dx 1 = 0 (1 is just a constant)

Answer 11

ok cool so then what the flutter is the formula for how i'm supposed to go about finding f'(2), let's say, for f(x) = 3x^2 - 5x ????

Answer 12

People generally don't like swearing on this site.

the derivative operation is linear:

d/dx ( f(x) + g(x)) = d/dx f(x) + d/dx g(x)

so d/dx (3x^2 - 5x) = 2*3 x^1 -5 x^0, or simply 6x-5

Answer 13

like do i use flutterin the uh DERIVATIVE as in d/dx or like do i use fluttering f prime of a and then fluttering plug that pellet in one of them i don't know to the slope m except like the flutterin slope m IS the fluttering derivative or what the flutter ever or at least their fluttering formulas are the fluttering SAME so like what in the flutterING WORLD DO I DO here cuz then i gotta use it (whatever the flutter THAT IS) to find the tangent line to the parabola at the pt (2,2). right. except like. they say to find f'(2). but like. isn't that the same. as fluttering everything else. that i'm supposed to even like plug it into. like wow. what the hell is the fluttering explanation for all this.

Answer 14

Here are a lot of videos. Khan explains things pretty well.
http://www.khanacademy.org/math/calculus/v/calculus--derivatives-1--new-hd-version

Answer 15

okay i'll check it out thanks man

Answer 16

oh see i thought it was 6x-5 to begin with but then i was like oh wow but then what about all the other fluttering formulas is it all gonna pan out equally i don't know that's where that rant came from but you seem to understand and i appreciate your patience for real; some people just get an attitude (holier than thou) and fluttering cut you out like you dissed their flutterin mothers or something. so thank you.

Answer 17

and sorry abotu the swearing heat of the moment