Question

a particle moving in a straight line is given by the equation of motion s=1/t^2. find the velocity of the particle at t=a

Answer 1

Using
\[\lim_{h \rightarrow 0} \frac{ F(a +h) -F(a) }{ h }\]

Answer 2

It doesn't look like a particularly hard limit

Answer 3

You want to find the derivative (by first principle, apparently) of s(t) because v = ds/dt.

\[\frac{ds}{dt} = \lim_{x \rightarrow 0}{\frac{1}{h}(\frac{1}{(t+h)^2} - \frac{1}{t^2} )}\]

Once you've ground that out, plug in t = a.

Answer 4

Sorry, that limit is obviously as h->0, not x!

Answer 5

what happened to the H in the denominator?

Answer 6

I wrote it as (1/h) times the numerator

Answer 7

ok I am still a bit confused because I thought is was times the reciprocal so it would be h/1?

Answer 8

brain food is done be back in 2 min

Answer 9

No, the messy thing in the parentheses is just the numerator of the original fraction. Here, I'll write it out in the original form, so you can see why I don't do that!

\[\frac{ds}{dt} = \lim_{h\rightarrow 0} \frac{\frac{1}{(t+h)^2}-\frac{1}{t^2}}{h}\]

Answer 10

I think it's easier to deal with it written like this:
\[\frac{ds}{dt} = \lim_{h \rightarrow 0} (\frac{1}{h})(\frac{1}{(t+h)^2} - \frac{1}{t^2} )\]
After you multiply both top and bottom of the left fraction in the parentheses by t^2 and the top and bottom of the right fraction by (t+h)^2 you can factor an h out in the numerator to cancel the 1/h, and then just take the limit.

Answer 11

@adam32885 Is the problem that you don't know how to subtract fractions with variables in them?

Answer 12

ok I think i was trying to make things harder than it was meant to be let me see what i get

Answer 13

Okay I am still messing up somewhere

Answer 14

Do you know how to subtract fractions?

Answer 15

yes find a common denominator

Answer 16

In this case what is it?

Answer 17

\[t ^{2}(t+h)^{2}\]

Answer 18

okay now you have to change the fractions

Answer 19

so it would be
\[\frac{ t ^{2}- (t+h)^{2} }{ t ^{2}(t+h)^{2} }\]

Answer 20

Yeah that's it. Now simplify

Answer 21

-1?

Answer 22

No, start by expanding the numerator.

Answer 23

remember foil?

Answer 24

yes

Answer 25

actually....

Answer 26

yeah basically you need to expand the numerator

Answer 27

just the numerator?

Answer 28

so I get
\[\frac{ -2th+h ^{2} }{ t ^{2}(t+h)^{2} }\]

Answer 29

yeah just the numerator for now

Answer 30

hurry, I gotta go soon.

Answer 31

go go go

Answer 32

Now you can divide by \(h\)

Answer 33

and evaluate the limit

Answer 34

I am still coming up with a 0 in the numerator

Answer 35

hell no.

Answer 36

\[
t^2(t+0)^2 \neq 0
\]

Answer 37

I was just about to say i see what i did thank you very much

Answer 38

\[
h-2t \neq 0
\]

Answer 39

done?

Answer 40

yes thank you