Differentiation: Tangent Lines and Velocity. Use position function s find velocity

Use the position function s (in meters) to find the velocity at time t= a seconds.

s(t) = 4/t, (a) a=2; (b) a=4

I know the formula starts out something like this.....

the limit as h approaches 0....... [s(2+h) - s(2)] / [(2+h) - 2]

but after that I am quite lost.....I don't understand what it is I am supposed to do next. I have looked at my calculus book, looked at some of the solutions to similar problems, but I just don't get it. None of the examples deal with this exact problem.

Any help or guidance would be greatly appreciated!!

Re: Differentiation: Tangent Lines and Velocity. Use position function s find velocit

Do you have to motivate the derivative with a difference quotient or are you allowed to use the "known" rules which follows from the definition? If you know the rules you have

s(t) = 4/t = 4*t^-1

s'(t) = -1*4*t^-2 = -4/t^2

If you have to use the limit of the difference quotient then we have the definition

s'(t) = lim(h->0) (s(t+h)-s(t))/h = lim(h->0) ((4/(t+h))-4/t)/h = lim(h->0) ((4t-4(t+h))/(t(t+h)))/h = lim(h->0) ((-4h))/(t(t+h)))/h = lim(h->0) ((-4))/(t(t+h)))/1 = -4/t^2

So we have the derivative

s'(t) = -4/t^2

Calculating the velocity at a = 2 were t = a

s'(t) = -4/t^2 => s'(2) = -4/2^2 = -1 m/s

I leave a = 4 as exercise for you.

Re: Differentiation: Tangent Lines and Velocity. Use position function s find velocit

Thanks for the quick reply, yes I have to use the limit.....I am reading through your response now, and If I do not understand any portion of it, I will post with further questions. Thank you so much for the reply though, I really appreciate it!!

Re: Differentiation: Tangent Lines and Velocity. Use position function s find velocit

Write if you don't understand, maybe I can write in LaTeX. But try to write it down with pen and paper, and you will get the idea.

Re: Differentiation: Tangent Lines and Velocity. Use position function s find velocit

Okay, so as I understand it......

A=4

s'(t) = lim(h->0) (s(t+h)-s(t))/h

= lim(h->0) ((4/(t+h))-4/t)/h

= lim(h->0) ((4t-4(t+h))/(t(t+h)))/h

= lim(h->0) ((-4h))/(t(t+h)))/h

= lim(h->0) ((-4))/(t(t+h)))/1

= -4/t^2

s'(t) = -4/t^2

=> s'(4)

= -4/4^2

= -0.25 m/s

By all means correct me If I am wrong.

Re: Differentiation: Tangent Lines and Velocity. Use position function s find velocit

Quote:

Originally Posted by

**JDS** Thanks for the quick reply, yes I have to use the limit.....I am reading through your response now, and If I do not understand any portion of it.

You don't because it is so damn hard to read.

Why not use LaTeX coding? **It is not helpful if it cannot be read.**

$\displaystyle \frac{{\frac{4}{{t + h}} - \frac{4}{t}}}{h} = \frac{{4(t) - 4(t + h)}}{{h(t + h)t}} = \frac{{ - 4}}{{(t + h)t}}$

Now you finish the limit.

Re: Differentiation: Tangent Lines and Velocity. Use position function s find velocit

You are right, Yours is much cleaner and I can understand it more clearly. However, shouldnt the end function there be......

-4/[2+(t+h)t]

Perhaps I am looking at it wrong though?

Re: Differentiation: Tangent Lines and Velocity. Use position function s find velocit

nevermind, im an idiot, I see what I did, lOl

Re: Differentiation: Tangent Lines and Velocity. Use position function s find velocit

grrr, actually the more I look at this the more I confuse myself.

Re: Differentiation: Tangent Lines and Velocity. Use position function s find velocit

Quote:

Originally Posted by

**Plato** You don't because it is so damn hard to read.

Why not use LaTeX coding? **It is not helpful if it cannot be read.**

$\displaystyle \frac{{\frac{4}{{t + h}} - \frac{4}{t}}}{h} = \frac{{4(t) - 4(t + h)}}{{h(t + h)t}} = \frac{{ - 4}}{{(t + h)t}}$

Now you finish the limit.

So I Did it with the way you have it set up here, and I still get the same answer I got above which is - 1/4 m/s or -0.25 m/s

Hope I understood it correctly!

Re: Differentiation: Tangent Lines and Velocity. Use position function s find velocit

But looking at it for A=2, I am coming up with -1/2 or -0.5 m/s

Am I correct, or am I missing something?

Re: Differentiation: Tangent Lines and Velocity. Use position function s find velocit

You said earlier

Quote:

s'(t) = lim(h->0) (s(t+h)-s(t))/h

= lim(h->0) ((4/(t+h))-4/t)/h

= lim(h->0) ((4t-4(t+h))/(t(t+h)))/h

= lim(h->0) ((-4h))/(t(t+h)))/h

= lim(h->0) ((-4))/(t(t+h)))/1

= -4/t^2

s'(t) = -4/t^2

When t= 2, that is $\displaystyle -4/2^2= -1$, not -1/2.

Re: Differentiation: Tangent Lines and Velocity. Use position function s find velocit

Yes you are correct but in the portion you quoted, that was based on the info I obtained from "fkf", When I reposted the answer of -1/2, I was going off of the formula version that Plato posted.....Was fkf and myself correct or is Plato correct?

Re: Differentiation: Tangent Lines and Velocity. Use position function s find velocit

Quote:

Originally Posted by

**JDS** Was fkf and myself correct or is Plato correct?

$\displaystyle \lim _{h \to 0} \left( {\frac{{ - 4}}{{(t + h)t}}} \right) = \frac{{ - 4}}{{t^2 }}$

So the derivative is $\displaystyle \frac{-4}{t^2}$ which is $\displaystyle -1$ if $\displaystyle t=2$.

Re: Differentiation: Tangent Lines and Velocity. Use position function s find velocit

I see, So then in that case my second answer is still , -(1/4)........Correct?

P.S. Thanks for all the help!!