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!!

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.

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!!

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.

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.

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.



Now you finish the limit.

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?

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

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

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.



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!

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?

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 , not -1/2.

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?

Quote:

---

Originally Posted by JDS

Was fkf and myself correct or is Plato correct?

---

So the derivative is which is if .

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

P.S. Thanks for all the help!!