## anonymous 4 years ago Let f(t)=4 t^2 + 4/t^2. Find the slope of the curve at t = -3.

f(t)'=8t -(8)/t^3 for t=-3 -24-8/27 -640/27=-20

Ooop it is not -20 -640/27=-23.7037

3. anonymous

yeah I got df/dt=8t-2/t^2 so t=-3 => 8*(-3)-2/(-3)^2 -24-2/9 ! but i dont trust my work

the second term when you differentiated;$4 \over t ^{2}$is$4 t ^{-2}$becomes $-8 t ^{-3}$

or$-8\over t ^{3}$

6. anonymous

but I would still have to plug in -3 for t

Yes indeed.

8. anonymous

wait was I on the right track to getting the answer?

that would be -8/27 for that term and 24 for the first term I see what you mean it can't be negative with 24 being positive!! lol, I better take a nap.

10. anonymous

11. anonymous

right?

Oh yes, thats right as t=-3 so it would be -24 and a negative 8/27 that means we have to combine the -24 and the -8/27 giving -648/27 - 8/27=-656/27= -24.29

Looking at your next to last post. How did you get -640/27? Wouldn't -24 X 27 =-648 ??

It would appear so, if I have kept track of the signs.

16. anonymous

I think I did mess up on that 2nd term when I plugged in the minus 3$-8\over t ^{3}$$-8\over (-3)^{3}$$-8\over -27$ or POSITIVE 8/27 lol there it is.

So we ADD the 8/27 to the negative 648/27 giving -640/27 So sorry paunic88, I did stray off the path of keeping track of signs.

19. anonymous

ok so the final anwser is -640/27 right? (ill compute it later)

-3 X -3 X -3 = -27 lol

Yes.

Good luck with your studies,and don't err with signs lol

23. anonymous

ok thank you very much for the help