ok, here's another exponent problem! If you guys don't mind...cube root of s over s^2 = ?
\[\sqrt[3]{s}\div s ^{2}\]
I believed this was asked just a little while ago today...
why is to the -1/3 and not positive 1/3?
sorry, my bad!
\[s ^{1/3} . s ^{-2}\] = s\[s ^{-5/6}\]
\[{s^{1/3} \over s^2} = {1 \over s^{2-\frac{1}{3}}}\]
1/s^(5/3)
thank you, jgeorge! Thanks so much!
any time :)
Amistre64, where in the heck have you been? I have calculus problems to solve, you know!!! ; )
i had to do laundry; and air up my tire; put oil in the car... you know, nonmathical stuff lol
Well, next time do those things when I DON'T need your help! Seriously, how have you been? Can you help me now with the antiderivative of that exponent you gave me? And wait a minute, why is it to the -5//6 and not -5/3?
he confused 3*2
2-(1/3) = (3*2 -1)/3
so it is s^-5/3, then? right?
yes
and you want to derive that?
Need help finding the antiderivative of this if you don't mind then. Let me try to attach an equation just to look professional, like i know what I'm doing! ; )
no keeping secrets now lol
\[1+s ^{-5/3}\]
Need to find the antid of that! I know it's s-something. but it's the "something" i get stuck on, sort of!
\[(-5/3)s^{(-5/3 - 3/3)}\]
\[\frac{-5}{3 s^2 \sqrt[3]{s^2}}\]
\(*\sqrt[3]{s}\) top and bottom to rationalise the denm
\[\frac{-5 \sqrt[3]{s}}{3s^3}\]
does that help?
Well I am trying to figure out the first step where had -5 over all that gobbledy-gook. I don't get that part at all!
its the same steps as derive \(x^4\)
\(\ 'exp' * x^{(exp-1)}\)
Where did the s^2 come fronm in the denominator? I see about the cube root, but not that.
lets take this one step at a time.... \[s^{(a)} \iff a*s^{a-1}\] \[\frac{-5}{3}s^{(\frac{-5}{3}-\frac{3}{3})} \iff \frac{-5 s^{-8/3}}{3}\]
\[\frac{-5}{3s^{8/3}} \iff \frac{-5}{3s^{6/3}s^{2/3}}\]
\[\frac{-5}{3s^2 s^{2/3}}*\frac{s^{1/3}}{s^{1/3}}=\frac{-5s^(1/3)}{3s^2 s}\]
Ok, I see where you are getting those numbers, but why do you have to do the 6/3 2/3 thing in the denominator in the first place? Why do you have to break it up like that? You can't just leave it? Why 2 "s" 's?
\[\frac{-5 \sqrt[3]{s}}{3s^3}\]
for the same reason why we dont leave \(\sqrt{36}\) as it is :)
ok, but why the s^3? I though it was s^2 in the denominator! aaarrrgh!
\[\sqrt[3]{s.s.s.s.s.s.s.s} \iff s^2 \sqrt[3]{s^2}\]
\[\sqrt[3]{s^3 s^3 s^2} \iff \sqrt[3]{s^3}\sqrt[3]{s^3}\sqrt[3]{s^2}\] \[s.s.\sqrt[3]{s^2} \iff s^2 \sqrt[3]{s^2}\]
ok, go back up if you would to where I said,"ok, I see where you get those numbers, but why do you have to do the 6/3 2/3 thing in the denominator." See that post of mine? the one right before it is yours. THATS what I don't understand!
The s^1/3 part.
and the posts i just did explain that :)
ok, then I am just an idiot, because I don't get it at all. What to do?
\(\frac{-5}{3s^{8/3}}\) is what we get in the denominator right?
your question is about the s^(8/3) and how we play with it
Is that 3s^8/3? It's quite small! But yes that is my question. How to play with it. But i have other things I would rather play with. Exponents AINT one of em!
Does \(s^{8/3}\) equal \(\sqrt[3]{s^8}\)?
yes, i would have to say it does. ok, go on... please!
Does \(s^8\) mean: \(s.s.s.s.s.s.s.s\) ?
yes.
Does \(s.s.s.s.s.s.s.s\) equal \((s.s.s).(s.s.s).(s.s)\) ? Does \((s.s.s).(s.s.s).(s.s) \iff s^3 s^3 s^2\) ?
yes, but is there a reason you split it up like that and not like (s.s.s.s).(s.s).(s.s)?
yes theres; our radical is a \(\sqrt[3]{...}\) which means we want to group these into \(s^3\) .
OH!!!!
and heres why: \(\sqrt[3]{s^8} \iff \sqrt[3]{s^3s^3s^2} \iff \sqrt[3]{s^3}\sqrt[3]{s^3}\sqrt[3]{s^2}\)
\[\sqrt[3]{s^3} = s\] \[s.s.\sqrt[3]{s^2} \iff s^2 \sqrt[3]{s^2 }\]
OMGosh! Ok, so back to the problem...I don't remember where we were even!
these are the basic steps that you tend to do in your head to avoid all the work of reproving them time and again lol
Maybe in YOUR head, but not MINE!
yes, it does help to actually step thru them to gain insight ;)
ok so we are left with: \[\frac{-5}{3s^2\sqrt[3]{s^2}}\] and thats fine but, we maybe want to get rid of the radical in the bottom part
we 'know' \(\sqrt[3]{s^3}=s\) so we want to multily top and bttom by \(\sqrt[3]{s}\) to get rid of it in the bottom right?
right.
\[\frac{-5 \sqrt[3]{s}}{3s^3}\]
wait, if you multiply the numerator by that radical and that radical =s, why do you still have the radical sign there?
\[-5 * \sqrt[3]{s} = -5\sqrt[3]{s}\]
yeah, but it's the cube root of s^3. That just equals s I thought?
\[3s^2\sqrt[3]{s^2}*\sqrt[3]{s}\iff3s^2\sqrt[3]{s^2.s}\iff3s^2\sqrt[3]{s^3}\iff 3s^2.s\] \[\implies 3s^3 \]
Are they all this hard? How in the heck do they expect newcomers here to calculus to know all this crap? I really don't understand! It's so hard!!!!! : (
they expect you to have learned this in college algebra; since thats all it is
if they have to reteach you what you should have already learned... then it gets rather ...... misses the point of dividing mathinto teachable sections lol
BTW your answers are always the best and easiest to work with! As are you! Thank you so much! I am going to go and try towrok this problem...will you be around? Oh, and the college algebra thing? I got a LOW C! I couldn't do it then; I aced calc in college 20 years ago, and now don't know a thing, but could help my daughter with high school algebra II like a pro (sort of).
:) youre doing fine :)
I get the point! Do you know of a good website with the exponent rules on it?
Actually, somehow in calc I'm getting a 97%!
my sirttes are all books; but i know a good practie site for math: interactmath.com
sirttes eh.... my uiniversal translator is on the fritz i think lol
Yeah, what is a sirttes anyways?
it my brain cells jockeying for position lol
Will you be around for some of today? Can I visit you here again if I need to?
ill be here as long as the weather holds up; im under the veranda since the libraries are closed
Ok, then again...thanks from the bottom of my heart for all your wonderful help! You're the best! Giving me all that time and patience...you're great, thanks
Join our real-time social learning platform and learn together with your friends!