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Mathematics 18 Online
OpenStudy (anonymous):

ok, here's another exponent problem! If you guys don't mind...cube root of s over s^2 = ?

OpenStudy (anonymous):

\[\sqrt[3]{s}\div s ^{2}\]

OpenStudy (mathteacher1729):

I believed this was asked just a little while ago today...

OpenStudy (anonymous):

why is to the -1/3 and not positive 1/3?

OpenStudy (anonymous):

sorry, my bad!

OpenStudy (anonymous):

\[s ^{1/3} . s ^{-2}\] = s\[s ^{-5/6}\]

OpenStudy (amistre64):

\[{s^{1/3} \over s^2} = {1 \over s^{2-\frac{1}{3}}}\]

OpenStudy (amistre64):

1/s^(5/3)

OpenStudy (anonymous):

thank you, jgeorge! Thanks so much!

OpenStudy (anonymous):

any time :)

OpenStudy (anonymous):

Amistre64, where in the heck have you been? I have calculus problems to solve, you know!!! ; )

OpenStudy (amistre64):

i had to do laundry; and air up my tire; put oil in the car... you know, nonmathical stuff lol

OpenStudy (anonymous):

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?

OpenStudy (amistre64):

he confused 3*2

OpenStudy (amistre64):

2-(1/3) = (3*2 -1)/3

OpenStudy (anonymous):

so it is s^-5/3, then? right?

OpenStudy (amistre64):

yes

OpenStudy (amistre64):

and you want to derive that?

OpenStudy (anonymous):

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! ; )

OpenStudy (amistre64):

no keeping secrets now lol

OpenStudy (anonymous):

\[1+s ^{-5/3}\]

OpenStudy (anonymous):

Need to find the antid of that! I know it's s-something. but it's the "something" i get stuck on, sort of!

OpenStudy (amistre64):

\[(-5/3)s^{(-5/3 - 3/3)}\]

OpenStudy (amistre64):

\[\frac{-5}{3 s^2 \sqrt[3]{s^2}}\]

OpenStudy (amistre64):

\(*\sqrt[3]{s}\) top and bottom to rationalise the denm

OpenStudy (amistre64):

\[\frac{-5 \sqrt[3]{s}}{3s^3}\]

OpenStudy (amistre64):

does that help?

OpenStudy (anonymous):

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!

OpenStudy (amistre64):

its the same steps as derive \(x^4\)

OpenStudy (amistre64):

\(\ 'exp' * x^{(exp-1)}\)

OpenStudy (anonymous):

Where did the s^2 come fronm in the denominator? I see about the cube root, but not that.

OpenStudy (amistre64):

lets take this one step at a time.... \[s^{(a)} \iff a*s^{a-1}\] \[\frac{-5}{3}s^{(\frac{-5}{3}-\frac{3}{3})} \iff \frac{-5 s^{-8/3}}{3}\]

OpenStudy (amistre64):

\[\frac{-5}{3s^{8/3}} \iff \frac{-5}{3s^{6/3}s^{2/3}}\]

OpenStudy (amistre64):

\[\frac{-5}{3s^2 s^{2/3}}*\frac{s^{1/3}}{s^{1/3}}=\frac{-5s^(1/3)}{3s^2 s}\]

OpenStudy (anonymous):

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?

OpenStudy (amistre64):

\[\frac{-5 \sqrt[3]{s}}{3s^3}\]

OpenStudy (amistre64):

for the same reason why we dont leave \(\sqrt{36}\) as it is :)

OpenStudy (anonymous):

ok, but why the s^3? I though it was s^2 in the denominator! aaarrrgh!

OpenStudy (amistre64):

\[\sqrt[3]{s.s.s.s.s.s.s.s} \iff s^2 \sqrt[3]{s^2}\]

OpenStudy (amistre64):

\[\sqrt[3]{s^3 s^3 s^2} \iff \sqrt[3]{s^3}\sqrt[3]{s^3}\sqrt[3]{s^2}\] \[s.s.\sqrt[3]{s^2} \iff s^2 \sqrt[3]{s^2}\]

OpenStudy (anonymous):

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!

OpenStudy (anonymous):

The s^1/3 part.

OpenStudy (amistre64):

and the posts i just did explain that :)

OpenStudy (anonymous):

ok, then I am just an idiot, because I don't get it at all. What to do?

OpenStudy (amistre64):

\(\frac{-5}{3s^{8/3}}\) is what we get in the denominator right?

OpenStudy (amistre64):

your question is about the s^(8/3) and how we play with it

OpenStudy (anonymous):

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!

OpenStudy (amistre64):

Does \(s^{8/3}\) equal \(\sqrt[3]{s^8}\)?

OpenStudy (anonymous):

yes, i would have to say it does. ok, go on... please!

OpenStudy (amistre64):

Does \(s^8\) mean: \(s.s.s.s.s.s.s.s\) ?

OpenStudy (anonymous):

yes.

OpenStudy (amistre64):

Does \(s.s.s.s.s.s.s.s\) equal \((s.s.s).(s.s.s).(s.s)\) ? Does \((s.s.s).(s.s.s).(s.s) \iff s^3 s^3 s^2\) ?

OpenStudy (anonymous):

yes, but is there a reason you split it up like that and not like (s.s.s.s).(s.s).(s.s)?

OpenStudy (amistre64):

yes theres; our radical is a \(\sqrt[3]{...}\) which means we want to group these into \(s^3\) .

OpenStudy (anonymous):

OH!!!!

OpenStudy (amistre64):

and heres why: \(\sqrt[3]{s^8} \iff \sqrt[3]{s^3s^3s^2} \iff \sqrt[3]{s^3}\sqrt[3]{s^3}\sqrt[3]{s^2}\)

OpenStudy (amistre64):

\[\sqrt[3]{s^3} = s\] \[s.s.\sqrt[3]{s^2} \iff s^2 \sqrt[3]{s^2 }\]

OpenStudy (anonymous):

OMGosh! Ok, so back to the problem...I don't remember where we were even!

OpenStudy (amistre64):

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

OpenStudy (anonymous):

Maybe in YOUR head, but not MINE!

OpenStudy (amistre64):

yes, it does help to actually step thru them to gain insight ;)

OpenStudy (amistre64):

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

OpenStudy (amistre64):

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?

OpenStudy (anonymous):

right.

OpenStudy (amistre64):

\[\frac{-5 \sqrt[3]{s}}{3s^3}\]

OpenStudy (anonymous):

wait, if you multiply the numerator by that radical and that radical =s, why do you still have the radical sign there?

OpenStudy (amistre64):

\[-5 * \sqrt[3]{s} = -5\sqrt[3]{s}\]

OpenStudy (anonymous):

yeah, but it's the cube root of s^3. That just equals s I thought?

OpenStudy (amistre64):

\[3s^2\sqrt[3]{s^2}*\sqrt[3]{s}\iff3s^2\sqrt[3]{s^2.s}\iff3s^2\sqrt[3]{s^3}\iff 3s^2.s\] \[\implies 3s^3 \]

OpenStudy (anonymous):

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!!!!! : (

OpenStudy (amistre64):

they expect you to have learned this in college algebra; since thats all it is

OpenStudy (amistre64):

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

OpenStudy (anonymous):

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).

OpenStudy (amistre64):

:) youre doing fine :)

OpenStudy (anonymous):

I get the point! Do you know of a good website with the exponent rules on it?

OpenStudy (anonymous):

Actually, somehow in calc I'm getting a 97%!

OpenStudy (amistre64):

my sirttes are all books; but i know a good practie site for math: interactmath.com

OpenStudy (amistre64):

sirttes eh.... my uiniversal translator is on the fritz i think lol

OpenStudy (anonymous):

Yeah, what is a sirttes anyways?

OpenStudy (amistre64):

it my brain cells jockeying for position lol

OpenStudy (anonymous):

Will you be around for some of today? Can I visit you here again if I need to?

OpenStudy (amistre64):

ill be here as long as the weather holds up; im under the veranda since the libraries are closed

OpenStudy (anonymous):

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

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