Question

S(t)=(At^p)e^-kt
find the intervals when it's increasing. Help!

Answer 1

t>=0, A,P,k are all positive constants. I got \[S'(t)=APt^{p-1}e^{-kt} -AKt ^{p}e^{-kt}\]
but I can't solve it when equate to zero to get the extrema.

Answer 2

your derivative looks good

So I would begin by factoring as much as I could

Answer 3

\[S'=At^{p-1}e^{-kt} \cdot P-At^{p-1}e^{-kt} \cdot Kt\]
I rewrote S'
I hope you can see the hint I'm trying to provide as to what you need to factor

Answer 4

\[S'(t) = Ae^{-kt}(Pt^{P-1} - kt^P)\]

I got this when I factored. My problem now is this
\[0 = Ae^{-kt}(Pt^{P-1} - kt^P)\]
I dont know how to isolate t.

Answer 5

Ok you need to all factor out t^{p-1}

Answer 6

also* (not all)

Answer 7

only one term has t^p-1

Answer 8

did you see my hint above

Answer 9

both terms have t^{p-1}

Answer 10

I suppose you could factor out t^p instead
I just thought it looked nicer with the t^{p-1} factored out

Answer 11

how did you get both terms to have t^{P-1}...when I solved it for derivative only one term has it. Could you recheck it please.

Answer 12

So I take it you didn't look at how I rewrote S'?

Answer 13

i did. I will again. wait

Answer 14

recall
t^p=t^{p-1} \times t

Answer 15

omg. i see it now. wth..how could i miss that :(

Answer 16

It happens
so many numbers to look at

Answer 17

I got \[At^{P-1}e^{-kt}(P-kt)\]
I will now equate it to zero

Answer 18

ok

Answer 19

looks great

Answer 20

\[At^{P-1}e^{-kt}(P-kt)=0\]
\[(P-kt)=0\]
\[P=kt\]
\[t=K/P\]

Answer 21

is that right? lol now I doubt my math lol

Answer 22

lol t=P/K

Answer 23

You have P=kt
So divide both sides by k

Answer 24

true. lol

Answer 25

I want you to know we are going the long way about this.

Answer 26

We can.

Answer 27

what's the fastest?

Answer 28

So you want to find when S'>0
We don't really need to find when S'=0

So we want to know for what values t that we have
\[At^{p-1}e^{-kt}(P-kt)>0\]

We know A is positive
WE know t^{p-1} is positive because t>=0
We know e^{-kt} is positive because e is positive
The only thing that decides if this will be negative or positive is the factor that is P-kt

I'm saying solve P-kt>0 instead of P-kt=0

Answer 29

Then we will be done

Answer 30

True. You are right. I thought of that but I wasn't sure if my thoughts are right. So I can do that in any functions? if I am just looking for increasing intervals?

Answer 31

so t<k/P

Answer 32

Well normally I would find when S'=0 and then plot any critical numbers/endpoints and test the intervals around them

But I decided it would be unnecessary here as so many factors of S' were positive


You could go about it like you were

Answer 33

you did the same thing again

Answer 34

t<p/k

Answer 35

dang. I like the sound of k/p than p/k that's why I keep messing it lol

Answer 36

Right
S is increasing when t<p/k

Answer 37

well good job @jorea143
You done it
Give yourself a pat on the back

Answer 38

so how do I write that in interval?

Answer 39

well numbers that are less than p/k go to the left of the number line