Find all critical points of R(t)=3t+77-15ln(t^2-3)

- anonymous

Find all critical points of R(t)=3t+77-15ln(t^2-3)

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- amistre64

R' = 3 -(30t)/(t^2-3) i think

- amistre64

3t^2 -30t -9
-------------
t^2- 3

- amistre64

3(t^2 -10 -3) would be the top if I did it right....

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- amistre64

-10t....

- anonymous

I Got R'=3-15/(t^2-3)(2t)

- amistre64

yep, as long as thats just screen clutter and not a glaring error :)

- amistre64

add the parts like fractions,
3(t^2 -3) -30t
------------
t^2 -3

- anonymous

Sorry how did you get 3(t^2-3) on top?

- amistre64

the top becomes a quadratic.... a fancy name for "highest power is a 2"

- amistre64

ok...
you pretty much have the form:
30
3 - ---- now how do we add fractions togther?
b

- anonymous

You have to have a common denominator?

- anonymous

But if R'(t)=3- 15 (2t)
-----
(t^2 -3)

- anonymous

That would be 3 - 30t
-------
(t^2 - 3)

- amistre64

30t
3 - ------- is what we got right?
(t^2 -3)

- anonymous

That's what I got

- amistre64

then to add the "fractions" mulitply the left part by:
3 t^2-3
-- x ----- right?
1 t^2-3

- anonymous

Oh! Okay

- amistre64

:) deriving...no problem.... fractions? ehhhh :)

- amistre64

3t^2 -30t -9
-----------
t^2 -3

- amistre64

quadratic formula for the top to find the critical points...

- amistre64

5 +- [sqrt(900 - (4)(3)(-9))]/6

- anonymous

Where is the 5 from?

- amistre64

5 +- [sqrt(1008)]/6

- amistre64

30/6 -b/2a

- amistre64

you familiar with the quadratic formula?

- anonymous

yes

- amistre64

good
heres what I get:
t = 5 +- 2sqrt(63)/3

- anonymous

Did you do quadratic formula of 3t^2-30t-9?

- amistre64

yep..... as long as I kept track of myself

- amistre64

30 sqrt(900-(4)(3)(-9))
--- +- --------------
6 6

- anonymous

So 30 +- sqrt(900+108)
----------------------
6

- amistre64

we can simplify that by factoring out a 3 first.... sneaky little bugger it is :)

- amistre64

3(t^2 -10t -3) now quad the left side

- amistre64

10 +- sqrt(100 -(4)(-3)(1))
-----------------------
2

- amistre64

10 +- sqrt(112)
--- -------- same result, just easier to see
2 2

- amistre64

112 = 16*7
4sqrt(7)

- anonymous

So 5 +- 2sqrt(7)

- amistre64

thats better :)

- amistre64

you know how to test for min and max?

- anonymous

So those are the critical points? The neg and pos value from that?

- anonymous

No

- amistre64

yep... those are the critical points.
to see how they are behaving, you take a second derivative....derive the derivative.

- anonymous

I don't think I have to do that.

- amistre64

you sure? its fun :) if you have to determine if they are MIN MAX or inflections, then we would, but if you just gotta find criticals, thats it.... dont forget to include the end points of the graph as well, they might be important

- anonymous

Wow, thanks! Can you help me with one more?

- amistre64

i dunno....... i feel a bout of stupidy brewing amongst me brain cells :) ........sure why not :)

- anonymous

Find all critical points of P(t)=-cos(4t)+39-2t

- anonymous

Holy cow sorry my computer spazzed out

- amistre64

thats just the cosine function moved about.... should be rather simple to determine

- anonymous

I have found P'(t) if that helps

- amistre64

P' = 4sin(4t) - 2 right?

- anonymous

Yes that's what I got

- amistre64

good, then we are either both genuises or idiots :) Ill go with genuisesss

- anonymous

(=

- amistre64

0 = 4sins(4t) -2 solve for sin(4t)

- anonymous

sin(4t)=1/2

- amistre64

what does 4t have to equal for sin to be 1/2?

- amistre64

there are 2 angles, that will satisfy this

- anonymous

I have no idea.

- amistre64

what? you gotta have SOME idea.... its just a circle :)

- amistre64

sin(30) = 1/2 and sin(150) = 1/2

- anonymous

How do I figure the two angles that will satisfy that?

- amistre64

you can either remember back to trig that a 30-60-90 triangle has sides of 1-2-sqrt(3) and interpolate from that, OR use the sin inverse function on a calculator

- anonymous

so sin^-1(1/2)=4t

- amistre64

yep

- anonymous

what from that tells me that 30 and 150 are what I'm looking for?

- anonymous

Sorry I want to understand

- amistre64

sin^-1 will only give you one angle tho...30. You have to remember to cross over the yaxis to get to the other one...150

- amistre64

4t = 30 and 4t = 150 solve for t :)

- amistre64

might be better to use radians instead of degrees, half dozen of one or 6 of the other

- anonymous

Oh! Okay thanks. Moving on

- amistre64

4t = pi/6
4t = 5pi/6

- anonymous

whoa where did that come from?

- amistre64

what!!! you see a bee??

- anonymous

No I mean the 4t=pi/6 and 4t=5pi/6

- anonymous

Wait is that 30 degrees and 150 degrees in radians?

- amistre64

:) there are a few ways to express angles. you can use degrees: 30 and 150 are degrees. OR you can use "radians" which measures the angle using the radius itself as a measuring stick.
a circle is 360 degrees around OR 2pi radians
pi/6 = 30 degrees
5pi/6 = 150 degrees

- anonymous

Okay

- amistre64

To get a number for "t" we should use radians :)

- anonymous

Okay

- amistre64

t = pi/24 t = 5pi/24 is what I come up with

- amistre64

30/4 or 150/4 depending on what your answers are looking for

- anonymous

so t=.1309 + pin/2

- amistre64

t = pi/24 = .1309
or
t = 5pi/24 = .6545

- amistre64

for degrees...
t = 7.5 degrees or t = 37.5 degrees

- anonymous

Okay. I have a question.

- amistre64

go ahead and ask :)

- anonymous

usually before solving for t my professor asks "which quadrants is sin negative in"
why did we not have to look for that in this problem?

- amistre64

sin(angle) = 1/2
sin is only positive in Q1 and Q2 .... no need to look any where else, unless you can think of a reason to..

- anonymous

Okay. That makes sense. Thank you sooo much!

- amistre64

when sin = -1/2, it puts a different value into the equation that is not a critical point

- amistre64

you are soooo welcome :)

- amistre64

if you have any more questions, just start a new posting :)

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