Question

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

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

-10t....

Answer 5

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

Answer 6

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

Answer 7

add the parts like fractions,

3(t^2 -3) -30t
------------
t^2 -3

Answer 8

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

Answer 9

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

Answer 10

ok...

you pretty much have the form:

30
3 - ---- now how do we add fractions togther?
b

Answer 11

You have to have a common denominator?

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

That's what I got

Answer 16

then to add the "fractions" mulitply the left part by:


3 t^2-3
-- x ----- right?
1 t^2-3

Answer 17

Oh! Okay

Answer 18

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

Answer 19

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

Answer 20

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

Answer 21

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

Answer 22

Where is the 5 from?

Answer 23

5 +- [sqrt(1008)]/6

Answer 24

30/6 -b/2a

Answer 25

you familiar with the quadratic formula?

Answer 26

yes

Answer 27

good

heres what I get:

t = 5 +- 2sqrt(63)/3

Answer 28

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

Answer 29

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

Answer 30

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

Answer 31

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

Answer 32

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

Answer 33

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

Answer 34

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

Answer 35

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

Answer 36

112 = 16*7

4sqrt(7)

Answer 37

So 5 +- 2sqrt(7)

Answer 38

thats better :)

Answer 39

you know how to test for min and max?

Answer 40

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

Answer 41

No

Answer 42

yep... those are the critical points.

to see how they are behaving, you take a second derivative....derive the derivative.

Answer 43

I don't think I have to do that.

Answer 44

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

Answer 45

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

Answer 46

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

Answer 47

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

Answer 48

Holy cow sorry my computer spazzed out

Answer 49

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

Answer 50

I have found P'(t) if that helps

Answer 51

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

Answer 52

Yes that's what I got

Answer 53

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

Answer 54

(=

Answer 55

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

Answer 56

sin(4t)=1/2

Answer 57

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

Answer 58

there are 2 angles, that will satisfy this

Answer 59

I have no idea.

Answer 60

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

Answer 61

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

Answer 62

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

Answer 63

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

Answer 64

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

Answer 65

yep

Answer 66

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

Answer 67

Sorry I want to understand

Answer 68

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

Answer 69

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

Answer 70

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

Answer 71

Oh! Okay thanks. Moving on

Answer 72

4t = pi/6

4t = 5pi/6

Answer 73

whoa where did that come from?

Answer 74

what!!! you see a bee??

Answer 75

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

Answer 76

Wait is that 30 degrees and 150 degrees in radians?

Answer 77

:) 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

Answer 78

Okay

Answer 79

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

Answer 80

Okay

Answer 81

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

Answer 82

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

Answer 83

so t=.1309 + pin/2

Answer 84

t = pi/24 = .1309

or

t = 5pi/24 = .6545

Answer 85

for degrees...

t = 7.5 degrees or t = 37.5 degrees

Answer 86

Okay. I have a question.

Answer 87

go ahead and ask :)

Answer 88

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?

Answer 89

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..

Answer 90

Okay. That makes sense. Thank you sooo much!

Answer 91

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

Answer 92

you are soooo welcome :)

Answer 93

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