Question

Calc questions anyone?

Answer 1

@math&ing001 Could you take a look at any of these?

Answer 2

@PaxPolaris Could you take a look at any of these?

Answer 3

Or @zepdrix :) even though you already did a lot for me

Answer 4

which one you doing, first or second?

Answer 5

All of them but I'll start on the first.

Answer 6

ok first one
you got the derivative of \[-4\sin(t)-\frac{t}{2}+10\]?

Answer 7

"no" is a fine answer, just asking

Answer 8

Wouldn't it just be 0?

Answer 9

?? oh no

Answer 10

the derivative is a function, not a number

Answer 11

Oh ok I see what you mean

Answer 12

do you know what the derivative of sine is?

Answer 13

Cos?

Answer 14

yes

Answer 15

so the derivative of \(-4\sin(t)\) is \(-4\cos(t)\)
the derivative of \(\frac{t}{2}\) is \(\frac{1}{2}\) and the derivative of \(10\) is \(0\)

Answer 16

so your derivative is \[s'(t)=-4\cos(t)-\frac{1}{2}\]

Answer 17

the derivative of the position function is velocity, can rewrite as \[v(t)=-4\cos(t)-\frac{1}{2}\]

Answer 18

next job, set that sucker equal to zero and solve
i.e. solve \[-4\cos(t)-\frac{1}{2}=0\]

Answer 19

any ideas on that one?

Answer 20

Hmmm

Answer 21

add one half, divide by \(-4\) then use a calculator

Answer 22

-.125

Answer 23

did i lose you? \[\cos(t)=-\frac{1}{8}\] but now you need a calculator to find that number \[t=\cos^{-1}(-\frac{1}{8})\]

Answer 24

i use this
http://www.wolframalpha.com/input/?i=arccos%28-1%2F8%29

Answer 25

btw we are still not done
you have to take the derivative one more time and plug in the result above

Answer 26

I got 1.69612

Answer 27

you know the derivative of \[-4\cos(t)\]?

Answer 28

yes, that is what i got too
now we need the second derivative for the acceleration , then we plug in the number you wrote above

Answer 29

So for -4cos(t)?

Answer 30

yeah the derivative of that one

Answer 31

I don't know the derivative, no. All I got was 0 from the calc but I don't think that's right

Answer 32

this is not a calculator exercise at this step
the derivative of cosine is minus sine

Answer 33

so the derivative of \[-4\cos(t)\] is \[4\sin(t)\]
plug in the number you got above e.e .compute \[4\sin(1.69612)\]

Answer 34

0.118394220192

Answer 35

that is not what i got

Answer 36

http://www.wolframalpha.com/input/?i=4sin%281.69612%29

Answer 37

3.96863*

Answer 38

yeah

Answer 39

Sorry my mistake

Answer 40

no problem
next question is totally different

Answer 41

let me know when you are ready, it is actually much easier

Answer 42

Ready

Answer 43

before we begin, i have a feeling that you are actually pretty lost on this stuff
i don't care, just asking
do you have any idea what the next question is asking you to do?
"no" again is a fine answer, i can try to explain

Answer 44

LOL ouch. Yes I am pretty lost.

Answer 45

my clue was when you thought the derivative of a function was zero

Answer 46

That's because I was multitasking lol I'm not super slow

Answer 47

i can show you how to do the next question , which is actually integral calculus (sort of) but i can just show you the mechanics of it
it is just a computation
your interval is from 0 to 8 and they want you to break it up in to 4 pieces, so each piece will have length \(8\div4=2\)

Answer 48

the are also asking for left hand endpoints, so the endpoints of the interval will be \[0,2,4,6\] if they were asking for right hand endpoints it would be \[2,4,6,8\]

Answer 49

Makes sense, yes

Answer 50

take \[g(0)\times 2+g(2)\times 2+g(4)\times 2+g(6)\times 2\] or \[\left(g(0)+g(2)+g(4)+g(6)\right)\times 2\]

Answer 51

that would be base (2) times height

Answer 52

Oh boy I'm not good at this.

Answer 53

you have all the numbers you need, it is just an addition is all

Answer 54

you can see from the table what \(g(0)\) is right?

Answer 55

Yes

Answer 56

and also \(g(2), g(4), g(6)\) yes?

Answer 57

Yes

Answer 58

add that mess up, then multiply the result by 2, that is all you need to do for this one

Answer 59

let me know when you get it

Answer 60

-9?

Answer 61

Wait no

Answer 62

yeah, no

Answer 63

-11

Answer 64

yes

Answer 65

Yay :)

Answer 66

lol yeah, yay, but not clear why we did it right? you can read about it later

Answer 67

how about the third one, any ideas?

Answer 68

Yeah, I won't need this class in the future anyway

Answer 69

actually you have no idea what you will need in the future
trust and old head when i say that
you might even need is soon, say for a test

Answer 70

I didn't put much thought into the third one. I'd just say the first two look similar.

Answer 71

yeah, but no

Answer 72

unless the function is always positive, it is not the same as its absolute value

Answer 73

so not the first two
it is the first and third by a u - substitution
whatever that is

Answer 74

Oh wow. Ok thank you lol :)