Question

Thoroughbred Bus Company finds that its monthly costs for one particular year were given by C(t) = 10,000 + t2 dollars after t months. After t months the company had P(t) = 1,000 + t2 passengers per month. How fast is its cost per passenger changing after 8 months? HINT [See Example 8(b).] (Round your answer to two decimal places.)

$_____________ per month

Answer 1

so plug 8 into both equations

Answer 2

i did but it's a negative #

Answer 3

C(t)/P(t)

Answer 4

i think i should also take the derivative of that

Answer 5

depends what does 8(b) look like?

Answer 6

the question ask "how fast" so it's asking the change of rate

Answer 7

so yes you would

Answer 8

ok. thank you, though.

Answer 9

you got it from here

Answer 10

nope, the answer was not right

Answer 11

here is a similar problem http://answers.yahoo.com/question/index?qid=20111107225609AARAMOb

Answer 12

Haha, i have looked that too. but it wasn't right

Answer 13

-2250/17689

Answer 14

that was the same as mine -0.127

Answer 15

hmmm dont know what to tell ya

Answer 16

@hartnn he has all the answers

Answer 17

really? can i ask @hartnn, then?

Answer 18

ya he should be here i a few he is helping someone else

Answer 19

he don't know how to do it

Answer 20

did you ask?

Answer 21

yap

Answer 22

lets try @Jonask he might be able to

Answer 23

ok

Answer 24

\[\frac{ C }{ month }=10000+t^2\]
\[\frac{ P }{ month }=1000+t^2\]
\[\frac{ C }{ P }=?\]
\[\frac{ C }{ month } \div \frac{ P }{ month }=\frac{ C }{ P }\]

Answer 25

\[\frac{ C }{ P }=(10000+t^2)\div(1000+t^2)\]
t=8

Answer 26

i am thinking about take the derivative of C(t)/P(t)

Answer 27

what i got is a negative #

Answer 28

\[\frac{ dC }{ dt }\div\frac{ dP }{ dt}=\frac{ dC }{ dP }\]

Answer 29

yap, the answer is -0.13, but it's wrong

Answer 30

\[\frac{ dC }{ dP }=(10000+t^2)(1000+t^2)\]
\[\frac{ 10000+8^2 }{ 1000+8^2 }=\frac{ 10064 }{ 1064 }\]

Answer 31

\[\frac{ dC }{ dP }=(10000+t^2) \div (1000+t^2)\]

Answer 32

which is also wrong

Answer 33

so\[\frac{ dC }{ dP }=\frac{ 10000+t^2 }{ 1000+t^2 }\]

Answer 34

whats your answer

Answer 35

-0.13

Answer 36

1258/133

Answer 37

still wrong

Answer 38

c(t) has to be derived

Answer 39

yap, that's how i got the negative #

Answer 40

2t but remember cost is naturally negetive

Answer 41

but what i got is wrong

Answer 42

ok guys good luck hope you get it but i g2g to bed

Answer 43

thanks

Answer 44

\[C(t)=\frac{ 10000+t^2 }{ 12 }\] per month

Answer 45

wish i could have helped more, but atleast i found someone to help so we will call it even. thanks jonask

Answer 46

but the question doesn't ask for each month

Answer 47

the function given is for one year,so for a month it will be /12

Answer 48

didnt see that good catch

Answer 49

\[\frac{ dC }{ dP }=\frac{ 2(8) }{ 1000+64 }\]
anyway how is this -

Answer 50

do you know the answer @laulungyuen

Answer 51

why is the ansewr in the form
$...month instead of
$...per passenger

Answer 52

i think i should use quotient rule

Answer 53

\[(\frac{ 10000+64 }{ 12 })\div(1000+64)=\frac{ 10064 }{ 12(1064) }\]

Answer 54

0,788

Answer 55

but the equation you gave me is not related to the derivative, though

Answer 56

we have to derive this\[\frac{ dC }{ dP }\]because question says
"How fast "
\[\frac{ dC }{ dP }=\frac{ 10000+t^2 }{ 1000+t^2 }\]
\[\frac{ d(\frac{ dC }{ dP } )}{ dt }=\frac{ 2t(1000+t^2)-2t(10000+t^2) }{ (1000+t^2) }\]

Answer 57

0,127

Answer 58

-0,127

Answer 59

yap, but it's wrong

Answer 60

how much is the answer

Answer 61

do you have 8(b)

Answer 62

nope

Answer 63

how do you know its wrong if you dont have the answer

Answer 64

im doing online assignment, so if the answer is wrong it will tell me

Answer 65

try -0,01059
i have to go to school ,,,later

Answer 66

ok, thanks

Answer 67

morning class?

Answer 68

yes,its 9am

Answer 69

oh ok
thanks