Question

Question with limits below

Answer 1

Use the position function
\[s(t)=-16t^2+113\]

Answer 2

to find the velocity in feet/second at time t=2 seconds. The velocity at time t=c seconds is given by
\[\lim_{t \rightarrow c}(\frac{ s(c)-s(t) }{ c-t})\]

Answer 3

I've got it simplified to the point
\[\frac{ 16t^2-16c^2 }{ c-t }\]
but im not sure what to do from there

Answer 4

@xapproachesinfinity

Answer 5

I also know the answer is -64 feet/second, but I don't know how to get there.

Answer 6

Take the derivative of your pos. function in order to get your velocity function

Answer 7

I think I already did that though

Answer 8

v(t)= what

Answer 9

eh no derivative is allowed since it is set up already as the limit definition of derivative

Answer 10

go further simplifying the experession you got there is work left still

Answer 11

wait

Answer 12

and you need to redo it since there is a math error

Answer 13

ok

Answer 14

you got some negative sign missed up
redo it again

Answer 15

well the easier way is to do derivative and sub c and you are done
but then they want you do to it this way

Answer 16

\[\frac{ (-16c^2+113)-(-16t^2+113) }{ (c-t) }\]

Answer 17

then it become 16t^2-16c^2 on top yes

Answer 18

yeah I had that

Answer 19

factor 16
so 16(t^2-c^2)=? what else can you do

Answer 20

factor (t^2-c^2) ?

Answer 21

yeah! why would i want to factor?

Answer 22

to get out a c-t so you can cancel

Answer 23

but the part that I got stuck on is how to factor out a (c-t)

Answer 24

yeah!

Answer 25

ok, divide the (t^2-c^2) by t and by -c

Answer 26

no

Answer 27

by -t and c

Answer 28

hmm how ? think

Answer 29

\[\frac{ (16)(c-t)(-t-c) }{ c-t }\]

Answer 30

t^2-c^2=(t-c)(t+c)

Answer 31

so 16(-4)

Answer 32

hmm how did you get that
in case you don't know t^2-c^2 is a difference of two squares
have a factored form

Answer 33

i wrote it above

Answer 34

did you actually wrote the limit correctly?
just check here before we proceed

Answer 35

yeah
limit as t approaches c

Answer 36

i'm asking about the rest f(c) -f(t)

Answer 37

yeah thats right

Answer 38

alright just checking, that doesn'[t change anything anyway

Answer 39

ok

Answer 40

so we are left with -16(t+c) yes

Answer 41

yeah

Answer 42

so since t=2 its -16(4)

Answer 43

-64

Answer 44

- because i canceled -(t-c) on the bottom not (c-t) as it was

Answer 45

yeah

Answer 46

t>>>c means t disappears yes?

Answer 47

so f'(c)=-16c

Answer 48

yeah

Answer 49

eh hold on something is wrong

Answer 50

yeah

Answer 51

the limit is the velocity at time (t=c)

Answer 52

so c=2 because it says t=2

Answer 53

so as t approaches c

Answer 54

t is 2 because c is 2

Answer 55

so -16(t+c)

Answer 56

-16(4)

Answer 57

no at t >>> c so evaluate t=c
-16(2+2)

Answer 58

yeah that's correct

Answer 59

yeah

Answer 60

just lost focus at some point and thought of t>>>0

Answer 61

yeah it was a more difficult question because it didn't directly say a variable approaching a number, you had to determine that from the question.

Answer 62

i guess we are done !

Answer 63

yup :)

Answer 64

thanks again

Answer 65

not really! just limit definition of derivative

Answer 66

c is a number why you find it difficult that way? hehehe
think of c a a number and it should be lol

Answer 67

and you are welcome