Question

How does lim as h goes to 0 (3+h) - s(3)/h = lim as h goes to 0 (3+h)^2 - 9/h?

Instantaneous velocity for s(t)=t^2 at t=3.

Answer 1

you don't want to plug in values until you've simplified it so you can take the limit without any problems...

Answer 2

I don't understand how to simplify it :)

Answer 3

f(t+h) = (t+h)^2 =t^2 +2th +h^2
-f(t) = -t^2

Answer 4

( f(h+t) -f(t) ) / h = ( t^2 +2th +h^2 -t^2) / h

Answer 5

you still can't take the limit as h->0 because h is in the denominator

Answer 6

see any simplifications you can make in the numerator?

Answer 7

factor an s?

Answer 8

there are no s's there is a t^2 and a -t^2

Answer 9

I just dont see it :/

Answer 10

t^2 +2th +h^2 -t^2

Answer 11

where? hold on maybe I didnt subtitute t^2 in?

Answer 12

I don't know where to put the t^2 in the equation

Answer 13

\[\frac{ s(t+h) -s(t) }{ h } = \frac{ t ^{2 } +2th +h^{2} -t ^{2}}{h }\]

Answer 14

agree so far?

Answer 15

hey what do I use to get the math formatted like that?

Answer 16

equation editor

Answer 17

thanks ok Im looking

Answer 18

see the
\[t ^{2} \]
and the \[-t ^{2}\]
in the numerator?

Answer 19

Im not sure of the steps you took to get the equation on the right side of the =

Answer 20

I plugged t+h into into s(t)

Answer 21

subtracted s(t) from that

Answer 22

and divided the whole thing by h

Answer 23

s(t+h) = (t+h)^2 =t^2 +2th +h^2
-s(t) = -t^2

Answer 24

ok Im screwing up on the pluggin in part :)

Answer 25

Im confused by the t+h thing and trying to substitute

Answer 26

put 't+h' in where ever there's a 't' in the equation
this equation is simply 't^2' so it's easy: "(t+h)^2"

Answer 27

yes Im seeing that now

Answer 28

cool. you got it from here?

Answer 29

and the 'S" just dissapears?

Answer 30

because its not a number
its like f(x)

Answer 31

you're confused about function notation
when they say s(t) = t^2
it's just a shorthand way of saying 's depends on t like t^2'
ie if t=2 s=4 if t=9 s=81 and so forth

Answer 32

the function is t^2.... that's what you want to deal with

Answer 33

do you get t^2+3-t^2/3?

Answer 34

no.

Answer 35

that was just me substituting in t^2 and h=3

Answer 36

\[\frac{ t^2+2th+h^2−t^2 }{ h }\]

Answer 37

that's where we left off...

Answer 38

need to simplify... t^2 and -t^2 cancel, so:
\[\frac{2th+h^2 }{ h } \]

Answer 39

the common term is 'h'
\[ \frac{ h(2t+h) }{h }\]

Answer 40

I dont understand how you get the first equation

Answer 41

which?

Answer 42

count 5 up from you

Answer 43

I need to see from the substitution and then the VERY next step

Answer 44

\[ \frac{t^2+2th+h^2−t^2 }{ h }\]
that?

Answer 45

yes
im not substituting correctly

Answer 46

I can't see what you're doing

Answer 47

show me your steps or follow my steps.

Answer 48

what does substituting t^2 into the equation look like?

Answer 49

before ANY simplification :)

Answer 50

naw man. you want to evaluate:
\[\lim_{h \rightarrow 0} \frac{ s(t+h) -s(t) }{ h}\]

Answer 51

yes ok thats the equation for instantaneous velocity

Answer 52

find s(t+h) : put (t+h) into the equation wherever there's a 't'
s(t) is simply t^2
so s(t+h) is (t+h)^2

Answer 53

I get T^2 +h-t^2/h

Answer 54

no.

Answer 55

\[(t+h)^2 = t^2 +2th +h^2\]

Answer 56

for some reason its hard for me to see this

Answer 57

foil it out and verify for yourself (t+h)(t+h) = ?

Answer 58

(t+h)^2 makes sense Im used to seeing that but this?
find s(t+h) : put (t+h) into the equation wherever there's a 't'
s(t) is simply t^2
so s(t+h) is (t+h)^2

Answer 59

that is killing me right now

Answer 60

when you see s(4) what do you do?

Answer 61

if s(t) = t^2

Answer 62

you put 4 in for t

Answer 63

s(4) =4^2 =16

Answer 64

it's just notation to tell you what goes into the function s( whatever) = ( whatever)^2

Answer 65

if we need s(t+h) then t+h goes in where ever there is a 't'
and since s(t) is simply t^2 s(t+h) is simply (t+h)^2

Answer 66

mull it over a bit

Answer 67

internet just went out I mean power ewnt out