Question

The position of an object at time t is given by s(t) = -8 - 9t. Find the instantaneous velocity at t = 1 by finding the derivative.

Answer 1

So v=s'

Answer 2

can you find s'?

Answer 3

i'm guessing you need to do this by the definition and no short cuts

Answer 4

?

Answer 5

though you know the derivative=slope

Answer 6

and s is just a line

Answer 7

is s = -17?

Answer 8

you don't need to find the position at t=1

Answer 9

you need to find the instantaneous velocity at t=1

Answer 10

so how would I do that exactly?

Answer 11

well i was asking you if you need to use the definition of derivative or could you use short cuts
or could you just notice s is a line and and then notice the slope of the line

Answer 12

\[v(t) =\frac{d}{dx}[s(t)]\]

Answer 13

\[v(1) = s'(1)\]

Answer 14

im just unsure on how to do this...

Answer 15

Take the derivative of \(s(t)\)

Answer 16

err
can you tell me how you are allowed to find the derivative?
by definition
by short cut ways
by just noticing it is a line and the slope of y=mx+b is m so therefore the derivative is m

Answer 17

by the way if we have s(t)=mt+b
then s'(t)=m for all t

Answer 18

that is what i'm saying with third third way there

Answer 19

well any way you could show me tbh. my lesson is really confusing and it doesn't make any sense to me

Answer 20

ok well you aren't really telling me what way you are suppose to find the derivative

but i actually already did if you realize that s(t)=-8-9t is in the form s(t)=mt+b

Answer 21

I would start by understanding what the slope of a line represents in calculus terms. Did you learn this yet?

Answer 22

well then is it just -9?

Answer 23

yep the slope of y=-9t-8 is -9
therefore the derivative of y
is y'=-9

Answer 24

\(\color{blue}{\text{Originally Posted by}}\) @freckles
by the way if we have s(t)=mt+b
then s'(t)=m for all t
\(\color{blue}{\text{End of Quote}}\)
That's what was said :)

Answer 25

But I'm kinda wondering since this was an easy function
if the directions were to find the derivative by using the formal definition of derivative

Answer 26

You could do that too.

Answer 27

\[\lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}\]

Answer 28

well could you try and show me the other way as well?:)

Answer 29

Ok, \[ s(t) = -8 - 9t\] Find :
1. \(s(x+h)\)
2. \(s(x)\)

Answer 30

Instead of the t use h or convenience, it just works to fit the definition, HAH

Answer 31

what would be h? just 1?

Answer 32

\[s(x+h) = -8-9(x+h)\]\[s(x) = -8-9(x)\]

Answer 33

h is a representation for a small change in the value of your function.

Answer 34

so \[\lim_{x\rightarrow 0} \frac{-8-9x-9h-(-8-9x)}{h}\]\[\lim_{x\rightarrow 0}\frac{-9h}{h} = -9\]

Answer 35

ok, where do I go from there? you see they give us some equation f(x+h)-f(x)/h = lim h->0, which doesn't make sense

Answer 36

lol wait you just showed me:p sorry im having connection problems with this website rn and idk why

Answer 37

I've got to go, hopefully @freckles can provide you with a better explanation than me :P Good luck!

Answer 38

ok so h will cancel out and with the x plugged in as 1, the answer will equal -9?:)

Answer 39

\[v(t)=s'(t)=\lim_{h \rightarrow 0}\frac{s(t+h)-s(t)}{h}\]
h is a small change as said above
we want this small change to approach 0 so our slope of our secant lines from (t+h,s(t+h)) to (t,s(t)) will approach the slope of the tangent line

Answer 40

\[s(t)=-8-9t \\ \text{ so we can find } s(t+h) \text{ by replacing the } t \text{ above with } t+h \\ s(t+h)=-8-9(t+h)\]
inputting into our formula:
\[v(t)=\lim_{h \rightarrow 0} \frac{[-8-9(t+h)]-[-8-9t]}{h}\]
simplify the top as @Jhannybean did above

Answer 41

then you will be allowed to rid yourself of the h on bottom by cancelling a common factor of h on top with the one on bottom
and therefore you will be able to plug in 0 for h after that if any h's remain that is

Answer 42

ahh ok thanks so much:)))

Answer 43

there is also another form of definition you can use
\[s'(t)=\lim_{z \rightarrow t} \frac{s(z)-s(t)}{z-t}\]
that from right there is probably what you have more commonly associated slope with
\[s'(t)=\lim_{z \rightarrow t}\frac{[-8-9z]-[-8-9t]}{z-t} \\ =\lim_{z \rightarrow t}\frac{-9z+9t}{z-t} =\lim_{z \rightarrow t}-9 \frac{z-t}{z-t}=-9(1)=-9\]

Answer 44

these forms are equivalent
you can show they are equivalent by substitution

Answer 45

let h=z-t
if z->t then h->z-z=0
so h->0
\[s'(t)=\lim_{z \rightarrow t}\frac{s(z)-s(t)}{z-t} \\= \lim_{h \rightarrow 0}\frac{s(t+h)-s(t)}{h}\]
since if h=z-t then h+t=z