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. PLEASE HELP

Answer 1

Will we be using the `Limit Definition of the Derivative`?
Or have you learned the shortcuts at this point?

Answer 2

I know I have to use the formula lim=f(x+h)-f(x)/h but I just have no idea how to use it for this example.

Answer 3

Ah ok :)

Answer 4

yeah :D

Answer 5

\[\large \lim_{h\to0}\frac{f(x+h)-f(x)}{h}\]

So the notation is a little different for the function we were given.
Lemme write the limit definition we'll be using so it matches our function.\[\large \lim_{h\to0}\frac{s(t+h)-s(t)}{h}\]

Answer 6

ah okay :)

Answer 7

So we were given \(\large s(t) = -8 - 9t\).
Do you see where we'll be able to plug that in for our formula?\[\large \lim_{h\to0}\frac{s(t+h)-\color{orangered}{s(t)}}{h}\]

We just need to figure out what s(t+h) is, then we can simplify.

Answer 8

\[\large s(\color{royalblue}{t})=-8-9(\color{royalblue}{t})\]Lemme know if this is confusing, function notation can take a little getting used to.\[\large s(\color{royalblue}{t+h})=-8-9(\color{royalblue}{t+h})\]Understand how we got our s(t+h)? :o

Answer 9

Yeah I understand how you got that :D it's just like if we were using x then we'd have 9(x+h) right?? lawd math sucks sometimes

Answer 10

Yah that sounds right.
lol math is awesome XD it's just a big jump switching from x,y to function notation :D
Can be tricky sometimes.

Answer 11

SOMETIMES it's awesome.. sometimes xD bahaha and yeah it's a bit confusing :( what would be the next step after we did this? :D

Answer 12

\[\large \lim_{h\to0}\frac{s(t+h)-s(t)}{h}\]So it looks like our limit, when we plug everything in, will give us,\[\large \lim_{h\to0}\frac{\left[-8-9(t+h)\right]-(-8-9t)}{h}\]
Our goal is to get it to a point where we can simply plug in h=0 without having a problem.
Right now if we tried to do that, we would end up dividing by zero.
So we'll have to do some simplification first.

Answer 13

Ok ok ok now you do some work :U
Try to simplify it down!!

Answer 14

so we'd have -8-9t-9h+8+9t/h which would then simplify to -9h/h then giving me -9 since both h's would cancel out?? I hope that's right :(

Answer 15

So you divide the h's out, and you're left with \(\large lim_{h\to0}-9\) right?
And since we have no h's, the limit thingy isn't really affecting us anymore :o

\[\large \lim_{h\to0}-9=-9\]

Yay good job \:D/

Answer 16

Oh they wanted the rate of change at a particular point.
We should have approached this a little bit differently.

It will step end up the same.. but meh..

Answer 17

The wanted the instantaneous rate of change at `t=1`.
So we actually wanted to set up our limit like this,\[\large \lim_{h\to0}\frac{\left[-8-9(1+h)\right]-(-8-9\cdot1)}{h}\]With 1's in place of all the t's.
But as you saw, when we did this particular problem, all the t's cancelled out anyway.
So our rate of change is still -9.

Answer 18

Bahh I hope I didn't confuse you there :U

Answer 19

LOL I was just about to say what do we do with the 1 now xD hehehe no problemo I understand!! THANK YOU S FRIGGIN MUCHHH!!!! <3 it makes a lot more sense now!

Answer 20

yay team \c:/

Answer 21

Woo! :D

Answer 22

Oh that was your first question? :OOO

\[\huge \color{royalblue}{\text{Welcome To OpenStudy! :)}}\]