Question

The position of an object at time t is given by s(t) = -2 - 6t. Find the instantaneous velocity at t = 2 by finding the derivative.

Answer 1

@zepdrix Hey!! I remember you helped me with a question similar to this before!I get the basic idea of it but I forgot how to turn the s(t) setup into the normal like function, like -6t is after -2? can you help me out with how I set this up to solve :D

Answer 2

Hmm I forget,
Have you learned the shortcuts for taking derivatives?
Or do we need to use the `Limit Definition of the Derivative`?

Answer 3

Usually I was using limh->0 f(x+h)-f(x)/h :D

Answer 4

ds/dt = -6

Answer 5

\[\Large \lim_{h\to0}\frac{s(t+h)-s(t)}{h}\]Using limit? Ok cool.
\[\Large s(t)=-2-6t\]\[\Large s(t+h)=-2-6(t+h)\]

Answer 6

Oh oh right, they said at t=2.

Answer 7

the t value doesn't matter, because the object is moving at constant velocity

Answer 8

So we actually wanna use,
\[\Large s(2)=-2-6(2)\]\[\Large s(2+h)=-2-6(2+h)\]


\[\Large s'(2)=\lim_{h\to0}\frac{s(2+h)-s(2)}{h}\]

Understand how to plug these in and simplify? :o

Answer 9

umm not really :( I never used it that way before with 2 in () afte s

Answer 10

Function notation is tricky D: Gotta try and get comfortable with it.
So like...

Answer 11

Limit definition of the derivative:
\[\Large f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}\]
But we can evaluate it at specific points.\[\Large f'(3)=\lim_{h\to0}\frac{f(3+h)-f(3)}{h}\]

Is that the part that's confusing you? :o

Answer 12

no when it's f(x) I completely understand but now that it's s(t) and the -6t comes after the -2 I'm in a brain tizzy :'(

Answer 13

We don't have to call it s(t) if that's confusing.
We can instead call our function,\[\Large s(t)=-2-6t \qquad\to\qquad f(x)=-2-6x\]

Answer 14

Let's try to fill in the pieces.\[\Large \color{royalblue}{f(x)=-2-6x}\]\[\Large \color{orangered}{f(x+h)=-2-6(x+h)}\]

\[\large f'(x)=\lim_{h\to0}\frac{\color{orangered}{f(x+h)}-\color{royalblue}{f(x)}}{h} \quad\to\quad \lim_{h\to0}\frac{\color{orangered}{-2-6(x+h)}-\color{royalblue}{(-2-6x)}}{h}\]

Answer 15

ahhh so -2-6x-6h+2+6x/h !?

Answer 16

mm yah that sounds right.

Answer 17

Keep simplifying, it should clean up nicely.

Answer 18

so then we have -6h/h and h and h cancel out so the answer is -6?

Answer 19

\[\Large f'(x)=-6 \qquad\to\qquad s'(t)=-6\]Ok good, but they wanted you to evaluate the function at t=2, what do you get? ;)

Answer 20

-3!??...(:?

Answer 21

\[\Large s'(2)=?\]Plug 2 into all of your t's in the equation :X *giggles*

Answer 22

wait so am I still keepin -6?? oh lawwwwd :) is it -12?

Answer 23

OH into the original!?

Answer 24

There are no t's, right? :o
The function is just a constant -6 at all times.

\[\Large s'(2)=-6\]

Answer 25

YOU TRICKED ME!!! xD LOL that's why I was confused xDDD

Answer 26

you're evil :3

Answer 27

hah :3

Answer 28

but thank yuhh <3 @zepdrix :D