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

@Mertsj Please?

Answer 2

Take the derivative of the position function.

Answer 3

Apply t=2 to -2-6(2)?

Answer 4

@kokopuffs890 no that would be wrong. You need to take the derivative first.

Answer 5

ok i know the derivative for this is -6, i just dont know how to show it

Answer 6

@Isaiah.Feynman

Answer 7

I'm working on it. I used this \[\lim_{h \rightarrow 0} \frac{ f(t+h)-f(t) }{ h }\]. But I got 6 which is not correct yet.

Answer 8

OH i see... i got it from here: http://www.wolframalpha.com/input/?i=derivative%3A+-2-6t%2C+t%3D2

Answer 9

Yes the derivative is indeed -6 and you can use that formula I used to show it. That formula is called the definition of a derivative.

Answer 10

So by just showing to solve that, would be correct the answer?

Answer 11

Yeah, use the definition to solve it.

Answer 12

ok lets see...

Answer 13

what is h in the equation? -6?

Answer 14

Okay, lets start afresh, discard everything I already said. Now the definition of a derivative is \[\lim_{h \rightarrow 0} \frac{ f(a+h) - f(a) }{ h }.\]. In that equation a is 2, and t=a+h. Substitute these values and see what happens.

Answer 15

ok, f(2+h)-f(2) / h

Answer 16

Yes and remember t=2+h.

Answer 17

s(t)=2-6t
\[\frac{ ds }{dt }=-6,at t=2,velocity = -6\]

Answer 18

so t-2=h?

Answer 19

oh wait! that formula, right, that would work too right @Isaiah.Feynman ?

Answer 20

The formula I just gave would work!

Answer 21

ok cool, so t-2 = h? if we substitute?

Answer 22

Yeah, but you don't have to substitute though. \[\lim_{h \rightarrow 0} \frac{ -2-6(2+h)-(2-12) }{ h }\], simplifying gives us \[\lim_{h \rightarrow 0}\frac{ -2 -12-6h+2+12 }{ h }\] 2's and 12's cancel out leaving us with

Answer 23

\[\lim_{h \rightarrow 0} \frac{ -6h }{ h } = -6\]

Answer 24

Oh wow ok, now i get it, and this methods always works?

Answer 25

Yeah, but when the function looks wingspan eyed, you can use the differentiation rules.

Answer 26

right i see, ok great, thanks for the help friend :D

Answer 27

No probs.