anonymous
  • anonymous
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.
Precalculus
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
@Mertsj Please?
Isaiah.Feynman
  • Isaiah.Feynman
Take the derivative of the position function.
anonymous
  • anonymous
Apply t=2 to -2-6(2)?

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Isaiah.Feynman
  • Isaiah.Feynman
@kokopuffs890 no that would be wrong. You need to take the derivative first.
anonymous
  • anonymous
ok i know the derivative for this is -6, i just dont know how to show it
anonymous
  • anonymous
@Isaiah.Feynman
Isaiah.Feynman
  • Isaiah.Feynman
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.
anonymous
  • anonymous
OH i see... i got it from here: http://www.wolframalpha.com/input/?i=derivative%3A+-2-6t%2C+t%3D2
Isaiah.Feynman
  • Isaiah.Feynman
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.
anonymous
  • anonymous
So by just showing to solve that, would be correct the answer?
Isaiah.Feynman
  • Isaiah.Feynman
Yeah, use the definition to solve it.
anonymous
  • anonymous
ok lets see...
anonymous
  • anonymous
what is h in the equation? -6?
Isaiah.Feynman
  • Isaiah.Feynman
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.
anonymous
  • anonymous
ok, f(2+h)-f(2) / h
Isaiah.Feynman
  • Isaiah.Feynman
Yes and remember t=2+h.
anonymous
  • anonymous
s(t)=2-6t \[\frac{ ds }{dt }=-6,at t=2,velocity = -6\]
anonymous
  • anonymous
so t-2=h?
anonymous
  • anonymous
oh wait! that formula, right, that would work too right @Isaiah.Feynman ?
Isaiah.Feynman
  • Isaiah.Feynman
The formula I just gave would work!
anonymous
  • anonymous
ok cool, so t-2 = h? if we substitute?
Isaiah.Feynman
  • Isaiah.Feynman
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
Isaiah.Feynman
  • Isaiah.Feynman
\[\lim_{h \rightarrow 0} \frac{ -6h }{ h } = -6\]
anonymous
  • anonymous
Oh wow ok, now i get it, and this methods always works?
Isaiah.Feynman
  • Isaiah.Feynman
Yeah, but when the function looks wingspan eyed, you can use the differentiation rules.
anonymous
  • anonymous
right i see, ok great, thanks for the help friend :D
Isaiah.Feynman
  • Isaiah.Feynman
No probs.

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