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anonymous

  • one year ago

The position of an object at time t is given by s(t) = -4 - 2t. Find the instantaneous velocity at t = 6 by finding the derivative.

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  1. anonymous
    • one year ago
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    pleeeaaaaaase help

  2. amistre64
    • one year ago
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    well, what is our derivative?

  3. anonymous
    • one year ago
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    I have no idea how to start it that's the problem @amistre64

  4. amistre64
    • one year ago
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    your course materials should have a basic guideline, what have you covered?

  5. amistre64
    • one year ago
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    how do you define a deriative?

  6. anonymous
    • one year ago
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    I know but none of them look like this. They never went over a problem like this with velocity

  7. amistre64
    • one year ago
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    lets start with how you define the derivative, what does your material say

  8. amistre64
    • one year ago
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    does slope play into it?

  9. anonymous
    • one year ago
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    yes

  10. amistre64
    • one year ago
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    then give me your best understanding of the concept of a derivative

  11. anonymous
    • one year ago
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    this is what we are suppose to use but it's confusing http://www.sosmath.com/calculus/diff/der00/der00.html

  12. amistre64
    • one year ago
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    i agree that the technical part of it may be confusing, but the concept is pretty simple ... most concepts are. the derivative can tell us the slope of a tangent line to a curve at any given point. and it is said that a line is tangent to itself ... does this mean anything to you?

  13. anonymous
    • one year ago
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    Okay somewhat

  14. anonymous
    • one year ago
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    If i followed the formula somewhat it would be -2 but that makes no sense to me

  15. amistre64
    • one year ago
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    mathically, there is a long process that can be worked out with the limit of the difference quotient. \[\lim_{h\to 0}\frac{f(x+h)-f(x)}{(x+h)-x}\] let f(x) be a line ... mx+b \[\lim_{h\to 0}\frac{m(x+h)+b-(mx+b)}{(x+h)-x}\] \[\lim_{h\to 0}\frac{mx+mh+b-mx-b}{h}\] \[\lim_{h\to 0}\frac{mh+b-b}{h}\] \[\lim_{h\to 0}\frac{mh}{h}\] \[\lim_{h\to 0}m\frac{h}{h}\] \[\lim_{h\to 0}m\] the derivative of a line, is just its slope, m

  16. anonymous
    • one year ago
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    f(x)=−4−2x f(x+h)=−4−2(x+h) f′(x)=limh→0f(x+h)−f(x)h→limh→0−4−2(x+h)−(−4−2x)h

  17. amistre64
    • one year ago
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    -2 is the derivative of your line equation. so yes. for any value of t, the derivative is a constant -2

  18. anonymous
    • one year ago
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    ohh geez. I overcomplicated it

  19. amistre64
    • one year ago
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    s(t) = -4-2t v(t) = -2

  20. anonymous
    • one year ago
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    Thank you!

  21. amistre64
    • one year ago
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    good luck :)

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