Word Question Involving the Definition of the Derivative

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The speed is the derivative of the distance with respect to time. You used the correct formula to work out the derivative but applied it incorrectly.

So, given the distance $s(t)=2200-4.9t^2$, we have speed $v$ given by:$$v=\frac{ds}{dt}=\lim_{h\to0}\frac{s(t+h)-s(t)}{h}=\lim_{h\to0}\frac{(2200-4.9(t+h)^2)-(2200-4.9t^2)}{h}$$$$\therefore v=\lim_{h\to0}\frac{2200-4.9(t^2+2th+h^2)-2200+4.9t^2}{h}$$Hopefully you can complete this from here...

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McB
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Updated on March 01, 2020

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  • McB
    McB over 3 years

    A skydiver jumps out of a plane from a height of $2200$m. The skydiver's height, $h$ meters above the ground, can be modeled by the function $$s = 2200 - 4.9t^2$$

    How fast is the skydiver falling at $t = 4$s?

    If the skydiver is approaching a height of $2200$m, I reasoned that I could use the following definition to find the solution $${f(x+h)-f(x)\over h}$$

    I entered the function into this definition and got stuck when I got to this$$f'(x) = \lim_{x\to 0} {2200-4.9t^2-9.8th-4.9h^2\over h}$$

    Where can I go from here? Can I continue with this method or did I make a mistake a few steps back?

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    • Admin
      Admin over 8 years
      Should the problem say "the skydiver's height, $s$" or should the equation be $h = 2200 - 4.9t^2$? And if not, you shouldn't use $h$ as the change in $x$ in your difference quotient because it already has a meaning. Also, why do you need to go back to the limit definition? Nowhere in the problem does it say to use the limit definition of the derivative so you should just use your differentiation rules.
    • McB
      McB over 8 years
      @Bye_World The question does read h meters. Good point, I didn't think of that. In what way could I use the differentiation rules?
    • Admin
      Admin over 8 years
      How would you find the derivative of $y=2200 - 4.9x^2$? Would you evaluate a limit or could you do it an easier way?
  • McB
    McB over 8 years
    Okay, thanks. I will try again. So why is h approaches infinity?
  • Mufasa
    Mufasa over 8 years
    Sorry I meant approaches zero - let me correct this...
  • McB
    McB over 8 years
    Ah, thank-you! Makes sense now...
  • McB
    McB over 8 years
    Think I've got it.. How does the edit look?
  • Mufasa
    Mufasa over 8 years
    Looks right to me :) - BTW: I wouldn't use s in your answer as they use that symbol to represent the skydivers height in your question. You could instead conclude with "speed=..."