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Thread: Implicit Derivatives?

  1. #1
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    Implicit Derivatives?

    I was wondering how this involves using implicit differentiation?

    v = v(t)
    a= a(t)
    v=sqrt(8s+16)


    since v(t) = sqrt(8s+16) do i have to use implicit differentiation? How would finding it help me show that acceleration is constant?


    Also, for Part C, do i just graph v(t) = sqrt(8s+16)

    I'm lost in how i can use implicit differentiation to solve this.


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  2. #2
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    Re: Implicit Derivatives?

    What is the question? You define the functions v and a but don't say what it is you want to do with them! And you say "Also, for Part C, do i just graph v(t) = sqrt(8s+16)" but don't say what "part C" says! How could we possibly say know what you are to do if we don't know what the question is?
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  3. #3
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    Re: Implicit Derivatives?

    (a) $a=\dfrac{dv}{dt} = \dfrac{d}{dt}\bigg[v=\sqrt{8s+16}\bigg] = \dfrac{4}{\sqrt{8s+16}} \cdot \dfrac{ds}{dt} = \dfrac{4}{\sqrt{8s+16}} \cdot \sqrt{8s+16} = 4$

    (b) you have $v$ as a function of $s$ ... do the graph

    (c) $a=4 \implies v = 4t+C$. $s(0) = 6 \implies v(0) = 8 \implies v = ?$ ... then graph $v$ as a function of time.
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  4. #4
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    Re: Implicit Derivatives?

    i can use this as v as a function of s ? --> v(s) = sqrt(8s+16)

    and then for the v as function of t ? How do you know that it's linear?
    Last edited by lc99; Oct 4th 2017 at 09:05 AM.
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  5. #5
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    Re: Implicit Derivatives?

    Quote Originally Posted by lc99 View Post
    i can use this as v as a function of s ? --> v(s) = sqrt(8s+16)
    yes

    and then for the v as function of t ? How do you know that it's linear?
    because acceleration is a constant
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  6. #6
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    Re: Implicit Derivatives?

    Oh my goodness, thanks so much. It seems obvious but not to me! It makes sense now. Thanks!
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