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marigirl

  • one year ago

Topic: Related rates of change- Differentiation Please explain what the question is actually asking me :) show that: \[\frac{ d }{ dx }(\frac{ v^2 }{ 2 })=\frac{ d^2x }{ dt^2 }\] Where x represents the distance traveled in t seconds by a particle moving with velocity v.

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  1. baru
    • one year ago
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    its just asking you to show that the left hand side of the equation and the right hand side can be simplified to the same thing.

  2. marigirl
    • one year ago
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    show that: \[\frac{ d }{ dx }(\frac{ v^2 }{ 2 })=\frac{ d^2x }{ dt^2 }\] Where x represents the distance traveled in t seconds by a particle moving with velocity v.

  3. ParthKohli
    • one year ago
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    \[\frac{d}{dx}\left(\frac{v^2}{2}\right) = \frac{d}{dv}\left(\frac{v^2}{2}\right)\cdot \frac{dv}{dx} = v \cdot \frac{dv}{dx} = \frac{dx}{dt}\cdot \frac{dv}{dx} = \frac{dv}{dt}\]

  4. ParthKohli
    • one year ago
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    In general,\[\frac{dy}{dx} = \frac{dy}{ds}\cdot \frac{ds}{dx}\]This is the chain-rule.

  5. anonymous
    • one year ago
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    \(\color{#0cbb34}{\text{Originally Posted by}}\) @ParthKohli \[\frac{d}{dx}\left(\frac{v^2}{2}\right) = \frac{d}{dv}\left(\frac{v^2}{2}\right)\cdot \frac{dv}{dx} = \color{red}{v} \cdot \frac{dv}{dx} = \color{red}{\frac{dx}{dt}}\cdot \frac{dv}{dx} = \frac{dv}{dt}\] \(\color{#0cbb34}{\text{End of Quote}}\) That's where my problem was at first.

  6. ParthKohli
    • one year ago
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    \[v := \frac{dx}{dt}\]

  7. anonymous
    • one year ago
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    Inooooo

  8. marigirl
    • one year ago
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    @ParthKohli Thank you for your response. lets see if I make sense now: So from chain rule we can form that: \[\frac{ d }{ dx }(\frac{ v^2 }{ 2 })=\frac{ d }{ dv }(\frac{ v^2 }{ 2 }) \times \frac{ dv }{ dx }\] because the dv on right side will cancel which will give us our left side

  9. ParthKohli
    • one year ago
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    Yeah, that's a way to think about the chain-rule.

  10. marigirl
    • one year ago
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    ok so then next step: we can actually differentiate \[\frac{ d }{ dv }(\frac{ v^2 }{ 2 }) \] because we are differentiating v^2/2 with respect to v so we get: \[\frac{ d }{ dx }(\frac{ v^2 }{ 2 })= v \times \frac{ dv }{ dx }\]

  11. ParthKohli
    • one year ago
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    yeah, you're following along well so far.

  12. marigirl
    • one year ago
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    Question states that x represents the distance traveled in t seconds by a particle moving with a velocity v so we can state \[\frac{ dx }{ dt }=v\] so now we have \[\frac{ d }{ dx }(\frac{ v^2 }{ 2 })=\frac{ dx }{ dt } \times \frac{ dv }{ dx } \] which cancels to \[\frac{ dv }{ dt }\]\\which is the second derivative of \[\frac{ dx }{ dt }\]

  13. ParthKohli
    • one year ago
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    Great! Though it'd suit much better to say that \(dv/dt\) is the second-derivative of \(x\) with respect to \(t\).

  14. marigirl
    • one year ago
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    oh right. Thank you so much

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