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marigirl

  • one year ago

Need clarification please Topic: Related rates of change- Differentiation show that: \[\frac{ d }{ dx }(\frac{ v^2 }{ 2 })=\frac{ d^2x }{ dt^2 }\] Where x represents the distance travelled in t seconds by a particle moving with velocity v This is what i have so far:

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  1. marigirl
    • one year ago
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    \[\frac{ dx }{ dt }=v\] Differentiate this with respect to time and get \[\frac{ d^2x }{ dt^2 }=\frac{ dv }{ dt } \]

  2. marigirl
    • one year ago
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    we know: \[\frac{ dv }{ dt }=\frac{ dv }{ dx }\times \frac{ dx }{ dt }\]

  3. marigirl
    • one year ago
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    which is \[\frac{ dv }{ dt }=\frac{ dv }{ dx } \times v\]

  4. marigirl
    • one year ago
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    im lost now

  5. anonymous
    • one year ago
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    By the chain rule, the left hand side is \[\frac{d}{dx}\left[\frac{v^2}{2}\right]=\frac{1}{2}\left(2v\frac{dv}{dx}\right)=v\frac{dv}{dx}\]which you showed is equal to \(\dfrac{dv}{dt}\).

  6. marigirl
    • one year ago
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    I am not sure how you for the left side of the equation you wrote above

  7. anonymous
    • one year ago
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    Do you know what the chain rule is?

  8. marigirl
    • one year ago
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    i think the v and t are confusing me because i keep writing x and y. i might try doing it again

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