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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.
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.

This Question is Closed

baru
 one year ago
Best ResponseYou've already chosen the best response.0its 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.

marigirl
 one year ago
Best ResponseYou've already chosen the best response.3show 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.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.5\[\frac{d}{dx}\left(\frac{v^2}{2}\right) = \frac{d}{dv}\left(\frac{v^2}{2}\right)\cdot \frac{dv}{dx} = v \cdot \frac{dv}{dx} = \frac{dx}{dt}\cdot \frac{dv}{dx} = \frac{dv}{dt}\]

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.5In general,\[\frac{dy}{dx} = \frac{dy}{ds}\cdot \frac{ds}{dx}\]This is the chainrule.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\(\color{#0cbb34}{\text{Originally Posted by}}\) @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.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.5\[v := \frac{dx}{dt}\]

marigirl
 one year ago
Best ResponseYou've already chosen the best response.3@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

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.5Yeah, that's a way to think about the chainrule.

marigirl
 one year ago
Best ResponseYou've already chosen the best response.3ok 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: \[\frac{ d }{ dx }(\frac{ v^2 }{ 2 })= v \times \frac{ dv }{ dx }\]

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.5yeah, you're following along well so far.

marigirl
 one year ago
Best ResponseYou've already chosen the best response.3Question 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 \[\frac{ dx }{ dt }\]

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.5Great! Though it'd suit much better to say that \(dv/dt\) is the secondderivative of \(x\) with respect to \(t\).

marigirl
 one year ago
Best ResponseYou've already chosen the best response.3oh right. Thank you so much
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