Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

\[\frac{d}{dx}\left(\frac{v^2}{2}\right) = \frac{d}{dv}\left(\frac{v^2}{2}\right)\cdot \frac{dv}{dx} = v \cdot \frac{dv}{dx} = \frac{dx}{dt}\cdot \frac{dv}{dx} = \frac{dv}{dt}\]

Answer 4

In general,\[\frac{dy}{dx} = \frac{dy}{ds}\cdot \frac{ds}{dx}\]This is the chain-rule.

Answer 5

\(\color{#0cbb34}{\text{Originally Posted by}}\) @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.

Answer 6

\[v := \frac{dx}{dt}\]

Answer 7

Inooooo

Answer 8

@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

Answer 9

Yeah, that's a way to think about the chain-rule.

Answer 10

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:
\[\frac{ d }{ dx }(\frac{ v^2 }{ 2 })= v \times \frac{ dv }{ dx }\]

Answer 11

yeah, you're following along well so far.

Answer 12

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 \[\frac{ dx }{ dt }\]

Answer 13

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

Answer 14

oh right. Thank you so much