## 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.

1. baru

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

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

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

4. ParthKohli

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

5. anonymous

$$\color{#0cbb34}{\text{Originally Posted by}}$$ @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

$v := \frac{dx}{dt}$

7. anonymous

Inooooo

8. marigirl

@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

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

10. marigirl

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

yeah, you're following along well so far.

12. marigirl

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

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

14. marigirl

oh right. Thank you so much