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There is a derivation for kinetic energy using calculus:

\begin{align} \Delta E_k&=\int_{x_0}^{x_1} F \ {\rm d}x \\ &= \int_{x_0}^{x_1} ma \ {\rm d}x \\ &= m \int_{x_0}^{x_1} \frac{{\rm d}v}{{\rm d}t} \ {\rm d}x \\ &= m \int_{x_0}^{x_1} \frac{{\rm d}v}{{\rm d}x}\frac{{\rm d}x}{{\rm d}t} \ {\rm d}x \\ &= m \int_{v_0}^{v_1} \frac{{\rm d}x}{{\rm d}t} \ {\rm d}v \\ &= m \frac{v^2}{2}\bigg\vert_{v_0}^{v_1} \end{align}

How to justify change of limits in this case: $t_1 \rightarrow v_1$ and $t_0 \rightarrow v_0$? I suppose it's a simple change of variables, but I'm not sure how to do it.

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    $\begingroup$ This is a question about mathematics, not physics. (Why were the limits originally $t_0, t_1$ when the integral was wrt $x$?) $\endgroup$ Commented Dec 8, 2017 at 19:51
  • $\begingroup$ You changed the integrand, so you need to change the boundaries. Nothing to see here! The derivation is 100 % correct. $\endgroup$
    – Gert
    Commented Dec 8, 2017 at 20:41

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I think it would be more helpful if you would go back to "integration by parts". When you change the $dx$ to $dv$, you're changing what you have "respect to" in terms of the integral. Going back to basic calculus, remember how when you let $u$ equal some variables and input it into an integral, you had to change the bounds. This is the same situation.

Let $T = \frac{1}{2}mv^2$,

Therefore, $$\frac{dT}{dt} = \frac{1}{2}m\frac{d}{dt}(v^2)$$ $$=\frac{1}{2}m[\dot{v} \cdot v + v \cdot \dot{v}]$$ $$=\frac{1}{2}m[2(\dot{v} \cdot v)] = m\dot{v} \cdot v = F \cdot v$$

Now for the integration by parts, bring the dt to the right and integrate,

$$dT = F \cdot v \ dt$$ $$T = \int\limits_{t_0}^{t_1} F \cdot v \ dt$$

Recall $v \ dt = dr$,

$$T = \int\limits_{t_0}^{t_1} F \cdot dr$$ $$=\int\limits_{t_0}^{t_1} m \dot{v} \cdot dr$$ $$=m\int\limits_{t_0}^{t_1} \frac{dv}{dt} dr$$ $$=m\int\limits_{t_0}^{t_1} \frac{dv}{dr}\frac{dr}{dt} dr$$ $$=m\int\limits_{v_0}^{v_1}dv \ \frac{dr}{dt} $$ It's at this point where you must change the bound variables because $dv$ is what you're integrating with respect to. Thinking about it in the $u$ and $du$ calculus, here, I used $v$ instead of $u$.

Recall $\frac{dr}{dt} = \dot{v}$,

$$T = m \int\limits_{v_0}^{v_1} \dot{v} \ dv$$ $$= m[\frac{v_1^2}{2} - \frac{v_0^2}{2}]$$

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B.C.'s answer is interesting. I never would have thought of using integration by parts to answer this question.

Here's an alternative way to derive it just using the chain rule.

$$T = \frac{1}{2}m{v^2}$$

Assuming the mass is constant: $$\frac{{dT}}{{dx}} = \frac{m}{2}\frac{d}{{dx}}\left( {{v^2}} \right)$$

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

So from the previous two equations we get

$$\frac{{dT}}{{dx}} = \frac{m}{2}\frac{d}{{dx}}\left( {{v^2}} \right) = \frac{m}{2} \cdot 2a = ma = F$$

And finally: $$\int_{{x_0}}^{{x_1}} {Fdx} = \int_{{x_0}}^{{x_1}} {\frac{{dT}}{{dx}}dx = T\left( {v\left( {{x_1}} \right)} \right)} - T\left( {v\left( {{x_0}} \right)} \right) = \frac{1}{2}mv_1^2 - \frac{1}{2}mv_0^2$$

So basically... the reason we are able to substitute $v_0$ and $v_1$ for $x_0$ and $x_1$ as our limits of integration is because $$F = \frac{{dT\left( {v\left( x \right)} \right)}}{{dx}} = \frac{{dT}}{{dv}}\frac{{dv}}{{dx}}$$

For more info on integration by substitution, plus some practice problems, check out these notes

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