# Deriving Work-Kinetic Energy Theorem

I am currently reading Physics for Scientists and Engineers (Ninth Edition) by Serway and Jewett and in Chapter 7.5, a derivation of the work-kinetic energy theorem was shown.

To give context, consider a system consisting of an object of mass $$m$$ moving through a displacement directed to the right due to a net force $$\sum F$$ , also directed to the right.

The derivation process was then, \begin{aligned} W_{\text{ext}} &= \int_{x_i}^{x_f} \sum F \ dx \\ &= \int_{x_i}^{x_f} ma \ dx \\ &= \int_{x_i}^{x_f} m \frac{dv}{dt} \ dx \\ &= \int_{x_i}^{x_f} m \frac{dv}{dx} \frac{dx}{dt} \ dx \\ &= \int_{v_i}^{v_f} mv \ dv \end{aligned}

Unfortunately, I'm having trouble understanding how the author arrived at the final step from the previous one.

I have tried equating $$v = dx/dt$$ and using substitution rule with the same variable $$v = x$$ and $$dv = dx$$ to arrive at

$$W_{\text{ext}}= \int_{v_i}^{v_f} mv \frac{dv}{dv} \ dv$$

and then cancelling the $$dv$$'s as if they were a fraction to get the same result as the book but I think I'm terribly mistaken.

Can someone perhaps shed insight on how to proceed with this problem?

• Think of it this way: if $f(v) = v$, what is $df/dv$? Commented Jun 16, 2022 at 14:55
• @MichaelSeifert is it $df/dv = 1$?
– Niko
Commented Jun 16, 2022 at 15:28
• Let v=x? Surely that is not a valid thing to do. You cant say v=x when v is defined as dx/dt.... Commented Jun 16, 2022 at 15:52
• You want to solidify your understanding that, $$\frac{dv}{dx} dx = dv$$ the rate at which dv changes for a tiny increase dx, (dv per unit dx)multiplied by dx, IS the tiny increase in v, aka dv Commented Jun 16, 2022 at 15:57
• @jensenpaull while I do recognize that, wouldn't that mean treating derivatives as fractions? I'm a little skeptical since I've heard that they shouldn't be treated as such but my understanding as to why that is is quite shaky, unfortunately.
– Niko
Commented Jun 16, 2022 at 16:06

A better way if you're not comfortable with the method shown:

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

$$dx = v dt$$

$$W = \int_{a}^{b} F dx$$ $$\int_{t_{0}}^{t} [m \frac{dv}{dt}] [v dt]$$

Rewriting:

$$m \int_{t_{0}}^{t} [v \frac{dv}{dt}] dt$$

Instead of "cancelling" dt, simply use the inverse chain rule to integrate this expression with respect to time directly, (raise by the power, divide by that power, divide by the derivative of the inside function!)

or in reverse : $$\frac{d}{dt} \frac{1}{2}v^2 = v \frac{dv}{dt}$$ $$W = m [\frac{1}{2} v(t)^2 - \frac{1}{2} v(t_{0})^2]$$

• A great alternative derivation, though I think the root of my problem is that why were the derivatives originally treated like fractions (i.e. can be "cancelled out") if it is said that they aren't supposed to be treated as such?
– Niko
Commented Jun 16, 2022 at 16:22
• @Niko as I recall, the highest-voted question on Math.SE is whether/when dy/dx can be treated as a fraction. You may want to check it out. Commented Jun 18, 2022 at 1:54

\begin{aligned} W_{\text{ext}} &= \int_{x_i}^{x_f} \sum F \ dx \\ &= \int_{x_i}^{x_f} ma \ dx \\ &= \int_{x_i}^{x_f} m \frac{dv}{dt} \ dx \end{aligned} From this point, we note that $$x = x(t)$$, so: $$\int_{x_i}^{x_f} m \frac{dv}{dt} \ dx = \int_{t_i}^{t_f} m \frac{dv}{dt} \ \frac{dx}{dt} \ dt = \int_{t_i}^{t_f} m \frac{dv}{dt} \ v \ dt$$

Here we can use integration by parts, where both functions are $$v$$: $$\int_{t_i}^{t_f} m v \frac{dv}{dt} \ dt = m(vv)_{t_i}^{t_f} - \int_{t_i}^{t_f} m v\frac{dv}{dt} \ dt$$ Because the left and right integral are the same: $$\int_{t_i}^{t_f} m v \frac{dv}{dt} \ dt = \Delta \left(\frac{1}{2}m v^2 \right)$$

From the second step, $$ma.dx = m.\frac{dv}{dt}.dx = m.dv.\frac{dx}{dt}$$ And with $$\frac{dx}{dt} = v$$, the integrand becomes $$mv.dv$$, as written in the last step.

Your substitution $$v = x$$ does not make sense as $$v$$ is defined to be $$\frac{dx}{dt} ≠ x$$, which is why it does not work.

Hope this helps.

• I believe my main issue here is that doing so would mean that derivatives are like fractions, which I've heard are not. Surely treating them like fractions that can be cancelled out makes the solution simpler but I'm just hesitating to go down that path.
– Niko
Commented Jun 16, 2022 at 16:11
• The chain rule relies explicitly on treating derivatives as fractions, which you seem to be comfortable using. Check the answers out on this MSE question on when it's okay to treat derivatives as fractions : math.stackexchange.com/questions/1784671/… Commented Jun 16, 2022 at 16:46

You have overcomplicated your derivation by trying to derive within the integral. Start with the differential forms first then integrate at the end

$$W = Fdx$$ $$W = m \frac{dv}{dt} dx$$ $$W = m \frac {dx}{dt} dv$$ $$W = mvdv$$

We start with the definition of work: $$W = \int_{x_{i}}^{x_{f}} F\cdot dx$$ From Newton's second law: $$W = \int_{x_{i}}^{x_{f}} m\frac{dv}{dt}\cdot dx$$ $$W = \int_{t_{i}}^{t_{f}} m\frac{dv}{dt}\cdot vdt$$ $$W = \underbrace{\int_{t_{i}}^{t_{f}}\left( m\frac{dv}{dt}\cdot v\right)dt}_{I}$$

We use integration by parts: $$\int f'g=fg-\int fg'$$

$$W = m v^2\biggr|_{t_{i}}^{t_{f}} - \underbrace{\int_{t_{i}}^{t_{f}}\left( mv \cdot\frac{dv}{dt}\right)dt}_{I}$$

From this we get that: $$W=\frac{1}{2}m\left[v\left(t_{f}\right)^2-v\left(t_{i}\right)^2\right]$$

$$$$a=\frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}$$$$ via chain rule, therefore $$$$a=\frac{dv}{dx}v=v\frac{dv}{dx} \ \xrightarrow{} \ a\,dx= v\, dv$$$$ Now you should be able to understand.

• I'm afraid I still have trouble understanding how you got to the last step. Wouldn't it be $a \ dx = v \ dv$?
– Niko
Commented Jun 16, 2022 at 15:32
• Yes and I apologize,that was a typographical error. Thanks for noticing. Commented Jun 16, 2022 at 15:37