If $v=dx/dt$ what is $v^2$ equal to?

Question

Just working on mechanics problems right now and I realized that I never actually dealt with a problem like this before.

Usually when I'm solving a differential equation, I would often be able to convert variable $v$ to $dx/dt$. However, this question requires me to convert $v$ into a form related to $dx/dt$.

Answer 1

Well... like that? $$(v = \frac{\mathrm{d}S }{\mathrm{d} t}) \xrightarrow{f: x \mapsto x^2} v^2 = \frac{\mathrm{d}S }{\mathrm{d} t} \frac{\mathrm{d}S }{\mathrm{d} t} \Leftrightarrow v^2 = \frac{\mathrm{d}S }{\mathrm{d} t^2} \mathrm{d}S $$

which give us $$v^2 = a * \mathrm{d}S$$

I'm probably wrong, but I though I could give it a try.

Answer 2

If you were asking for $(\frac{d}{dt})^2$, then you were looking at the derivation operator, that is the function which its domain is the set of differentiable functions with respect to $t$, and its codomain is also a set of functions. $$\begin{array}{lll} \frac{d}{dt}\colon & \lbrace\text{functions in }t\text{ which are differentiable}\rbrace & \rightarrow \lbrace\text{functions in }t\rbrace\\ & f(t) &\mapsto \frac{d f(t)}{dt} \end{array}$$ Now assuming that by multiplication, you mean multiplication of operators, which is just the "function composition", then $(\frac{d}{dt})^2$ means $\frac{d}{dt}\circ\frac{d}{dt}=\frac{d}{dt}(\frac{d}{dt})$. Applying it on a function, say $f(t)$ will give you this; $$(\frac{d}{dt})^2\big(f(t)\big)=\frac{d(\frac{d f(t)}{dt})}{dt}=\frac{d^2 f(t)}{dt^2}$$ Which is the second derivative of $f(t)$ with respect to $t$. But you wrote $(\frac{dx}{dt})^2$ which means you are looking at $\frac{dx}{dt}$, this has the derivation already done and you are looking at the result of the derivation. So let's say $x$ was a function on $t$, then $\frac{dx}{dt}$ is a function, not an operator. Therefore, $(\frac{dx}{dt})^2$ is just square of the obtained function.