book:chap3:3.2_time_derivatives_of

3.2 Time derivatives of vectors

In this section we consider the motion of a particle with mass $m$ that is at position $\mathbf q(t)$ at time $t$. Its average velocity ${\mathbf v}_{\text{av}}(t, \Delta t)$ during the time interval $[t, t+\Delta t]$ is \begin{align*} {\mathbf v}_{\text{av}} (t, \Delta t) = \frac{ \mathbf q(t + \Delta t) - \mathbf q(t) }{\Delta t} \end{align*} When the limit $\lim_{\Delta t \to 0} \mathbf v_\text{av}(t, \Delta t)$ exists ¹⁾ we can define the velocity of the particle at time $t$, \begin{align} \tag{3.2.1} \mathbf v (t) = \lim_{\Delta t \to 0} \frac{ \mathbf q(t + \Delta t) - \mathbf q(t) }{\Delta t} \end{align} The velocity is then the time derivative of the position, and in an immediate generalization of the time derivative of scalar functions we also write \begin{align*} \dot{\mathbf q} (t) = \mathbf v (t) = \frac{\mathrm{d} \mathbf q (t) }{\mathrm{d} t} \end{align*} Finally, we point out that the components of the time derivative of a vector amount to the derivatives of the components.

Theorem 3.1 Time derivatives of vectors

Let $\mathbf a(t)$ be a vector with time-dependent components $a_i(t)$ with respect to orthonormal basis $\{ \hat{\boldsymbol e}_i, \; i = 1 \cdots D \}$ that is fixed in time. Then $\dot{\mathbf a}(t) = \sum_i \dot a_i(t) \, \hat{\boldsymbol e}_i$. The components of $\dot{\mathbf a}(t)$ amount to the time derivatives of the components of $\mathbf a(t)$.

Proof. For each time we have ${\mathbf a}(t) = \sum_i a_i(t) \, \hat{\boldsymbol e}_i$ where it is understood that the sum runs over $i=1\cdots D$. We insert this into the definition, Definition 3.2.1, of the the time derivative and use the linearity of scalar products with vectors to obtain \begin{align*} \dot{\mathbf a}(t) &= \lim_{\Delta t \to 0} \frac{ \mathbf a(t + \Delta t) - \mathbf a(t) }{\Delta t} = \lim_{\Delta t \to 0} \frac{ \sum_i a_i(t + \Delta t) \: \hat{\boldsymbol e}_i - \sum_i a_i(t) \: \hat{\boldsymbol e}_i}{\Delta t} \\ &= \lim_{\Delta t \to 0} \sum_i \hat{\boldsymbol e}_i \frac{ a_i(t + \Delta t) - a_i(t) }{\Delta t} = \sum_i \hat{\boldsymbol e}_i \; \lim_{\Delta t \to 0} \frac{ a_i(t + \Delta t) - a_i(t) }{\Delta t} \\ &= \sum_i \hat{\boldsymbol e}_i \: \dot a_i(t) \end{align*} The subtle step here, from a mathematical point of view, is the swapping of the limit and the sum in the second line of the argument. Courses on vector calculus will spell out the assumptions needed to justify this step (or, more interestingly from a physics perspective, under which conditions it fails). qed

The change of the velocity will be denoted as acceleration. Based on an analogous argument as for the velocity, it will be written as a time derivative

Definition 3.1 Acceleration

The time derivative of the velocity $\mathbf v(t) = \dot{\mathbf q}(t)$ is denoted as acceleration, and written as \begin{align*} \frac{\mathrm{d} \mathbf v(t)}{\mathrm{d} t} = \dot{\mathbf v}(t) = \ddot{\mathbf q}(t) \end{align*}

In the next section it will be related to the action of forces $\mathbf F ( \mathbf q, t )$ acting on a particle that resides at the position $\mathbf q$ at time $t$.

3.2.1 Self Test

Problem 3.1: Derivatives of elementary functions
Recall that \begin{align*} \frac{\mathrm{d}}{\mathrm{d} x} \sin x &= \cos x & \frac{\mathrm{d}}{\mathrm{d} x} \mathrm{e}^x &= \mathrm{e}^x & \frac{\mathrm{d}}{\mathrm{d} x} \ln x &= x^{-1} \end{align*} and use only the three rules for derivatives \begin{align*} \frac{\mathrm{d}}{\mathrm{d} x} \bigl( f(x) + g(x) \bigr) &= f'(x) + g'(x) \\ \frac{\mathrm{d}}{\mathrm{d} x} \bigl( f(x) g(x) \bigr) &= f'(x) \: g(x) + f(x) \: g'(x) \\ \frac{\mathrm{d}}{\mathrm{d} x} f\bigl( g(x) \bigr) &= g'(x) \: f'\bigl( g(x) \bigr) \end{align*} to work out the following derivatives

a) $\displaystyle \sinh x = \frac{1}{2} \left( \mathrm{e}^x - \mathrm{e}^{-x} \right) \quad$ and $\quad \displaystyle \cosh x = \frac{1}{2} \left( \mathrm{e}^x + \mathrm{e}^{-x} \right)$

b) $\cos x = \sin(\pi/2 + x) \qquad$

c) $x^a = \mathrm{e}^{a \, \ln x}$ for $a \in \mathbb{R}$.
$\quad$ What does this imply for the derivative of $f(x) = x^{-1}$?

d) Use the result from c) to proof the quotient rule: \begin{align*} \frac{\mathrm{d}}{\mathrm{d} x} \frac{ f(x) }{ g(x) } = \frac{ f'(x) \, g(x) - f(x) \, g'(x) }{ \bigl( g(x) \bigr)^2 } \end{align*}
e) $\displaystyle \tan x = \frac{ \sin x }{ \cos x } \quad $ and $\quad \displaystyle \tanh x = \frac{ \sinh x }{ \cosh x }$
f) Find the derivative of $\ln x$ solely based on $\displaystyle \frac{\mathrm{d}}{\mathrm{d} x} \mathrm{e}^x = \mathrm{e}^x$.
$\quad$ Hint: Use that $x = \mathrm{e}^{\ln x}$ and take the derivative of both sides.

Problem 3.2: Integrals of elementary functions
In a moment we will also perform integrals to determine the work performed on a body when it is moving subject to a force. Practice your skills by evaluating the following integrals.

a) $\displaystyle \int_{-1}^1 {\mathrm{d} x}\, ( a + x )^2$
b) $\displaystyle \int_{-5}^5 {\mathrm{d} q}\, ( a + b\, q^3 )$
c) $\displaystyle \int_0^B {\mathrm{d} k}\, \tanh^2( k x )$
d) $\displaystyle \int_0^\infty {\mathrm{d} x}\, \mathrm{e}^{-x/L}$
e) $\displaystyle \int_{-L}^L {\mathrm{d} y}\, \mathrm{e}^{-y/\xi}$

f) $\displaystyle \int_{0}^L {\mathrm{d} z}\, \frac{z}{a+b\,z^2}$
g) $\displaystyle \int_{0}^\infty {\mathrm{d} x} \, x \, \mathrm{e}^{-x^2/(2Dt)}$
h) $\displaystyle \int_{-\sqrt{Dt}}^{\sqrt{Dt}} {\mathrm{d} \ell} \, \ell \, \mathrm{e}^{-\ell^2/(2Dt)}$
i) $\displaystyle \int_{-\sqrt{Dt}}^{\sqrt{Dt}} {\mathrm{d} z}\, x \, \mathrm{e}^{-z\, x^2}$

Except for the integration variable all quantities are considered to be constant.
Hint: $\quad$ Sometimes symmetries can substantially reduce the work needed to evaluate an integral.

¹⁾

The discussion of this limit for general functions is a core topic of vector calculus. For our present purpose the intuitive understanding based on the idea that $ \mathbf q (t + \Delta t) \simeq \mathbf q (t) + \Delta t \; \mathbf v (t) $ provides the right idea. To provide a hint for the origin of the mathematical subtleties we point out that the approximation works unless there is an instantaneous collision with a wall at some point in the time interval $[t, t+\Delta t]$. In physics we try our luck, and fix the problem when we face it. Indeed, upon a close look there are no instantaneous collisions in physics, see Problem 3.17.