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Velocity and Differentiability
Definition The vector function $r(t) = \left\langle x(t), y(t), z(t) \right\rangle$ has a derivative (is differentiable) at $t$ if $x$, $y$, and $z$ have derivatives at $t$. The derivative is the vector function: $$r\prime (t) = \frac{dr}{dt} = \lim_{\Delta t\to 0} \frac{r(t + \Delta t) - r(t)}{\Delta t} = \left\langle x\prime (t), y\prime (t), z\prime (t) \right\rangle$$ For a differentiable vector valued function $r(t) = \left\langle x(t), y(t), z(t) \right\rangle$ representing the position vector of a particle at time $t$, its velocity vector is given as: $$ v(t) = \frac{dr}{dt} = r\prime (t) = \left\langle x\prime (t), y\prime (t), z\prime (t) \right\rangle$$ Bird's Eye View We will be discussing a very specific case of differentiation through the course of these notes: with respect to space curves (which, as we already know, are the set of points $(x(t), y(t), z(t))$, obtained as $t$ varies over a given parameter interval), to analyse how and when we can (and cannot!) differentiate on a given space curve. Context of the Definition Formally, the domain of $f$ is the set of points in the domain of $f$ for which the limit exists, and the domain may be the same or smaller than the domain of $f$. If $f$ exists at a particular $t$, we say that $f$ is differentiable (has a derivative) at $t$. If $f$ exists at every point in the domain of $f$, we call $f$ differentiable. Mathematically, $$r\prime (t) = \frac{dr}{dt} = \lim_{\Delta t\to 0} \frac{r(t + \Delta t) - r(t)}{\Delta t} = \left\langle x\prime (t), y\prime (t), z\prime (t) \right\rangle$$ As we have studied in Equations of Planes and Lines, we know that the slope of a line in $R^{3}$ is given by a direction, rather than a scalar value. The direction vector given by $r\prime (t)$ represents the slope of the curve at a given $t$. 0
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Figure 1: Interpretation of $r\prime (t)$
There are multiple interpretations of the derivative: 1. Rate of Change If $f(x)$ represents a quantity at any $x$ then the derivative $f\prime (a)$represents the instantaneous rate of change of $f(x)$ at $x = a$. Illustration: Suppose that the volume of water in a tank for $0\leq t \leq 6 $ is given by $Q(t)=10+5t−t^{2}$ Is the volume of water increasing or decreasing at $t=0$ ? Is the volume of water increasing or decreasing at $t=6$ ? Does the volume of water ever stop changing? If yes, at what times(s) does the volume stop changing? Hint: From Figure 1, we can infer that if the derivative of a function at a given point is positive, the function is increasing. 2. Slope of a Tangent Line The slope of the tangent line to $f(x)$ at $x=a$ is $f\prime (a)$. The tangent line then is given by, $$y = f(a) + f\prime (a)(x - a)$$ Illustration: What is the equation of the tangent line to $f(x)= \frac{5}{x}$ at $x=\frac{1}{2}$ ? 3. Velocity This is a special case of the rate of change interpretation. If the position of an object is given by $r(t)$ after t units of time the velocity of the object at $t=a$ is given by $r\prime (a)$. Illustration: The position of an object at any time $t$ is given by $s(t) = \frac{t + 1}{t + 4}$ Determine the velocity of the object at any time $t$. Does the object ever stop moving? If yes, at what time(s) does the object stop moving? Hint: From Figure 1, we can infer that if the derivative of a function at a given point is zero, the function neither increases nor decreases. Can velocity be obtained for any space curve? A mathematical version of the above question would be: " For a given curve $r(t)$, can the curve be differentiated for any and every value of $t$ in the parameter interval?". This property of the curve is called differentiability. A function can fail to be differentiable at a point if: The function is not continuous at that point. The graph has a sharp corner at the point. The graph has a vertical line at the point. The animation given below depicts all three cases. 0
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Figure 2: Non-differentiable curves
Smooth Curves A curve $x = f(t), y = g(t)$ is called a smooth curve if $f\prime (t), g\prime(t)$ are continuous and for no value of $t$ are $f\prime (t)$ and $g\prime (t)$ simultaneously equal to zero. The number of continuous derivatives of the position function for a curve to be considered smooth depends on the problem at hand and may vary from two to infinity. To understand this better, consider the following example: $$r(t) = \left\langle t^{2}e^{-t}, 2(t - 1)^{2}\right\rangle$$ Computing $r\prime (t)$, we get: $r\prime (t) = \left\langle 2te^{-t} - t^{2}e^{-t}, 4(t - 1)\right\rangle$ $r_{x}\prime = 0 \implies t = 0, 2$ $r_{y}\prime = 0 \implies t = 1$ Hence, since for a given value of $t$, $r_{x}\prime$ and $r_{y}\prime$ are not simultaneously equal to zero, i.e. $r(t) \neq \overrightarrow{0}$, the given curve $r(t)$ is a smooth curve. Visually, if a curve on zooming in infinitely, looks like a straight line, it is a smooth curve. Zooming in on $\left \| x \right \|$ at $x = 0$, will always show the sharp corner: 5
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Figure 3: Smooth curves \| source: Eric Duminil (https://math.stackexchange.com/users/386794/eric-duminil), Intuitively, why are the curves of exponential, log, and parabolic functions all smooth, even though the gradient is being changed at every point?, URL (version: 2018-05-23): https://math.stackexchange.com/q/2791202
Tangent Lines to Space Curves If $r(t) = \left\langle f(t), g(t), h(t) \right\rangle$ is a position vector along a curve in $\mathbb{R}^{3}$, then $r\prime (t) = \left\langle f\prime (t), g\prime (t), h\prime (t)\right\rangle$ is a vector in the direction of the tangent line to the curve. Would this mean the velocity vector of a point on a curve is always tangential to the curve? Something to think about. (or not!) Note that it is required that $r(t)\neq\overrightarrow{0}$ to have a tangent vector, as $r(t) = \overrightarrow{0}$ implies that the vector has no magnitude and hence would not give the direction of the tangent. The video below depicts a tangent line to the given space curve at the given point. 7
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Figure 4: Tangent line to a space curve
Speed and Acceleration As seen above, the first derivative of the position vector $r(t)$ is called velocity, often denoted by $v(t) = r\prime (t)$. Speed is defined as the length of velocity, i.e. $\left\|\left\| r\prime (t)\right\|\right\|$ and is a non-negative scalar. While studying motion of objects, the total acceleration is often broken up as a tangential component, $a_{T}$ and a normal component, $a_{N}$. The tangential component is that part of the acceleration which is tangential to the curve and the normal component is the part of the acceleration normal to the curve. $$a_{T} = v\prime (t) = \frac{r\prime(t)\cdot r\prime\prime(t)}{\left\|\left\|r\prime(t)\right\|\right\|}$$ $$a_{N} = k\hspace{0.2cm}\left\|\left\|r\prime (t)\right\|\right\|^{2} = \frac{\left\|\left\|r\prime(t)\times r\prime\prime(t)\right\|\right\|}{\left\|\left\|r\prime(t)\right\|\right\|}$$ where $k$ is the curvature of $r(t)$. The total acceleration is given by: $$a = a_{T}\overrightarrow{T} + a_{N}\overrightarrow{N}$$ where $\overrightarrow{T}$ and $\overrightarrow{N}$ are the unit tangent and unit normal vectors for $r(t)$ respectively. References [1] https://www.ugrad.math.ubc.ca/coursedoc/math100/notes/zoo/continuous.html [2] https://ximera.osu.edu/mklynn2/multivariable/content/02_2_velocity_and_speed/velocity_and_speed [3] https://opentextbc.ca/calculusv1openstax/chapter/the-derivative-as-a-function/ 9

Velocity and Differentiability

Definition

The vector function $r(t) = \left\langle x(t), y(t), z(t) \right\rangle$ has a derivative (is differentiable) at $t$ if $x$, $y$, and $z$ have derivatives at $t$. The derivative is the vector function: $$r\prime (t) = \frac{dr}{dt} = \lim_{\Delta t\to 0} \frac{r(t + \Delta t) - r(t)}{\Delta t} = \left\langle x\prime (t), y\prime (t), z\prime (t) \right\rangle$$

For a differentiable vector valued function $r(t) = \left\langle x(t), y(t), z(t) \right\rangle$ representing the position vector of a particle at time $t$, its velocity vector is given as: $$ v(t) = \frac{dr}{dt} = r\prime (t) = \left\langle x\prime (t), y\prime (t), z\prime (t) \right\rangle$$

Bird's Eye View

We will be discussing a very specific case of differentiation through the course of these notes: with respect to space curves (which, as we already know, are the set of points $(x(t), y(t), z(t))$, obtained as $t$ varies over a given parameter interval), to analyse how and when we can (and cannot!) differentiate on a given space curve.

Context of the Definition

Formally, the domain of $f$ is the set of points in the domain of $f$ for which the limit exists, and the domain may be the same or smaller than the domain of $f$. If $f$ exists at a particular $t$, we say that $f$ is differentiable (has a derivative) at $t$. If $f$ exists at every point in the domain of $f$, we call $f$ differentiable. Mathematically, $$r\prime (t) = \frac{dr}{dt} = \lim_{\Delta t\to 0} \frac{r(t + \Delta t) - r(t)}{\Delta t} = \left\langle x\prime (t), y\prime (t), z\prime (t) \right\rangle$$

As we have studied in Equations of Planes and Lines, we know that the slope of a line in $R^{3}$ is given by a direction, rather than a scalar value. The direction vector given by $r\prime (t)$ represents the slope of the curve at a given $t$.

Figure 1: Interpretation of $r\prime (t)$

There are multiple interpretations of the derivative:

1. Rate of Change

If $f(x)$ represents a quantity at any $x$ then the derivative $f\prime (a)$represents the instantaneous rate of change of $f(x)$ at $x = a$.

Illustration:
Suppose that the volume of water in a tank for $0\leq t \leq 6 $ is given by $Q(t)=10+5t−t^{2}$

Is the volume of water increasing or decreasing at $t=0$ ?
Is the volume of water increasing or decreasing at $t=6$ ?
Does the volume of water ever stop changing? If yes, at what times(s) does the volume stop changing?

Hint: From Figure 1, we can infer that if the derivative of a function at a given point is positive, the function is increasing.

2. Slope of a Tangent Line

The slope of the tangent line to $f(x)$ at $x=a$ is $f\prime (a)$. The tangent line then is given by, $$y = f(a) + f\prime (a)(x - a)$$

Illustration:
What is the equation of the tangent line to $f(x)= \frac{5}{x}$ at $x=\frac{1}{2}$ ?

3. Velocity

This is a special case of the rate of change interpretation. If the position of an object is given by $r(t)$ after t units of time the velocity of the object at $t=a$ is given by $r\prime (a)$.

Illustration:
The position of an object at any time $t$ is given by $s(t) = \frac{t + 1}{t + 4}$

Determine the velocity of the object at any time $t$.
Does the object ever stop moving? If yes, at what time(s) does the object stop moving?

Hint: From Figure 1, we can infer that if the derivative of a function at a given point is zero, the function neither increases nor decreases.

Can velocity be obtained for any space curve?

A mathematical version of the above question would be: " For a given curve $r(t)$, can the curve be differentiated for any and every value of $t$ in the parameter interval?". This property of the curve is called differentiability. A function can fail to be differentiable at a point if:

The function is not continuous at that point.
The graph has a sharp corner at the point.
The graph has a vertical line at the point.

The animation given below depicts all three cases.

Figure 2: Non-differentiable curves

Smooth Curves

A curve $x = f(t), y = g(t)$ is called a smooth curve if $f\prime (t), g\prime(t)$ are continuous and for no value of $t$ are $f\prime (t)$ and $g\prime (t)$ simultaneously equal to zero.

The number of continuous derivatives of the position function for a curve to be considered smooth depends on the problem at hand and may vary from two to infinity.

To understand this better, consider the following example: $$r(t) = \left\langle t^{2}e^{-t}, 2(t - 1)^{2}\right\rangle$$

Computing $r\prime (t)$, we get:

$r\prime (t) = \left\langle 2te^{-t} - t^{2}e^{-t}, 4(t - 1)\right\rangle$

$r_{x}\prime = 0 \implies t = 0, 2$

$r_{y}\prime = 0 \implies t = 1$

Hence, since for a given value of $t$, $r_{x}\prime$ and $r_{y}\prime$ are not simultaneously equal to zero, i.e. $r(t) \neq \overrightarrow{0}$, the given curve $r(t)$ is a smooth curve.

Visually, if a curve on zooming in infinitely, looks like a straight line, it is a smooth curve. Zooming in on $\left | x \right |$ at $x = 0$, will always show the sharp corner:

Figure 3: Smooth curves | source: Eric Duminil (https://math.stackexchange.com/users/386794/eric-duminil), Intuitively, why are the curves of exponential, log, and parabolic functions all smooth, even though the gradient is being changed at every point?, URL (version: 2018-05-23): https://math.stackexchange.com/q/2791202

Tangent Lines to Space Curves

If $r(t) = \left\langle f(t), g(t), h(t) \right\rangle$ is a position vector along a curve in $\mathbb{R}^{3}$, then $r\prime (t) = \left\langle f\prime (t), g\prime (t), h\prime (t)\right\rangle$ is a vector in the direction of the tangent line to the curve.

Would this mean the velocity vector of a point on a curve is always tangential to the curve? Something to think about. (or not!)

Note that it is required that $r(t)\neq\overrightarrow{0}$ to have a tangent vector, as $r(t) = \overrightarrow{0}$ implies that the vector has no magnitude and hence would not give the direction of the tangent.

The video below depicts a tangent line to the given space curve at the given point.

Figure 4: Tangent line to a space curve

Speed and Acceleration

As seen above, the first derivative of the position vector $r(t)$ is called velocity, often denoted by $v(t) = r\prime (t)$. Speed is defined as the length of velocity, i.e. $\left|\left| r\prime (t)\right|\right|$ and is a non-negative scalar.

While studying motion of objects, the total acceleration is often broken up as a tangential component, $a_{T}$ and a normal component, $a_{N}$. The tangential component is that part of the acceleration which is tangential to the curve and the normal component is the part of the acceleration normal to the curve.

$$a_{T} = v\prime (t) = \frac{r\prime(t)\cdot r\prime\prime(t)}{\left|\left|r\prime(t)\right|\right|}$$

$$a_{N} = k\hspace{0.2cm}\left|\left|r\prime (t)\right|\right|^{2} = \frac{\left|\left|r\prime(t)\times r\prime\prime(t)\right|\right|}{\left|\left|r\prime(t)\right|\right|}$$

where $k$ is the curvature of $r(t)$.

The total acceleration is given by: $$a = a_{T}\overrightarrow{T} + a_{N}\overrightarrow{N}$$ where $\overrightarrow{T}$ and $\overrightarrow{N}$ are the unit tangent and unit normal vectors for $r(t)$ respectively.

References

[1] https://www.ugrad.math.ubc.ca/coursedoc/math100/notes/zoo/continuous.html

[2] https://ximera.osu.edu/mklynn2/multivariable/content/02_2_velocity_and_speed/velocity_and_speed

[3] https://opentextbc.ca/calculusv1openstax/chapter/the-derivative-as-a-function/

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