Vector Problems
More on Curvature and Arc Length

A function of type \boldsymbol{r}\left (t\right ) =\left \langle x\left (t\right ) ,y\left (t\right ) ,z\left (t\right )\right \rangle defined on a set of real numbers (with vector outputs) is said to be a vector-valued function. For example \boldsymbol{r}\left (t\right ) =\left \langle t ,t^{2} ,\log t\right \rangle is a vector-valued function.

Example 1.

Consider the function \boldsymbol{r}\left (s\right ) =\left \langle s +1 ,s ,1 -s\right \rangle. This function can be written as

(1)   \begin{equation*}x\left (s\right ) =1 +s ,y\left (s\right ) =s ,z\left (s\right ) =1 -s\end{equation*}

which represents a parametric equation of a line. Any vector output of this function lands on a point on the above line.

Example 2.

Consider \boldsymbol{r}\left (s\right ) =\left \langle \cos s ,\sin s\right \rangle. This equation can be written as x\left (s\right ) =\cos s ,y\left (s\right ) =\sin s. Using the fact that

(2)   \begin{equation*}x^{2} +y^{2} =\sin ^{2}s +\cos ^{2}s =1\end{equation*}

we see that the function \boldsymbol{r} provides the set of all points on the unit circle that is the terminal points of all vectors with initial point at the origin. The picture below shows that when s =\frac{3\pi }{4} the function \boldsymbol{r} provides the vector \left \langle -\frac{1}{\sqrt{2}} ,\frac{1}{\sqrt{2}}\right \rangle.

Example 3.

Consider \boldsymbol{r}\left (s\right ) =\left \langle \cos s ,\sin s ,s\right \rangle . This equation can be written as x\left (s\right ) =\cos s ,y\left (s\right ) =\sin s ,z\left (s\right ) =s . From previous example, we saw that in xy-plane, x^{2} +y^{2} =1 provides the unit circle and for this example, for any given z-value (or s-value), we obtain a circle. This will give rise to a cylinder which is upward parallel to the z-axis. Now, as z =s , we know that we are looking at exactly one point on each of these circles cut from the cylinder. For example, when s =\pi , the vector provided by \boldsymbol{r} is

(3)   \begin{equation*}\boldsymbol{r}\left (\pi \right ) =\left \langle \cos \pi ,\sin \pi ,\pi \right \rangle =\left \langle -1 , 0 ,\pi \right \rangle \end{equation*}

This identifies a vector with initial point at the origin and terminal point on the unit circle on the plane z =\pi and coordinate (-1 , 0) . This way we get a curve on the body of the cylinder x^{2} +y^{2} =1. The direction of travel as s increases will be in the direction of travel on the unit circle that is counterclockwise (obviously upward since s increases!)

Observation 1.

The concepts of limit, differentiation and integration are straight forward and are done component-wise with respect to vector-valued functions. Consider a vector-valued function as below \boldsymbol{r}\left (t\right ) . Take two points p and q on the curve and use \boldsymbol{r}\left (t\right ) and \boldsymbol{r}\left (t +h\right ) to show them. Of course a difference vector, shown in yellow, like \frac{\boldsymbol{r}(t +h) -\boldsymbol{r}\left (t\right )}{h} passes the points p and q. Now if one chooses h small, the difference vector which passes p and q becomes closer to the tangent vector (in the sense of direction shown in red) at p .

This may not always be the case; for example, a tangent vector may not exist but when it does and it’s a nonzero vector, we have an intuition how \boldsymbol{r}^{ \prime }\left (t\right ) is defined. In fact, it’s easy to show that the differentiation for a vector valued function is done component-wise. In application, the vector \boldsymbol{r} is called the position vector, \boldsymbol{r}^{ \prime } is called the velocity vector and \boldsymbol{r}^{ \prime \prime } is called the acceleration vector. This makes sense, since r shows the position of an article based on time t on the curve (analogous interpretation for \boldsymbol{r}^{ \prime } and \boldsymbol{r}^{ \prime \prime }). Speed of the particle is obtained by evaluating the magnitude of \boldsymbol{r}^{ \prime } at any time.

Example 4.

Consider two vector-valued functions \boldsymbol{r}_{1}\left (t\right ) = \langle x_{1} ,y_{1}\left (t\right ) ,z_{1}\left (t\right ) \rangle and \boldsymbol{r}_{2}\left (t\right ) = \langle x_{2}(t),y_{2}\left (t\right ) ,z_{2}\left (t\right ) \rangle. We would like to find the dot product of the two vectors that is

(4)   \begin{align*}\boldsymbol{r}_{1} \cdot \boldsymbol{r}_{2} =\left \langle x_{1} ,y_{1} ,z_{1}\right \rangle \cdot \langle x_{2} ,y_{2} ,z_{2} \\=x_{1}x_{2} +y_{1}y_{2} +z_{1}z_{2}\end{align*}

Next we compute the derivative of \boldsymbol{r}_{1} \cdot \boldsymbol{r}_{2} and note that this is a function depending on a variable t:

(5)   \begin{align*}\frac{d}{dt}(\boldsymbol{r}_{1} \cdot \boldsymbol{r}_{2}) =\frac{d}{dx}(x_{1}x_{2} +y_{1}y_{2} +z_{1}z_{2}) \\ =\frac{d}{dx}\left (x_{1}x_{2}\right ) +\frac{d}{dx}\left (y_{1}y_{2}\right ) +\frac{d}{dx}\left (z_{1}z_{2}\right ) \\ =\left (x_{1}^{ \prime }x_{2} +x_{1}x_{2}^{ \prime }\right ) +\left (y_{1}^{ \prime }y_{2} +y_{1}y_{2}^{ \prime }\right ) +\left (z_{1}^{ \prime }z_{2} +z_{1}z_{2}^{ \prime }\right ) \\ =\left (x_{1}^{ \prime }x_{2} +y_{1}^{ \prime }y_{2} +z_{1}^{ \prime }z_{2}\right ) +\left (x_{1}x_{2}^{ \prime } +y_{1}y_{2}^{ \prime } +z_{1}z_{2}^{ \prime }\right ) \\ =\boldsymbol{r}_{1}^{ \prime }\cdot\boldsymbol{r}_{2} +\boldsymbol{r}_{1}\cdot\boldsymbol{r}_{2}^{ \prime }\end{align*}

Example 5.

Given the vectors \boldsymbol{r}_{1} and \boldsymbol{r}_{2} as the above example, we would like to find \boldsymbol{r}_{1} \times \boldsymbol{r}_{2}

(6)   \begin{align*}\boldsymbol{r}_{1} \times \boldsymbol{r}_{2} =(y_{1}z_{2} -y_{2}z_{1})\boldsymbol{i} -(x_{1}z_{2} -x_{2}z_{1})\boldsymbol{j} +(x_{1}y_{2} -x_{2}y_{1})\boldsymbol{k} \\ \,\end{align*}

Next, we like to find the derivative of this cross product so we differentiate each component

(7)   \begin{align*}\frac{d}{dt}(\boldsymbol{r}_{1} \times \boldsymbol{r}_{2}) =(y_{1}^{ \prime }z_{2} +y_{1}z_{2}^{ \prime } -y_{2}^{ \prime }z_{1} -y_{2}z_{1}^{ \prime })\boldsymbol{i} -(x_{1}^{ \prime }z_{2} +x_{1}z_{2}^{ \prime } -x_{2}^{ \prime }z_{1} -x_{2}z_{1}^{ \prime })\boldsymbol{j} +(x_{1}^{ \prime }y_{2} +x_{1}y_{2}^{ \prime } -x_{2}^{ \prime }z_{1} -x_{2}z_{1}^{ \prime })\boldsymbol{k} \\ =\left (y_{1}^{ \prime }z_{2} -y_{2}z_{1}^{ \prime }\right )\boldsymbol{i} +\left (y_{1}z_{2}^{ \prime } -y_{2}^{ \prime }z_{1}\right )\boldsymbol{i} -\left (x_{1}^{ \prime }z_{2} -x_{2}z_{1}^{ \prime }\right )\boldsymbol{j} -\left (x_{1}z_{2}^{ \prime } -x_{2}^{ \prime }z_{1}\right )\boldsymbol{j} +\left (x_{1}^{ \prime }y_{2} -x_{2}z_{1}^{ \prime }\right )\boldsymbol{k} +\left (x_{1}y_{2}^{ \prime } -x_{2}^{ \prime }z_{1}\right )\boldsymbol{k} \\ =\left (\boldsymbol{r}_{1}^{ \prime } \times \boldsymbol{r}_{2}\right ) +\left (\boldsymbol{r}_{1} \times \boldsymbol{r}_{2}^{ \prime }\right )\end{align*}

Observation 2.

The vector \frac{\boldsymbol{r}^{ \prime }\left (t\right )}{\left \Vert \boldsymbol{r}^{ \prime }\left (t\right )\right \Vert } is said to be the unit tangent vector and shown by \boldsymbol{T}\left (t\right ). Suppose \boldsymbol{T}^{ \prime } exists. We want to show that \boldsymbol{T} and \boldsymbol{T}^{ \prime } are orthogonal to each other. Observe that if we show

(8)   \begin{equation*}\boldsymbol{T} \cdot \boldsymbol{T}^{ \prime } =0 \end{equation*}

we are done. We know

(9)   \begin{equation*}\left \Vert \boldsymbol{T}\right \Vert ^{2} =1 \end{equation*}


(10)   \begin{equation*}\frac{d}{dt}(\left \Vert \boldsymbol{T}\right \Vert ^{2}) =0 \end{equation*}


(11)   \begin{equation*}\left \Vert \boldsymbol{T}\right \Vert ^{2} =\boldsymbol{T} \cdot \boldsymbol{T} \end{equation*}


(12)   \begin{equation*}\frac{d}{dt}\left (\boldsymbol{T} \cdot \boldsymbol{T}\right ) =\boldsymbol{T}^{ \prime } \cdot \boldsymbol{T} +\boldsymbol{T} \cdot \boldsymbol{T}^{ \prime } =2\text{}\boldsymbol{T} \cdot \boldsymbol{T}^{ \prime } =0 \end{equation*}

Now we are done. Among many possibilities of vectors that could be orthogonal to \boldsymbol{T} , one is important for us. We define the principle unit normal vector by

(13)   \begin{equation*}\boldsymbol{N}\left (t\right ) =\frac{\boldsymbol{T}^{ \prime }\left (t\right )}{\left \Vert \boldsymbol{T}^{ \prime }\left (t\right )\right \Vert } \end{equation*}

Example 6.

Given \boldsymbol{r}\left (t\right ) =\left \langle 1 ,\sin t ,\cos t\right \rangle we compute the unit tangent vector

(14)   \begin{align*}\boldsymbol{r}^{ \prime } =\left \langle 0 ,\cos t , -\sin t\right \rangle \\ \left \Vert \boldsymbol{r}^{ \prime }\right \Vert =\sqrt{\cos ^{2}t +( -\sin t)^{2}} =1 \\ \boldsymbol{T} =\frac{\boldsymbol{r}^{ \prime }}{\left \Vert \boldsymbol{r}^{ \prime }\right \Vert } \\ =\left \langle 0 ,\cos t , -\sin t\right \rangle \end{align*}

Next, we find the principle unit normal vector

(15)   \begin{align*}\boldsymbol{T}^{ \prime }\left (t\right ) =\langle 0 , -\sin t , -\cos t\rangle ,\left \Vert \boldsymbol{T}^{ \prime }\right \Vert =1 \\ \boldsymbol{N}\left (t\right ) =\frac{\boldsymbol{T}^{ \prime }\left (t\right )}{\left \Vert \boldsymbol{T}^{ \prime }\left (t\right )\right \Vert } =\left \langle 0 , -\sin t , -\cos t\right \rangle \end{align*}


Suppose we have a smooth curve defined by \boldsymbol{r}. The fact that how quickly or how slowly \boldsymbol{r} bends (changing direction) at every point is determined by a concept called the curvature of \boldsymbol{r}; it’s shown by \kappa . There are many ways to define \kappa and to compute it. The formulas that we use for the next examples are

(16)   \begin{equation*}\kappa =\frac{\left \Vert \boldsymbol{T}^{ \prime }\right \Vert }{\left \Vert \boldsymbol{r}^{ \prime }\right \Vert } =\frac{\left \Vert \boldsymbol{r}^{ \prime } \times \boldsymbol{r}^{ \prime \prime }\right \Vert }{\left \Vert \boldsymbol{r}^{ \prime }\right \Vert ^{3}} \end{equation*}

and for curves in xy-plane, y=y(x),

(17)   \begin{equation*}\kappa =\frac{\left \vert y^{ \prime \prime }\right \vert }{\left (1 +\left [y^{ \prime }\right ]^{2}\right )^{3/2}}\end{equation*}

An interesting fact about curvature is its relationship and the osculating circle. Imagine a circle that is attached, in fact tangent, to a point of a given smooth curve. There could be many such circles but for the reason of understanding, the one that is the closest to be tangent at two points next to each other is called the osculating circle. Now suppose R is the radius of such circle. Then

(18)   \begin{equation*}\kappa =\frac{1}{R} \end{equation*}

This is a well-known relation between the curvature and radius of the osculating circle at a point on a smooth curve. Note how direction changes in the picture. The inverse relation between curvature and radius of the osculating circle dictates that the more quickly a curve bends, radius of the osculating circle will be smaller.

Example 7.

We find the curvature of \boldsymbol{r}\left (t\right ) =\left \langle 1 ,\sin t ,\cos t\right \rangle at t =\pi . Note that using the result of previous example

(19)   \begin{equation*}\kappa =\frac{\left \Vert \boldsymbol{T}^{ \prime }\right \Vert }{\left \Vert \boldsymbol{r}^{ \prime }\right \Vert } =\frac{1}{1} =1 \end{equation*}

Example 8.

We are given a vector-valued function \boldsymbol{r}\left (t\right ) =\left \langle 2t ,t^{2} ,\log t\right \rangle.To compute the curvature for this curve we use the formula

(20)   \begin{align*}\boldsymbol{r}^{ \prime } =\langle 2 ,2t ,\frac{1}{t}\rangle \\ \left \Vert \boldsymbol{r}^{ \prime }\right \Vert =\sqrt{2^{2} +(2t)^{2} +(\frac{1}{t})^{2}} \\ =\sqrt{4 +4t^{2} +\frac{1}{t^{2}}} \\ =\sqrt{\frac{\left (2t^{2} +1\right )^{2}}{t^{2}}} \\ =\frac{2t^{2} +1}{t} \\ \boldsymbol{T} =\frac{\boldsymbol{r}^{ \prime }}{\left \Vert \boldsymbol{r}^{ \prime }\right \Vert } \\ =\frac{t}{2t^{2} +1}\left \langle 2 ,2t ,\frac{1}{t}\right \rangle \\ =\left \langle \frac{2t}{2t^{2} +1} ,\frac{2t^{2}}{2t^{2} +1} ,\frac{1}{2t^{2} +1}\right \rangle \end{align*}

We need to find \left \Vert \boldsymbol{T}^{ \prime }\right \Vert

(21)   \begin{align*}\boldsymbol{T}^{ \prime } =\langle \frac{2(2t^{2} +1) -4t\left (2t\right )}{\left (2t^{2} +1\right )^{2}} ,\frac{4t\left (2t^{2} +1\right ) -4t\left (2t^{2}\right )}{\left (2t^{2} +1\right )^{2}} ,\frac{ -4t}{\left (2t^{2} +1\right )^{2}}\rangle =\langle \frac{2(1 -t^{2})}{\left (2t^{2} +1\right )^{2}} ,\frac{4t}{\left (2t^{2} +1\right )^{2}} ,\frac{ -4t}{\left (2t^{2} +1\right )^{2}}\rangle \\ \left \Vert\boldsymbol{T}^{ \prime }\right \Vert =\sqrt{\frac{4(1 -t^{2})^{2}}{\left (2t^{2} +1\right )^{4}} +\frac{16t^{2}}{\left (2t^{2} +1\right )^{4}} +\frac{16t^{2}}{\left (2t^{2} +1\right )^{4}}} =\frac{2\sqrt{6t^{4} +t^{2} +1}}{\left (2t^{2} +1\right )^{2}}\end{align*}

And finally the curvature is provided by

(22)   \begin{align*}\kappa =\frac{\left \Vert \boldsymbol{T}^{ \prime }\right \Vert }{\left \Vert \boldsymbol{r}^{ \prime }\right \Vert } \\ =\frac{2\sqrt{6t^{4} +t^{2} +1}}{\left (2t^{2} +1\right )^{2}}\genfrac{(}{)}{}{}{t}{2t^{2} +1} \\ =\frac{2t\sqrt{6t^{4} +t^{2} +1}}{\left (2t^{2} +1\right )^{3}}\end{align*}

Note that, at t =0 , the curvature is equal to zero and when t =1 , the curvature is equal to \frac{4\sqrt{2}}{27}.

Example 9.

Suppose we would like to find the curvature of the unit circle at any point. Of course, one can use the relationship of the osculating circle and curvature but we want to attempt with a different formula. A point \left (x ,y\right ) is on the unit circle when x^{2} +y^{2} =1. We need y^{ \prime } and y^{ \prime \prime }. So differentiate implicitly

(23)   \begin{align*}2x +2yy^{ \prime } =0 \\ y^{ \prime } = -\frac{x}{y} \\ y^{ \prime \prime } = -\frac{y -xy^{ \prime }}{y^{2}} \\ =\frac{ -x\frac{x}{y} -y}{y^{2}} \\ = -\frac{1}{y^{3}}\end{align*}

We know curvature is

(24)   \begin{align*}\kappa =\frac{\left \vert y^{ \prime \prime }\right \vert }{\left (1 +\left [y^{ \prime }\right ]^{2}\right )} \\ =\frac{\left \vert -\frac{1}{y^{3}}\right \vert }{\left (1 +\left [ -\frac{x}{y}\right ]^{2}\right )^{3/2}} \\ =1\end{align*}

Observation 3.

Consider a curve in space \boldsymbol{r}\left (t\right ) =\left \langle x\left (t\right ) ,y\left (t\right ) ,z\left (t\right )\right \rangle . Let use take two points on this curve P_{0} and P_{n} associated to t =a and t =b. Make a partition of this curve with respect to t , say all of equal size \bigtriangleup t . This can be shown as followed. We call each endpoint of the partition P_{1} ,P_{2} , . . . ,P_{n} .

Observe that as shown in the picture, the distance between P_{0} and P_{1} is

(25)   \begin{equation*}d\left (P_{0} ,P_{1}\right ) =\sqrt{\left (x_{1} -x_{0}\right )^{2} +(y_{1} -y_{0})^{2} +(z_{1} -z_{0})^{2}} \end{equation*}

The length of the curve \boldsymbol{r}\left (t\right ) can be approximated using a sum. In the third step we multiply and divide by \bigtriangleup t

(26)   \begin{align*}d\left (P_{0} ,P_{1}\right ) +d\left (P_{1} ,P_{2}\right ) + . . . +d\left (P_{n -1} ,P_{n}\right ) =\sum _{i =0}^{n -1}d\left (P_{i} ,P_{i +1}\right ) \\ =\sum _{i =0}^{n -1}\sqrt{\left (x_{i +1} -x_{i}\right )^{2} +(y_{i +1} -y_{i})^{2} +(z_{i +1} -z_{i})^{2}} \\ =\sum _{i =1}^{n -1}\sqrt{\left (x_{i +1} -x_{i}\right )^{2} +(y_{i +1} -y_{i})^{2} +\left (z_{i +1} -z_{i}\right )^{2}}\frac{ \bigtriangleup t}{ \bigtriangleup t} \\ =\sum _{i =1}^{n -1}\sqrt{\frac{\left (x_{i +1} -x_{i}\right )^{2} +\left (y_{i +1} -y_{i}\right )^{2} +\left (z_{i +1} -z_{i}\right )^{2}}{( \bigtriangleup t)^{2}}} \bigtriangleup t \\ =\sum _{i =1}^{n -1}\sqrt{\genfrac{(}{)}{}{}{x_{i +1} -x_{i}}{ \bigtriangleup t}^{2} +(\frac{y_{i +1} -y_{i}}{ \bigtriangleup t})^{2} +(\frac{z_{i +1} -z_{i}}{ \bigtriangleup t})^{2}} \bigtriangleup t\end{align*}

Note that using Mean Value Theorem guarantees that for some t values

(27)   \begin{equation*}\frac{x_{i +1} -x_{i}}{ \bigtriangleup t} =x^{ \prime }\left (t\right ) ,\frac{y_{i +1} -y_{i}}{ \bigtriangleup t} =y^{ \prime }\left (t\right ) ,\frac{z_{i +1} -z_{i}}{ \bigtriangleup t} =z^{ \prime }\left (t\right ) \end{equation*}

Now we can have n as large as we like to make this approximation, which call arc length AL to be exact

(28)   \begin{align*}AL =\lim _{n \rightarrow \infty }\sum _{i =1}^{n -1}\sqrt{\left (x^{ \prime }\left (t\right )\right )^{2} +(y^{ \prime }\left (t\right ))^{2} +(z^{ \prime }\left (t\right ))^{2}} \bigtriangleup t \\ =\int _{a}^{b}\sqrt{[x^{ \prime }\left (t\right )]^{2} +[y^{ \prime }\left (t\right )]^{2} +[z^{ \prime }\left (t\right )]^{2}}dt\end{align*}

Now looking closer we see that the notion \sqrt{\left [x^{ \prime }\left (t\right )\right ]^{2} +[y^{ \prime }\left (t\right )]^{2} +[z^{ \prime }\left (t\right )]^{2}} is the magnitude of \boldsymbol{r}^{ \prime }\left (t\right ) so

(29)   \begin{equation*}AL =\int _{a}^{b}\Vert \boldsymbol{r}^{ \prime }\left (t\right )\Vert dt \end{equation*}

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