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Arc Length and Curvature
Determine the length of a particle’s path in space by using the arc-length function.
Arc Length for Vector Functions
We have seen how a vector-valued function describes a curve in either two or three dimensions. Recall , which states that the formula for the arc length of a curve defined by the parametric functions \(x=x(t),y=y(t),{t}_{1}\le t\le {t}_{2}\) is given by
\[s={\int }_{{t}_{1}}^{{t}_{2}}\sqrt{{({x}^{'}(t))}^{2}+{({y}^{'}(t))}^{2}}dt.\]In a similar fashion, if we define a smooth curve using a vector-valued function \(\text{r}(t)=f(t)\ \text{i}+g(t)\ \text{j},\) where \(a\le t\le b,\) the arc length is given by the formula
\[s={\int }_{a}^{b}\sqrt{{({f}^{'}(t))}^{2}+{({g}^{'}(t))}^{2}}dt.\]In three dimensions, if the vector-valued function is described by \(\text{r}(t)=f(t)\ \text{i}+g(t)\ \text{j}+h(t)\ \text{k}\) over the same interval \(a\le t\le b,\) the arc length is given by
\[s={\int }_{a}^{b}\sqrt{{({f}^{'}(t))}^{2}+{({g}^{'}(t))}^{2}+{({h}^{'}(t))}^{2}}dt.\]The two formulas are very similar; they differ only in the fact that a space curve has three component functions instead of two. Note that the formulas are defined for smooth curves: curves where the vector-valued function \(\text{r}(t)\) is continuously differentiable with a non-zero derivative. The smoothness condition guarantees that the curve has no cusps (or corners) that could make the formula problematic.
We now return to the helix introduced earlier in this chapter. A vector-valued function that describes a helix can be written in the form
\[\text{r}(t)=R\ \text{cos}(\frac{2\pi Nt}{h})\ \text{i}+R\ \text{sin}(\frac{2\pi Nt}{h})\ \text{j}+t\ \text{k},\ 0\le t\le h,\]where R represents the radius of the helix, h represents the height (distance between two consecutive turns), and the helix completes N turns. Let’s derive a formula for the arc length of this helix using . First of all,
Therefore,
Condensed — the full section is in OpenStax Calculus Volume 3.
Arc-Length Parameterization
We now have a formula for the arc length of a curve defined by a vector-valued function. Let’s take this one step further and examine what an arc-length function is.
If a vector-valued function represents the position of a particle in space as a function of time, then the arc-length function measures how far that particle travels as a function of time. The formula for the arc-length function follows directly from the formula for arc length:
\[s(t)={\int }_{a}^{t}\sqrt{{({f}^{'}(u))}^{2}+{({g}^{'}(u))}^{2}+{({h}^{'}(u))}^{2}}du.\]If the curve is in two dimensions, then only two terms appear under the square root inside the integral. The reason for using the independent variable u is to distinguish between time and the variable of integration. Since \(s(t)\) measures distance traveled as a function of time, \({s}^{'}(t)\) measures the speed of the particle at any given time. Since we have a formula for \(s(t)\) in , we can differentiate both sides of the equation:
\[\begin{array}{ll}{s}^{'}(t) & =\frac{d}{dt}[{\int }_{a}^{t}\sqrt{{({f}^{'}(u))}^{2}+{({g}^{'}(u))}^{2}+{({h}^{'}(u))}^{2}}du] \\ & =\frac{d}{dt}[{\int }_{a}^{t}‖{r}^{'}(u)‖\ du] \\ & =‖{r}^{'}(t)‖.\end{array}\]If we assume that \(\text{r}(t)\) defines a smooth curve, then the arc length is always increasing, so \({s}^{'}(t)>0\) for \(t>a.\) Last, if \(\text{r}(t)\) is a curve on which \(‖{r}^{'}(t)‖=1\) for all t, then
\[s(t)={\int }_{a}^{t}‖{r}^{'}(u)‖\ du={\int }_{a}^{t}1\ du=t-a,\]which means that t represents the arc length as long as \(a=0.\)
A useful application of this theorem is to find an alternative parameterization of a given curve, called an arc-length parameterization. Recall that any vector-valued function can be reparameterized via a change of variables. For example, if we have a function \(\text{r}(t)=〈3\ \text{cos}\ t,3\ \text{sin}\ t〉,0\le t\le 2\pi\) that parameterizes a circle of radius 3, we can change the parameter from t to \(4t,\) obtaining a new parameterization \(\text{r}(t)=〈3\ \text{cos}\ 4t,3\ \text{sin}\ 4t〉.\) The new parameterization still defines a circle of radius 3, but now we need only use the values \(0\le t\le \pi \text{/}2\) to traverse the circle once.
Condensed — the full section is in OpenStax Calculus Volume 3.
Curvature
An important topic related to arc length is curvature. The concept of curvature provides a way to measure how sharply a smooth curve turns. A circle has constant curvature. The smaller the radius of the circle, the greater the curvature.
Think of driving down a road. Suppose the road lies on an arc of a large circle. In this case you would barely have to turn the wheel to stay on the road. Now suppose the radius is smaller. In this case you would need to turn more sharply to stay on the road. In the case of a curve other than a circle, it is often useful first to inscribe a circle to the curve at a given point so that it is tangent to the curve at that point and “hugs” the curve as closely as possible in a neighborhood of the point (). The curvature of the graph at that point is then defined to be the same as the curvature of the inscribed circle.
The formula in the definition of curvature is not very useful in terms of calculation. In particular, recall that \(\text{T}(t)\) represents the unit tangent vector to a given vector-valued function \(\text{r}(t),\) and the formula for \(\text{T}(t)\) is \(\text{T}(t)=\frac{{r}^{'}(t)}{‖{r}^{'}(t)‖}.\) To use the formula for curvature, it is first necessary to express \(\text{r}(t)\) in terms of the arc-length parameter s, then find the unit tangent vector \(\text{T}(s)\) for the function \(\text{r}(s),\) then take the derivative of \(\text{T}(s)\) with respect to s. This is a tedious process. Fortunately, there are equivalent formulas for curvature.
Condensed — the full section is in OpenStax Calculus Volume 3.
The Normal and Binormal Vectors
We have seen that the derivative \({r}^{'}(t)\) of a vector-valued function is a tangent vector to the curve defined by \(\text{r}(t),\) and the unit tangent vector \(\text{T}(t)\) can be calculated by dividing \({r}^{'}(t)\) by its magnitude. When studying motion in three dimensions, two other vectors are useful in describing the motion of a particle along a path in space: the principal unit normal vector and the binormal vector.
Note that, by definition, the binormal vector is orthogonal to both the unit tangent vector and the normal vector. Furthermore, \(\text{B}(t)\) is always a unit vector. This can be shown using the formula for the magnitude of a cross product
\[‖\text{B}(t)‖=‖\text{T}(t)\ \times \ \text{N}(t)‖=‖\text{T}(t)‖‖\text{N}(t)‖\text{sin}\ \theta ,\]where \(\theta\) is the angle between \(\text{T}(t)\) and \(\text{N}(t).\) Since \(\text{N}(t)\) is the derivative of a unit vector, property (vii) of the derivative of a vector-valued function tells us that \(\text{T}(t)\) and \(\text{N}(t)\) are orthogonal to each other, so \(\theta =\pi \text{/}2.\) Furthermore, they are both unit vectors, so their magnitude is 1. Therefore, \(‖\text{T}(t)‖‖\text{N}(t)‖\text{sin}\ \theta =(1)(1)\text{sin}(\pi \text{/}2)=1\) and \(\text{B}(t)\) is a unit vector.
The principal unit normal vector can be challenging to calculate because the unit tangent vector involves a quotient, and this quotient often has a square root in the denominator. In the three-dimensional case, finding the cross product of the unit tangent vector and the unit normal vector can be even more cumbersome. Fortunately, we have alternative formulas for finding these two vectors, and they are presented in Motion in Space.
To find the equation of an osculating circle in two dimensions, we need find only the center and radius of the circle.
Condensed — the full section is in OpenStax Calculus Volume 3.
Key Concepts
- The arc-length function for a vector-valued function is calculated using the integral formula \(s(t)={\int }_{a}^{t}‖{r}^{'}(u)‖\ du.\) This formula is valid in both two and three dimensions.
- The curvature of a curve at a point in either two or three dimensions is defined to be the curvature of the inscribed circle at that point. The arc-length parameterization is used in the definition of curvature.
- There are several different formulas for curvature. The curvature of a circle is equal to the reciprocal of its radius.
- The principal unit normal vector at t is defined to be
\[\text{N}(t)=\frac{{T}^{'}(t)}{‖{T}^{'}(t)‖}.\] - The binormal vector at t is defined as \(\text{B}(t)=\text{T}(t)\ \times \ \text{N}(t),\) where \(\text{T}(t)\) is the unit tangent vector.
- The Frenet frame of reference is formed by the unit tangent vector, the principal unit normal vector, and the binormal vector.
- The osculating circle is tangent to a curve at a point and has the same curvature as the tangent curve at that point.
Key Equations
| Arc length of space curve | \(s={\int }_{a}^{b}\sqrt{\ {[{f}^{'}(t)]}^{2}+{[{g}^{'}(t)]}^{2}+{[{h}^{'}(t)]}^{2}}\ dt={\int }_{a}^{b}‖{r}^{'}(t)‖dt\) |
| Arc-length function | \(s(t)={\int }_{a}^{t}\sqrt{{({f}^{'}(u))}^{2}+{({g}^{'}(u))}^{2}+{({h}^{'}(u))}^{2}}du\ \text{or}\ s(t)={\int }_{a}^{t}‖{r}^{'}(u)‖\ du\) |
| Curvature | \(κ=\frac{‖{T}^{'}(t)‖}{‖{r}^{'}(t)‖}\ \text{or}\ κ=\frac{‖{r}^{'}(t)\ \times \ {r}^{″}(t)‖}{{‖{r}^{'}(t)‖}^{3}}\ \text{or}\ κ=\frac{|{y}^{″}|}{{[1+{({y}^{'})}^{2}]}^{3\text{/}2}}\) |
| Principal unit normal vector | \(\text{N}(t)=\frac{{T}^{'}(t)}{‖{T}^{'}(t)‖}\) |
| Binormal vector | \(\text{B}(t)=\text{T}(t)\ \times \ \text{N}(t)\) |
Arc Length and Curvature
Find the arc length of the curve on the given interval.
A particle moves along the plane curve C described by \(\text{r}(t)=t\text{i}+{t}^{2}\text{j}.\) Solve the following problems.
The surface of a large cup is formed by revolving the graph of the function \(y=0.25{x}^{1.6}\) from \(x=0\) to \(x=5\) about the y-axis (measured in centimeters).
Practice (40)
Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.
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Calculate the arc length for each of the following vector-valued functions:
- \(\text{r}(t)=(3t-2)\ \text{i}+(4t+5)\ \text{j},1\le t\le 5\)
- \(\text{r}(t)=〈t\ \text{cos}\ t,t\ \text{sin}\ t,2t〉,0\le t\le 2\pi\)
Одкриј го одговорот
- Using , \({r}^{'}(t)=3\text{i}+4\text{j},\) so
\[\begin{array}{ll}s & ={\int }_{a}^{b}‖{r}^{'}(t)‖\ dt \\ & ={\int }_{a}^{5}\sqrt{{3}^{2}+{4}^{2}}dt \\ & ={\int }_{1}^{5}5\ dt={5t|}_{1}^{5}=20.\end{array}\] - Using , \({r}^{'}(t)=〈\text{cos}\ t-t\ \text{sin}\ t,\text{sin}\ t+t\ \text{cos}\ t,2〉,\) so
\[\begin{array}{ll}s & ={\int }_{a}^{b}‖{r}^{'}(t)‖\ dt \\ & ={\int }_{0}^{2\pi }\sqrt{{(\text{cos}\ t-t\ \text{sin}\ t)}^{2}+{(\text{sin}\ t+t\ \text{cos}\ t)}^{2}+{2}^{2}}dt \\ & ={\int }_{0}^{2\pi }\sqrt{({\text{cos}}^{2}t-2t\ \text{sin}\ t\ \text{cos}\ t+{t}^{2}{\text{sin}}^{2}t)+({\text{sin}}^{2}t+2t\ \text{sin}\ t\ \text{cos}\ t+{t}^{2}{\text{cos}}^{2}t)+4}\ dt \\ & ={\int }_{0}^{2\pi }\sqrt{{\text{cos}}^{2}t+{\text{sin}}^{2}t+{t}^{2}({\text{cos}}^{2}t+{\text{sin}}^{2}t)+4}\ dt \\ & ={\int }_{0}^{2\pi }\sqrt{{t}^{2}+5}\ dt.\end{array}\]
Here we can use a table integration formula
\[\int \sqrt{{u}^{2}+{a}^{2}}\ du=\frac{u}{2}\sqrt{{u}^{2}+{a}^{2}}+\frac{{a}^{2}}{2}\text{ln}\ |u+\sqrt{{u}^{2}+{a}^{2}}|+C,\]
so we obtain
\[\begin{array}{ll}{\int }_{0}^{2\pi }\sqrt{{t}^{2}+5}\ dt & =\frac{1}{2}{(t\sqrt{{t}^{2}+5}+5\ \text{ln}\ |t+\sqrt{{t}^{2}+5}|)}_{0}^{2\pi } \\ & =\frac{1}{2}(2\pi \sqrt{4{\pi }^{2}+5}+5\ \text{ln}\ (2\pi +\sqrt{4{\pi }^{2}+5}))-\frac{5}{2}\ \text{ln}\ \sqrt{5} \\ & \approx 25.343.\end{array}\]
-
Calculate the arc length of the parameterized curve
\[\text{r}(t)=〈2{t}^{2}+1,2{t}^{2}-1,{t}^{3}〉,0\le t\le 3.\]Одкриј го одговорот
\({r}^{'}(t)=〈4t,4t,3{t}^{2}〉,\) so \(s=\frac{1}{27}({113}^{3\text{/}2}-{32}^{3\text{/}2})\approx 37.785\)
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Find the arc-length parameterization for each of the following curves:
- \(\text{r}(t)=4\ \text{cos}\ t\ \text{i}+4\ \text{sin}\ t\ \text{j},t\ge 0\)
- \(\text{r}(t)=〈t+3,\ 2t-4,2t〉,t\ge 3\)
Одкриј го одговорот
- First we find the arc-length function using :
\[\begin{array}{ll}s(t) & ={\int }_{a}^{t}‖{r}^{'}(u)‖\ du \\ & ={\int }_{0}^{t}‖〈-4\ \text{sin}\ u,4\ \text{cos}\ u〉‖\ du \\ & ={\int }_{0}^{t}\sqrt{{(-4\ \text{sin}\ u)}^{2}+{(4\ \text{cos}\ u)}^{2}}\ du \\ & ={\int }_{0}^{t}\sqrt{16\ {\text{sin}}^{2}u+16\ {\text{cos}}^{2}u}\ du \\ & ={\int }_{0}^{t}4\ du=4t,\end{array}\]
which gives the relationship between the arc length s and the parameter t as \(s=4t;\) so, \(t=s\text{/}4.\) Next we replace the variable t in the original function \(\text{r}(t)=4\ \text{cos}\ t\ \text{i}+4\ \text{sin}\ t\ \text{j}\) with the expression \(s\text{/}4\) to obtain
\[\text{r}(t)=4\ \text{cos}(\frac{s}{4})\ \text{i}+4\ \text{sin}(\frac{s}{4})\ \text{j}\text{.}\]
This is the arc-length parameterization of \(\text{r}(t).\) Since the original restriction on t was given by \(t\ge 0,\) the restriction on s becomes \(s\text{/}4\ge 0,\) or \(s\ge 0.\) - The arc-length function is given by :
\[\begin{array}{ll}s(t) & ={\int }_{a}^{t}‖{r}^{'}(u)‖\ du \\ & ={\int }_{3}^{t}‖〈1,2,2〉‖\ du \\ & ={\int }_{3}^{t}\sqrt{{1}^{2}+{2}^{2}+{2}^{2}}\ du \\ & ={\int }_{3}^{t}3\ du \\ & =3t-9.\end{array}\]
Therefore, the relationship between the arc length s and the parameter t is \(s=3t-9,\) so \(t=\frac{s}{3}+3.\) Substituting this into the original function \(\text{r}(t)=〈t+3,\ 2t-4,2t〉\) yields
\[\text{r}(s)=〈(\frac{s}{3}+3)+3,\ 2(\frac{s}{3}+3)-4,2(\frac{s}{3}+3)〉=〈\frac{s}{3}+6,\ \frac{2s}{3}+2,\frac{2s}{3}+6〉.\]
This is an arc-length parameterization of \(\text{r}(t).\) The original restriction on the parameter \(t\) was \(t\ge 3,\) so the restriction on s is \((s\text{/}3)+3\ge 3,\) or \(s\ge 0.\)
-
Find the arc-length function for the helix
\[\text{r}(t)=〈3\ \text{cos}\ t,3\ \text{sin}\ t,4t〉,t\ge 0.\]Then, use the relationship between the arc length and the parameter t to find an arc-length parameterization of \(\text{r}(t).\)
Одкриј го одговорот
\(s=5t,\) or \(t=s\text{/}5.\) Substituting this into \(\text{r}(t)=〈3\ \text{cos}\ t,3\ \text{sin}\ t,4t〉\) gives
\(\text{r}(s)=〈3\ \text{cos}(\frac{s}{5}),\ 3\ \text{sin}(\frac{s}{5}),\frac{4s}{5}〉,s\ge 0.\)
-
Find the curvature for each of the following curves at the given point:
- \(\text{r}(t)=4\ \text{cos}\ t\ \text{i}+4\ \text{sin}\ t\ \text{j}+3t\ \text{k},t=\frac{4\pi }{3}\)
- \(f(x)=\sqrt{4x-{x}^{2}},x=2\)
Одкриј го одговорот
- This function describes a helix.
The curvature of the helix at \(t=(4\pi )\text{/}3\) can be found by using . First, calculate \(\text{T}(t)\text{:}\)
\[\begin{array}{ll}\text{T}(t) & =\frac{{r}^{'}(t)}{‖{r}^{'}(t)‖} \\ & =\frac{〈-4\ \text{sin}\ t,4\ \text{cos}\ t,3〉}{\sqrt{{(-4\ \text{sin}\ t)}^{2}+{(4\ \text{cos}\ t)}^{2}+{3}^{2}}} \\ & =〈-\frac{4}{5}\text{sin}\ t,\frac{4}{5}\text{cos}\ t,\frac{3}{5}〉.\end{array}\]
Next, calculate \({T}^{'}(t)\text{:}\)
\[{T}^{'}(t)=〈-\frac{4}{5}\text{cos}\ t,-\frac{4}{5}\text{sin}\ t,0〉.\]
Last, apply :
\[\begin{array}{ll}κ & =\frac{‖{T}^{'}(t)‖}{‖{r}^{'}(t)‖}=\frac{‖〈-\frac{4}{5}\text{cos}\ t,-\frac{4}{5}\text{sin}\ t,0〉‖}{‖〈-4\ \text{sin}\ t,4\ \text{cos}\ t,3〉‖} \\ & =\frac{\sqrt{{(-\frac{4}{5}\text{cos}\ t)}^{2}+{(-\frac{4}{5}\text{sin}\ t)}^{2}+{0}^{2}}}{\sqrt{{(-4\ \text{sin}\ t)}^{2}+{(4\ \text{cos}\ t)}^{2}+{3}^{2}}} \\ & =\frac{4\text{/}5}{5}=\frac{4}{25}.\end{array}\]
The curvature of this helix is constant at all points on the helix. - This function describes a semicircle.
To find the curvature of this graph, we must use . First, we calculate \({y}^{'}\) and \({y}^{″}\text{:}\)
\[\begin{array}{ll}y & =\sqrt{4x-{x}^{2}}={(4x-{x}^{2})}^{1\text{/}2} \\ {y}^{'} & =\frac{1}{2}{(4x-{x}^{2})}^{-1\text{/}2}(4-2x)=(2-x){(4x-{x}^{2})}^{-1\text{/}2} \\ {y}^{″} & =-{(4x-{x}^{2})}^{-1\text{/}2}+(2-x)(-\frac{1}{2}){(4x-{x}^{2})}^{-3\text{/}2}(4-2x) \\ & =-\frac{4x-{x}^{2}}{{(4x-{x}^{2})}^{3\text{/}2}}-\frac{{(2-x)}^{2}}{{(4x-{x}^{2})}^{3\text{/}2}} \\ & =\frac{{x}^{2}-4x-(4-4x+{x}^{2})}{{(4x-{x}^{2})}^{3\text{/}2}} \\ & =-\frac{4}{{(4x-{x}^{2})}^{3\text{/}2}}.\end{array}\]
Then, we apply :
\[\begin{array}{ll}κ & =\frac{|{y}^{″}|}{{[1+{({y}^{'})}^{2}]}^{3\text{/}2}} \\ & =\frac{|-\frac{4}{{(4x-{x}^{2})}^{3\text{/}2}}|}{{[1+{((2-x){(4x-{x}^{2})}^{-1\text{/}2})}^{2}]}^{3\text{/}2}}=\frac{|\frac{4}{{(4x-{x}^{2})}^{3\text{/}2}}|}{{[1+\frac{{(2-x)}^{2}}{4x-{x}^{2}}]}^{3\text{/}2}} \\ & =\frac{|\frac{4}{{(4x-{x}^{2})}^{3\text{/}2}}|}{{[\frac{4x-{x}^{2}+{x}^{2}-4x+4}{4x-{x}^{2}}]}^{3\text{/}2}}=|\frac{4}{{(4x-{x}^{2})}^{3\text{/}2}}|\cdot \frac{{(4x-{x}^{2})}^{3\text{/}2}}{8} \\ & =\frac{1}{2}.\end{array}\]
The curvature of this circle is equal to the reciprocal of its radius.
-
Find the curvature of the curve defined by the function
\[y=3{x}^{2}-2x+4\]at the point \(x=2.\)
Одкриј го одговорот
\(κ=\frac{6}{{101}^{3\text{/}2}}\approx 0.0059\)
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For each of the following vector-valued functions, find the principal unit normal vector. Then, if possible, find the binormal vector.
- \(\text{r}(t)=4\ \text{cos}\ t\ \text{i}-4\ \text{sin}\ t\ \text{j}\)
- \(\text{r}(t)=(6t+2)\ \text{i}+5{t}^{2}\ \text{j}-8t\ \text{k}\)
Одкриј го одговорот
- This function describes a circle.
To find the principal unit normal vector, we first must find the unit tangent vector \(\text{T}(t)\text{:}\)
\[\begin{array}{ll}\text{T}(t) & =\frac{{r}^{'}(t)}{‖{r}^{'}(t)‖} \\ & =\frac{-4\ \text{sin}\ t\ \text{i}-4\ \text{cos}\ t\ \text{j}}{\sqrt{{(-4\ \text{sin}\ t)}^{2}+{(-4\ \text{cos}\ t)}^{2}}} \\ & =\frac{-4\ \text{sin}\ t\ \text{i}-4\ \text{cos}\ t\ \text{j}}{\sqrt{16\ {\text{sin}}^{2}t+16\ {\text{cos}}^{2}t}} \\ & =\frac{-4\ \text{sin}\ t\ \text{i}-4\ \text{cos}\ t\ \text{j}}{\sqrt{16({\text{sin}}^{2}t+{\text{cos}}^{2}t)}} \\ & =\frac{-4\ \text{sin}\ t\ \text{i}-4\ \text{cos}\ t\ \text{j}}{4} \\ & =-\text{sin}\ t\ \text{i}-\text{cos}\ t\ \text{j}.\end{array}\]
Next, we use :
\[\begin{array}{ll}\text{N}(t) & =\frac{{T}^{'}(t)}{‖{T}^{'}(t)‖} \\ & =\frac{\text{-}\text{cos}\ t\ \text{i}+\text{sin}\ t\ \text{j}}{\sqrt{{(\text{-}\text{cos}\ t)}^{2}+{(\text{sin}\ t)}^{2}}} \\ & =\frac{\text{-}\text{cos}\ t\ \text{i}+\text{sin}\ t\ \text{j}}{\sqrt{{\text{cos}}^{2}t+{\text{sin}}^{2}t}} \\ & =-\text{cos}\ t\ \text{i}+\text{sin}\ t\ \text{j}.\end{array}\]
Notice that the unit tangent vector and the principal unit normal vector are orthogonal to each other for all values of t:
\[\begin{array}{ll}\text{T}(t)\cdot \text{N}(t) & =〈\text{-}\text{sin}\ t,-\text{cos}\ t〉\cdot 〈\text{-}\text{cos}\ t,\text{sin}\ t〉 \\ & =\text{sin}\ t\ \text{cos}\ t-\text{cos}\ t\ \text{sin}\ t \\ & =0.\end{array}\]
Furthermore, the principal unit normal vector points toward the center of the circle from every point on the circle. Since \(\text{r}(t)\) defines a curve in two dimensions, we cannot calculate the binormal vector.
- This function looks like this:
To find the principal unit normal vector, we first find the unit tangent vector \(\text{T}(t)\text{:}\)
\[\begin{array}{ll}\text{T}(t) & =\frac{{r}^{'}(t)}{‖{r}^{'}(t)‖} \\ & =\frac{6\ \text{i}+10t\ \text{j}-8\ \text{k}}{\sqrt{{6}^{2}+{(10t)}^{2}+{(-8)}^{2}}} \\ & =\frac{6\ \text{i}+10t\ \text{j}-8\ \text{k}}{\sqrt{36+100{t}^{2}+64}} \\ & =\frac{6\ \text{i}+10t\ \text{j}-8\text{k}}{\sqrt{100({t}^{2}+1)}} \\ & =\frac{3\ \text{i}+5t\ \text{j}-4\text{k}}{5\sqrt{{t}^{2}+1}} \\ & =\frac{3}{5}{({t}^{2}+1)}^{-1\text{/}2}\text{i}+t{({t}^{2}+1)}^{-1\text{/}2}\text{j}-\frac{4}{5}{({t}^{2}+1)}^{-1\text{/}2}\text{k}.\end{array}\]
Next, we calculate \({T}^{'}(t)\) and \(‖{T}^{'}(t)‖\text{:}\)
\[\begin{array}{ll}{T}^{'}(t) & =\frac{3}{5}(-\frac{1}{2}){({t}^{2}+1)}^{-3\text{/}2}(2t)\ \text{i}+({({t}^{2}+1)}^{-1\text{/}2}-t(\frac{1}{2}){({t}^{2}+1)}^{-3\text{/}2}(2t))\ \text{j} \\ & \ -\frac{4}{5}(-\frac{1}{2}){({t}^{2}+1)}^{-3\text{/}2}(2t)\ \text{k} \\ & =-\frac{3t}{5{({t}^{2}+1)}^{3\text{/}2}}\text{i}+\frac{1}{{({t}^{2}+1)}^{3\text{/}2}}\text{j}+\frac{4t}{5{({t}^{2}+1)}^{3\text{/}2}}\text{k} \\ ‖{T}^{'}(t)‖ & =\sqrt{{(-\frac{3t}{5{({t}^{2}+1)}^{3\text{/}2}})}^{2}+{(-\frac{1}{{({t}^{2}+1)}^{3\text{/}2}})}^{2}+{(\frac{4t}{5{({t}^{2}+1)}^{3\text{/}2}})}^{2}} \\ & =\sqrt{\frac{9{t}^{2}}{25{({t}^{2}+1)}^{3}}+\frac{1}{{({t}^{2}+1)}^{3}}+\frac{16{t}^{2}}{25{({t}^{2}+1)}^{3}}} \\ & =\sqrt{\frac{25{t}^{2}+25}{25{({t}^{2}+1)}^{3}}} \\ & =\sqrt{\frac{1}{{({t}^{2}+1)}^{2}}} \\ & =\frac{1}{{t}^{2}+1}.\end{array}\]
Therefore, according to :
\[\begin{array}{ll}\text{N}(t) & =\frac{{T}^{'}(t)}{‖{T}^{'}(t)‖} \\ & =(-\frac{3t}{5{({t}^{2}+1)}^{3\text{/}2}}\text{i}+\frac{1}{{({t}^{2}+1)}^{3\text{/}2}}\text{j}+\frac{4t}{5{({t}^{2}+1)}^{3\text{/}2}}\text{k})({t}^{2}+1) \\ & =-\frac{3t}{5{({t}^{2}+1)}^{1\text{/}2}}\text{i}+\frac{5}{5{({t}^{2}+1)}^{1\text{/}2}}\text{j}+\frac{4t}{5{({t}^{2}+1)}^{1\text{/}2}}\text{k} \\ & =-\frac{3t\ \text{i}-5\text{j}-4t\ \text{k}}{5\sqrt{{t}^{2}+1}}.\end{array}\]
Once again, the unit tangent vector and the principal unit normal vector are orthogonal to each other for all values of t:
\[\begin{array}{ll}\text{T}(t)\cdot \text{N}(t) & =(\frac{3\ \text{i}+5t\ \text{j}-4\text{k}}{5\sqrt{{t}^{2}+1}})\cdot (-\frac{3t\ \text{i}-5\text{j}-4t\ \text{k}}{5\sqrt{{t}^{2}+1}}) \\ & =\frac{3(-3t)-5t(-5)-4(4t)}{5\sqrt{{t}^{2}+1}} \\ & =\frac{-9t+25t-16t}{5\sqrt{{t}^{2}+1}} \\ & =0.\end{array}\]
Last, since \(\text{r}(t)\) represents a three-dimensional curve, we can calculate the binormal vector using :
\[\begin{array}{llllllll}\text{B}(t) & =\text{T}(t)\ \times \ \text{N}(t) \\ & =|\begin{array}{lll}\text{i} & \text{j} & \text{k} \\ \frac{3}{5\sqrt{{t}^{2}+1}} & +\frac{5t}{5\sqrt{{t}^{2}+1}} & -\frac{4}{5\sqrt{{t}^{2}+1}} \\ -\frac{3t}{5\sqrt{{t}^{2}+1}} & +\frac{5}{5\sqrt{{t}^{2}+1}} & \frac{4t}{5\sqrt{{t}^{2}+1}}\end{array}| \\ & =((+\frac{5t}{5\sqrt{{t}^{2}+1}})(\frac{4t}{5\sqrt{{t}^{2}+1}})-(-\frac{4}{5\sqrt{{t}^{2}+1}})(+\frac{5}{5\sqrt{{t}^{2}+1}}))\ \text{i} \\ & \ +((\frac{3}{5\sqrt{{t}^{2}+1}})(\frac{4t}{5\sqrt{{t}^{2}+1}})-(-\frac{4}{5\sqrt{{t}^{2}+1}})(-\frac{3t}{5\sqrt{{t}^{2}+1}}))\ \text{j} \\ & \ +((\frac{3}{5\sqrt{{t}^{2}+1}})(+\frac{5}{5\sqrt{{t}^{2}+1}})-(+\frac{5t}{5\sqrt{{t}^{2}+1}})(-\frac{3t}{5\sqrt{{t}^{2}+1}}))\ \text{k} \\ & =(\frac{20{t}^{2}+20}{25({t}^{2}+1)})\ \text{i}+(\frac{-15-15{t}^{2}}{25({t}^{2}+1)})\ \text{k} \\ & =20(\frac{{t}^{2}+1}{25({t}^{2}+1)})\ \text{i+}15(\frac{{t}^{2}+1}{25({t}^{2}+1)})\ \text{k} \\ & =\frac{4}{5}\text{i+}\frac{3}{5}\text{k}.\end{array}\]
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Find the unit normal vector for the vector-valued function \(\text{r}(t)=({t}^{2}-3t)\ \text{i}+(4t+1)\ \text{j}\) and evaluate it at \(t=2.\)
Одкриј го одговорот
\(\text{N}(2)=\frac{\sqrt{2}}{2}(\text{i}-\text{j})\)
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Find an equation of the osculating circle of the curve defined by the function \(y={x}^{3}-3x+1\) at \(x=1.\)
Одкриј го одговорот
shows the graph of \(y={x}^{3}-3x+1.\)
First, let’s calculate the curvature at \(x=1\text{:}\)
\[κ=\frac{|{f}^{″}(x)|}{{(1+{[{f}^{'}(x)]}^{2})}^{3\text{/}2}}=\frac{|6x|}{{(1+{[3{x}^{2}-3]}^{2})}^{3\text{/}2}}.\]This gives \(κ=6.\) Therefore, the radius of the osculating circle is given by \(R=\frac{1}{κ}=\frac{1}{6}.\) Next, we then calculate the coordinates of the center of the circle. When \(x=1,\) the slope of the tangent line is zero. Therefore, the center of the osculating circle is directly above the point on the graph with coordinates \((1,-1).\) The center is located at \((1,-\frac{5}{6}).\) The formula for a circle with radius r and center \((h,k)\) is given by \({(x-h)}^{2}+{(y-k)}^{2}={r}^{2}.\) Therefore, the equation of the osculating circle is \({(x-1)}^{2}+{(y+\frac{5}{6})}^{2}=\frac{1}{36}.\) The graph and its osculating circle appears in the following graph.
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Find an equation of the osculating circle of the curve defined by the vector-valued function \(y=2{x}^{2}-4x+5\) at \(x=1.\)
Одкриј го одговорот
\(κ=\frac{4}{{[1+{(4x-4)}^{2}]}^{3\text{/}2}}\)
At the point \(x=1,\) the curvature is equal to 4. Therefore, the radius of the osculating circle is \(\frac{1}{4}.\)
A graph of this function appears next:
The vertex of this parabola is located at the point \((1,3).\) Furthermore, the center of the osculating circle is directly above the vertex. Therefore, the coordinates of the center are \((1,\frac{13}{4}).\) The equation of the osculating circle is
\({(x-1)}^{2}+{(y-\frac{13}{4})}^{2}=\frac{1}{16}.\)
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\(\text{r}(t)={t}^{2}\text{i}+14t\text{j},\ 0\le t\le 7.\) This portion of the graph is shown here:
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\(\text{r}(t)={t}^{2}\text{i}+(2{t}^{2}+1)\text{j},\ 1\le t\le 3\)
Одкриј го одговорот
\(8\sqrt{5}\)
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\(\text{r}(t)=〈2\ \text{sin}\ t,5t,2\ \text{cos}\ t〉,0\le t\le \pi .\) This portion of the graph is shown here:
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\(\text{r}(t)=〈{t}^{2}+1,4{t}^{3}+3〉,\ -1\le t\le 0\)
Одкриј го одговорот
\(\frac{1}{54}({37}^{3\text{/}2}-1)\)
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\(\text{r}(t)=〈{e}^{\text{-}t}\text{cos}\ t,{e}^{\text{-}t}\text{sin}\ t〉\) over the interval \([0,\frac{\pi }{2}].\) Here is the portion of the graph on the indicated interval:
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Find the length of one turn of the helix given by \(\text{r}(t)=\frac{1}{2}\text{cos}\ t\text{i}+\frac{1}{2}\text{sin}\ t\text{j}+\sqrt{\frac{3}{4}}\ t\ \text{k}.\)
Одкриј го одговорот
Length \(=2\pi\)
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Find the arc length of the vector-valued function \(\text{r}(t)=-t\text{i}+4t\text{j}+3t\text{k}\) over \([0,1].\)
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A particle travels once around a circle with the equation of motion \(\text{r}(t)=3\ \text{cos}\ t\text{i}+3\ \text{sin}\ t\text{j}+0\text{k}.\) Find the distance traveled around the circle by the particle.
Одкриј го одговорот
\(6\pi\)
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Set up an integral to find the circumference of the ellipse with the equation \(\text{r}(t)=\text{cos}\ t\text{i}+2\ \text{sin}\ t\text{j}+0\text{k}.\)
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Find the length of the curve \(\text{r}(t)=〈\sqrt{2}t,{e}^{t},{e}^{\text{-}t}〉\) over the interval \(0\le t\le 1.\) The graph is shown here:
Одкриј го одговорот
\(e-\frac{1}{e}\)
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Find the length of the curve \(\text{r}(t)=〈2\ \text{sin}\ t,5t,2\ \text{cos}\ t〉\) for \(t\in [-10,10].\)
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The position function for a particle is \(\text{r}(t)=a\ \text{cos}(\omega t)\text{i}+b\ \text{sin}(\omega t)\text{j}.\) Find the unit tangent vector and the unit normal vector at \(t=0.\)
Одкриј го одговорот
\(\text{T}(0)=\text{j},\) \(\text{N}(0)=-\text{i}\)
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Given \(\text{r}(t)=a\ \text{cos}(\omega t)\text{i}+b\ \text{sin}(\omega t)\text{j},\) find the binormal vector \(\text{B}(0).\)
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Given \(\text{r}(t)=〈2{e}^{t},{e}^{t}\text{cos}\ t,{e}^{t}\text{sin}\ t〉,\) determine the tangent vector \(\text{T}(t).\)
Одкриј го одговорот
\(\text{T}(t)=〈\frac{2}{\sqrt{6}},\frac{\text{cos}\ t-\text{sin}\ t}{\sqrt{6}},\frac{\text{cos}\ t+\text{sin}\ t}{\sqrt{6}}〉\)
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Given \(\text{r}(t)=〈2{e}^{t},{e}^{t}\text{cos}\ t,{e}^{t}\text{sin}\ t〉,\) determine the unit tangent vector \(\text{T}(t)\) evaluated at \(t=0.\)
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Given \(\text{r}(t)=〈2{e}^{t},{e}^{t}\text{cos}\ t,{e}^{t}\text{sin}\ t〉,\) find the unit normal vector \(\text{N}(t)\) evaluated at \(t=0,\) \(\text{N}(0).\)
Одкриј го одговорот
\(\text{N}\left(0\right)=⟨0,-\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}⟩\)
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Given \(\text{r}(t)=〈2{e}^{t},{e}^{t}\text{cos}\ t,{e}^{t}\text{sin}\ t〉,\) find the unit binormal vector evaluated at \(t=0.\)
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Given \(\text{r}(t)=t\text{i}+{t}^{2}\text{j}+t\text{k},\) find the unit tangent vector \(\text{T}(t).\) The graph is shown here:
Одкриј го одговорот
\(\text{T}(t)=\frac{1}{\sqrt{4{t}^{2}+2}}<1,2t,1>\)
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Find the unit tangent vector \(\text{T}(t)\) and unit normal vector \(\text{N}(t)\) at \(t=0\) for the plane curve \(\text{r}(t)=〈{t}^{3}-4t,5{t}^{2}-2〉.\) The graph is shown here:
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Find the unit tangent vector \(\text{T}(t)\) for \(\text{r}(t)=3t\text{i}+5{t}^{2}\text{j}+2t\text{k}\)
Одкриј го одговорот
\(\text{T}(t)=\frac{1}{\sqrt{100{t}^{2}+13}}(3\text{i}+10t\text{j}+2\text{k})\)
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Find the principal normal vector to the curve \(\text{r}(t)=〈6\ \text{cos}\ t,6\ \text{sin}\ t〉\) at the point determined by \(t=\pi \text{/}3.\)
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Find \(\text{T}(t)\) for the curve \(\text{r}(t)=({t}^{3}-4t)\ \text{i}+(5{t}^{2}-2)\ \text{j}.\)
Одкриј го одговорот
\(\text{T}(t)=\frac{1}{\sqrt{9{t}^{4}+76{t}^{2}+16}}([3{t}^{2}-4]\text{i}+10t\text{j})\)
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Find \(\text{N}(t)\) for the curve \(\text{r}(t)=({t}^{3}-4t)\ \text{i}+(5{t}^{2}-2)\ \text{j}.\)
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Find the unit normal vector \(\text{N}(t)\) for \(\text{r}(t)=〈2\ \text{sin}\ t,5t,2\ \text{cos}\ t〉.\)
Одкриј го одговорот
\(\text{N}(t)=〈\text{-}\text{sin}\ t,0,-\text{cos}\ t〉\)
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Find the unit tangent vector \(\text{T}(t)\) for \(\text{r}(t)=〈2\ \text{sin}\ t,5t,2\ \text{cos}\ t〉.\)
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Find the arc-length function \(s(t)\) for the line segment given by \(\text{r}(t)=〈3-3t,4t〉.\) Write r as a parameter of s.
Одкриј го одговорот
Arc-length function: \(s(t)=5t;\) r as a parameter of s: \(\text{r}(s)=(3-\frac{3s}{5})\ \text{i}+\frac{4s}{5}\text{j}\)
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Parameterize the helix \(\text{r}(t)=\text{cos}\ t\text{i}+\text{sin}\ t\text{j}+t\text{k}\) using the arc-length parameter s, from \(t=0.\)
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Parameterize the curve using the arc-length parameter s, at the point at which \(t=0\) for \(\text{r}(t)={e}^{t}\text{sin}\ t\text{i}+{e}^{t}\text{cos}\ t\text{j}.\)
Одкриј го одговорот
\(\text{r}(s)=(1+\frac{s}{\sqrt{2}})\ \text{sin}(\text{ln}(1+\frac{s}{\sqrt{2}}))\ \text{i}+(1+\frac{s}{\sqrt{2}})\ \text{cos}[\text{ln}(1+\frac{s}{\sqrt{2}})]\text{j}\)
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Find the curvature of the curve \(\text{r}(t)=5\ \text{cos}\ t\text{i}+4\ \text{sin}\ t\text{j}\) at \(t=\pi \text{/}3.\) (Note: The graph is an ellipse.)
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Find the x-coordinate at which the curvature of the curve \(y=1\text{/}x\) is a maximum value.
Одкриј го одговорот
The maximum value of the curvature occurs at \(x=1.\)
Symbols used here
Antiderivative (indefinite) or signed area from a to b (definite).
The non-negative number whose square (n-th power) is x.
Ratio of a circle's circumference to its diameter, 3.14159…
The usual name for an angle.
Instantaneous rate of change; slope of the graph.
i² = −1.
Equal to the precision shown, not exactly.
Inequalities that allow equality; < and > exclude it.
The two sides are different.
Derivative with respect to x, holding the other variables fixed.
Vector of partial derivatives; points uphill.
Integral over a region of the plane; integral around a closed curve.
A quantity with magnitude and direction; a column of numbers.
Σ u_i v_i; the length of v, √(v·v).
How to: Arc Length and Curvature
- Determine the length of a particle’s path in space by using the arc-length function.
- Explain the meaning of the curvature of a curve in space and state its formula.
- Describe the meaning of the normal and binormal vectors of a curve in space.
- Using
- Using
- First we find the arc-length function using
- The arc-length function is given by
- This function describes a helix.
Questions people ask
What is a partial derivative?
The ordinary derivative with respect to one variable while every other variable is frozen — the slope of the surface in one coordinate direction.
What does the gradient point at?
Uphill: the direction of steepest increase, with length equal to that steepest slope. It is perpendicular to the level curves.
Обиди се со себе.
Parts of this page are adapted from OpenStax Calculus Volume 3 (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.
Повеќе во Multivariable Calculus
Functions of several variablesPartial derivatives and the gradientOptimisation in several variablesDouble and triple integralsVector fields, line integrals and the big theorems