Arc Length and Curvature

In this section, we study formulas related to curves in both two and three dimensions, and see how they are related to various properties of the same curve. For example, suppose a vector-valued function describes the motion of a particle in space. We would like to determine how far the particle has traveled over a given time interval, which can be described by the arc length of the path it follows. Or, suppose that the vector-valued function describes a road we are building and we want to determine how sharply the road curves at a given point. This is described by the curvature of the function at that point. We explore each of these concepts in this section.

Arc Length for Vector Functions

We have seen how a vector-valued function describes a curve in either two or three dimensions. Recall [link], 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} \leq t \leq t_{2}

In a similar fashion, if we define a smooth curve using a vector-valued function

r (t) = f (t) i + g (t) j,

In three dimensions, if the vector-valued function is described by

r (t) = f (t) i + g (t) j + h (t) k

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

r (t)

is 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

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 [link]. First of all,

This gives a formula for the length of a wire needed to form a helix with N turns that has radius R and height h.

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:

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 the speed of the particle at any given time. Since we have a formula for

s (t)

defines a smooth curve, then the arc length is always increasing, so

s^{'} (t) > 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

r (t) = 〈 3 cos t, 3 sin t 〉, 0 \leq t \leq 2 π

that parameterizes a circle of radius 3, we can change the parameter from t to

4 t,

The new parameterization still defines a circle of radius 3, but now we need only use the values

0 \leq t \leq π / 2

and are able to solve this function for t as a function of s. We can then reparameterize the original function

r (t)

The vector-valued function is now written in terms of the parameter s. Since the variable s represents the arc length, we call this an arc-length parameterization of the original function

r (t) .

One advantage of finding the arc-length parameterization is that the distance traveled along the curve starting from

s = 0

is now equal to the parameter s. The arc-length parameterization also appears in the context of curvature (which we examine later in this section) and line integrals, which we study in the Introduction to Vector Calculus.

Finding an Arc-Length Parameterization

Find the arc-length parameterization for each of the following curves:

$r (t) = 4 cos t i + 4 sin t j, t \geq 0$
$r (t) = 〈 t + 3, 2 t - 4, 2 t 〉, t \geq 3$

First we find the arc-length function using [link]:

$\begin{matrix} s (t) & = \int_{a}^{t} ‖ r^{'} (u) ‖ d u \\ = \int_{0}^{t} ‖ 〈 −4 sin u, 4 cos u 〉 ‖ d u \\ = \int_{0}^{t} \sqrt{{(−4 sin u)}^{2} + {(4 cos u)}^{2}} d u \\ = \int_{0}^{t} \sqrt{16 {sin}^{2} u + 16 {cos}^{2} u} d u \\ = \int_{0}^{t} 4 d u = 4 t, \end{matrix}$

which gives the relationship between the arc length s and the parameter t as
$s = 4 t;$
so,
$t = s / 4.$
Next we replace the variable t in the original function
$r (t) = 4 cos t i + 4 sin t j$
with the expression
$s / 4$
to obtain

$r (s) = 4 cos (\frac{s}{4}) i + 4 sin (\frac{s}{4}) j .$

This is the arc-length parameterization of
$r (t) .$
Since the original restriction on t was given by
$t \geq 0,$
the restriction on s becomes
$s / 4 \geq 0,$
or
$s \geq 0.$
The arc-length function is given by [link]:

$\begin{matrix} s (t) & = \int_{a}^{t} ‖ r^{'} (u) ‖ d u \\ = \int_{3}^{t} ‖ 〈 1, 2, 2 〉 ‖ d u \\ = \int_{3}^{t} \sqrt{1^{2} + 2^{2} + 2^{2}} d u \\ = \int_{3}^{t} 3 d u \\ = 3 t - 9. \end{matrix}$

Therefore, the relationship between the arc length s and the parameter t is
$s = 3 t - 9,$
so
$t = \frac{s}{3} + 3.$
Substituting this into the original function
$r (t) = 〈 t + 3, 2 t - 4, 2 t 〉$
yields

$r (s) = 〈 (\frac{s}{3} + 3) + 3, 2 (\frac{s}{3} + 3) - 4, 2 (\frac{s}{3} + 3) 〉 = 〈 \frac{s}{3} + 6, \frac{2 s}{3} + 2, \frac{2 s}{3} + 6 〉 .$

This is an arc-length parameterization of
$r (t) .$
The original restriction on the parameter
$t$
was
$t \geq 3,$
so the restriction on s is
$(s / 3) + 3 \geq 3,$
or
$s \geq 0.$

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 ([link]). 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

T (t)

in terms of the arc-length parameter s, then find the unit tangent vector

T (s)

with respect to s. This is a tedious process. Fortunately, there are equivalent formulas for curvature.

Proof

where s is the arc length along the curve C. Dividing both sides by

d s / d t,

In the case of a three-dimensional curve, we start with the formulas

T (t) = (r^{'} (t)) / ‖ r^{'} (t) ‖

are orthogonal. This means that

‖ T \times T^{'} ‖ = ‖ T ‖ ‖ T^{'} ‖ sin (π / 2) = ‖ T^{'} ‖,

This proves [link]. To prove [link], we start with the assumption that curve C is defined by the function

y = f (x) .

Finding Curvature

Find the curvature for each of the following curves at the given point:

$r (t) = 4 cos t i + 4 sin t j + 3 t k, t = \frac{4 π}{3}$
$f (x) = \sqrt{4 x - x^{2}}, x = 2$

This function describes a helix.

The curvature of the helix at
$t = (4 π) / 3$
can be found by using [link]. First, calculate
$T (t) :$

$\begin{matrix} T (t) & = \frac{r^{'} (t)}{‖ r^{'} (t) ‖} \\ = \frac{〈 −4 sin t, 4 cos t, 3 〉}{\sqrt{{(−4 sin t)}^{2} + {(4 cos t)}^{2} + 3^{2}}} \\ = 〈 - \frac{4}{5} sin t, \frac{4}{5} cos t, \frac{3}{5} 〉 . \end{matrix}$

Next, calculate
$T^{'} (t) :$

$T^{'} (t) = 〈 - \frac{4}{5} cos t, - \frac{4}{5} sin t, 0 〉 .$

Last, apply [link]:

$\begin{matrix} κ & = \frac{‖ T^{'} (t) ‖}{‖ r^{'} (t) ‖} = \frac{‖ 〈 - \frac{4}{5} cos t, - \frac{4}{5} sin t, 0 〉 ‖}{‖ 〈 −4 sin t, 4 cos t, 3 〉 ‖} \\ = \frac{\sqrt{{(- \frac{4}{5} cos t)}^{2} + {(- \frac{4}{5} sin t)}^{2} + 0^{2}}}{\sqrt{{(−4 sin t)}^{2} + {(4 cos t)}^{2} + 3^{2}}} \\ = \frac{4 / 5}{5} = \frac{4}{25} . \end{matrix}$

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 [link]. First, we calculate
$y^{'}$
and
$y ″ :$

$\begin{matrix} y & = \sqrt{4 x - x^{2}} = {(4 x - x^{2})}^{1 / 2} \\ y^{'} & = \frac{1}{2} {(4 x - x^{2})}^{- 1 / 2} (4 - 2 x) = (2 - x) {(4 x - x^{2})}^{- 1 / 2} \\ y ″ & = - {(4 x - x^{2})}^{- 1 / 2} + (2 - x) (- \frac{1}{2}) {(4 x - x^{2})}^{- 3 / 2} (4 - 2 x) \\ = - \frac{4 x - x^{2}}{{(4 x - x^{2})}^{3 / 2}} - \frac{{(2 - x)}^{2}}{{(4 x - x^{2})}^{3 / 2}} \\ = \frac{x^{2} - 4 x - (4 - 4 x + x^{2})}{{(4 x - x^{2})}^{3 / 2}} \\ = - \frac{4}{{(4 x - x^{2})}^{3 / 2}} . \end{matrix}$

Then, we apply [link]:

$\begin{matrix} κ & = \frac{\| y ″ \|}{{[1 + {(y^{'})}^{2}]}^{3 / 2}} \\ = \frac{\| - \frac{4}{{(4 x - x^{2})}^{3 / 2}} \|}{{[1 + {((2 - x) {(4 x - x^{2})}^{- 1 / 2})}^{2}]}^{3 / 2}} = \frac{\| \frac{4}{{(4 x - x^{2})}^{3 / 2}} \|}{{[1 + \frac{{(2 - x)}^{2}}{4 x - x^{2}}]}^{3 / 2}} \\ = \frac{\| \frac{4}{{(4 x - x^{2})}^{3 / 2}} \|}{{[\frac{4 x - x^{2} + x^{2} - 4 x + 4}{4 x - x^{2}}]}^{3 / 2}} = \| \frac{4}{{(4 x - x^{2})}^{3 / 2}} \| \cdot \frac{{(4 x - x^{2})}^{3 / 2}}{8} \\ = \frac{1}{2} . \end{matrix}$

The curvature of this circle is equal to the reciprocal of its radius. There is a minor issue with the absolute value in [link]; however, a closer look at the calculation reveals that the denominator is positive for any value of x.

The Normal and Binormal Vectors

of a vector-valued function is a tangent vector to the curve defined by

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,

B (t)

is always a unit vector. This can be shown using the formula for the magnitude of a cross product

is the derivative of a unit vector, property (vii) of the derivative of a vector-valued function tells us that

T (t)

Furthermore, they are both unit vectors, so their magnitude is 1. Therefore,

‖ T (t) ‖ ‖ N (t) ‖ sin θ = (1) (1) sin (π / 2) = 1

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.

Finding the Principal Unit Normal Vector and Binormal Vector

For each of the following vector-valued functions, find the principal unit normal vector. Then, if possible, find the binormal vector.

$r (t) = 4 cos t i - 4 sin t j$
$r (t) = (6 t + 2) i + 5 t^{2} j - 8 t k$

This function describes a circle.

To find the principal unit normal vector, we first must find the unit tangent vector
$T (t) :$

$\begin{matrix} T (t) & = \frac{r^{'} (t)}{‖ r^{'} (t) ‖} \\ = \frac{−4 sin t i - 4 cos t j}{\sqrt{{(−4 sin t)}^{2} + {(−4 cos t)}^{2}}} \\ = \frac{−4 sin t i - 4 cos t j}{\sqrt{16 {sin}^{2} t + 16 {cos}^{2} t}} \\ = \frac{−4 sin t i - 4 cos t j}{\sqrt{16 ({sin}^{2} t + {cos}^{2} t)}} \\ = \frac{−4 sin t i - 4 cos t j}{4} \\ = - sin t i - cos t j . \end{matrix}$

Next, we use [link]:

$\begin{matrix} N (t) & = \frac{T^{'} (t)}{‖ T^{'} (t) ‖} \\ = \frac{− cos t i + sin t j}{\sqrt{{(− cos t)}^{2} + {(sin t)}^{2}}} \\ = \frac{− cos t i + sin t j}{\sqrt{{cos}^{2} t + {sin}^{2} t}} \\ = - cos t i + sin t j . \end{matrix}$

Notice that the unit tangent vector and the principal unit normal vector are orthogonal to each other for all values of t:

$\begin{matrix} T (t) \cdot N (t) & = 〈 − sin t, - cos t 〉 \cdot 〈 − cos t, sin t 〉 \\ = sin t cos t - cos t sin t \\ = 0. \end{matrix}$

Furthermore, the principal unit normal vector points toward the center of the circle from every point on the circle. Since
$r (t)$
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
$T (t) :$

$\begin{matrix} T (t) & = \frac{r^{'} (t)}{‖ r^{'} (t) ‖} \\ = \frac{6 i + 10 t j - 8 k}{\sqrt{6^{2} + {(10 t)}^{2} + {(−8)}^{2}}} \\ = \frac{6 i + 10 t j - 8 k}{\sqrt{36 + 100 t^{2} + 64}} \\ = \frac{6 i + 10 t j - 8 k}{\sqrt{100 (t^{2} + 1)}} \\ = \frac{3 i - 5 t j - 4 k}{5 \sqrt{t^{2} + 1}} \\ = \frac{3}{5} {(t^{2} + 1)}^{- 1 / 2} i - t {(t^{2} + 1)}^{- 1 / 2} j - \frac{4}{5} {(t^{2} + 1)}^{- 1 / 2} k . \end{matrix}$

Next, we calculate
$T^{'} (t)$
and
$‖ T^{'} (t) ‖ :$

$\begin{matrix} T^{'} (t) & = \frac{3}{5} (- \frac{1}{2}) {(t^{2} + 1)}^{- 3 / 2} (2 t) i - ({(t^{2} + 1)}^{- 1 / 2} - t (\frac{1}{2}) {(t^{2} + 1)}^{- 3 / 2} (2 t)) j \\ - \frac{4}{5} (- \frac{1}{2}) {(t^{2} + 1)}^{- 3 / 2} (2 t) k \\ = - \frac{3 t}{5 {(t^{2} + 1)}^{3 / 2}} i - \frac{1}{{(t^{2} + 1)}^{3 / 2}} j + \frac{4 t}{5 {(t^{2} + 1)}^{3 / 2}} k \\ ‖ T^{'} (t) ‖ & = \sqrt{{(- \frac{3 t}{5 {(t^{2} + 1)}^{3 / 2}})}^{2} + {(- \frac{1}{{(t^{2} + 1)}^{3 / 2}})}^{2} + {(\frac{4 t}{5 {(t^{2} + 1)}^{3 / 2}})}^{2}} \\ = \sqrt{\frac{9 t^{2}}{25 {(t^{2} + 1)}^{3}} + \frac{1}{{(t^{2} + 1)}^{3}} + \frac{16 t^{2}}{25 {(t^{2} + 1)}^{3}}} \\ = \sqrt{\frac{25 t^{2} + 25}{25 {(t^{2} + 1)}^{3}}} \\ = \sqrt{\frac{1}{{(t^{2} + 1)}^{2}}} \\ = \frac{1}{t^{2} + 1} . \end{matrix}$

Therefore, according to [link]:

$\begin{matrix} N (t) & = \frac{T^{'} (t)}{‖ T^{'} (t) ‖} \\ = (- \frac{3 t}{5 {(t^{2} + 1)}^{3 / 2}} i - \frac{1}{{(t^{2} + 1)}^{3 / 2}} j + \frac{4 t}{5 {(t^{2} + 1)}^{3 / 2}} k) (t^{2} + 1) \\ = - \frac{3 t}{5 {(t^{2} + 1)}^{1 / 2}} i - \frac{5}{5 {(t^{2} + 1)}^{1 / 2}} j + \frac{4 t}{5 {(t^{2} + 1)}^{1 / 2}} k \\ = - \frac{3 t i + 5 j - 4 t k}{5 \sqrt{t^{2} + 1}} . \end{matrix}$

Once again, the unit tangent vector and the principal unit normal vector are orthogonal to each other for all values of t:

$\begin{matrix} T (t) \cdot N (t) & = (\frac{3 i - 5 t j - 4 k}{5 \sqrt{t^{2} + 1}}) \cdot (- \frac{3 t i + 5 j - 4 t k}{5 \sqrt{t^{2} + 1}}) \\ = \frac{3 (−3 t) - 5 t (−5) - 4 (4 t)}{5 \sqrt{t^{2} + 1}} \\ = \frac{−9 t + 25 t - 16 t}{5 \sqrt{t^{2} + 1}} \\ = 0. \end{matrix}$

Last, since
$r (t)$
represents a three-dimensional curve, we can calculate the binormal vector using [link]:

$\begin{matrix} B (t) & = T (t) \times N (t) \\ = \| \begin{matrix} i & j & k \\ \frac{3}{5 \sqrt{t^{2} + 1}} & - \frac{5 t}{5 \sqrt{t^{2} + 1}} & - \frac{4}{5 \sqrt{t^{2} + 1}} \\ - \frac{3 t}{5 \sqrt{t^{2} + 1}} & - \frac{5}{5 \sqrt{t^{2} + 1}} & \frac{4 t}{5 \sqrt{t^{2} + 1}} \end{matrix} \| \\ = ((- \frac{5 t}{5 \sqrt{t^{2} + 1}}) (\frac{4 t}{5 \sqrt{t^{2} + 1}}) - (- \frac{4}{5 \sqrt{t^{2} + 1}}) (- \frac{5}{5 \sqrt{t^{2} + 1}})) i \\ - ((\frac{3}{5 \sqrt{t^{2} + 1}}) (\frac{4 t}{5 \sqrt{t^{2} + 1}}) - (- \frac{4}{5 \sqrt{t^{2} + 1}}) (- \frac{3 t}{5 \sqrt{t^{2} + 1}})) j \\ + ((\frac{3}{5 \sqrt{t^{2} + 1}}) (- \frac{5}{5 \sqrt{t^{2} + 1}}) - (- \frac{5 t}{5 \sqrt{t^{2} + 1}}) (- \frac{3 t}{5 \sqrt{t^{2} + 1}})) k \\ = (\frac{−20 t^{2} - 20}{25 (t^{2} + 1)}) i + (\frac{−15 - 15 t^{2}}{25 (t^{2} + 1)}) k \\ = −20 (\frac{t^{2} + 1}{25 (t^{2} + 1)}) i - 15 (\frac{t^{2} + 1}{25 (t^{2} + 1)}) k \\ = - \frac{4}{5} i - \frac{3}{5} k . \end{matrix}$

For any smooth curve in three dimensions that is defined by a vector-valued function, we now have formulas for the unit tangent vector T, the unit normal vector N, and the binormal vector B. The unit normal vector and the binormal vector form a plane that is perpendicular to the curve at any point on the curve, called the normal plane. In addition, these three vectors form a frame of reference in three-dimensional space called the Frenet frame of reference (also called the TNB frame) ([link]). Lat, the plane determined by the vectors T and N forms the osculating plane of C at any point P on the curve.

Suppose we form a circle in the osculating plane of C at point P on the curve. Assume that the circle has the same curvature as the curve does at point P and let the circle have radius r. Then, the curvature of the circle is given by

1 / r .

We call r the radius of curvature of the curve, and it is equal to the reciprocal of the curvature. If this circle lies on the concave side of the curve and is tangent to the curve at point P, then this circle is called the osculating circle of C at P, as shown in the following figure.

To find the equation of an osculating circle in two dimensions, we need find only the center and radius of the circle.

Key Concepts

Key Equations

Find the arc length of the curve on the given interval.

r (t) = t^{2} i + 14 t j, 0 \leq t \leq 7.

This portion of the graph is shown here:

This figure is the graph of a curve beginning at the origin and increasing.

r (t) = t^{2} i + (2 t^{2} + 1) j, 1 \leq t \leq 3

8 \sqrt{5}

r (t) = 〈 2 sin t, 5 t, 2 cos t 〉, 0 \leq t \leq π .

This portion of the graph is shown here:

This figure is the graph of a curve in 3 dimensions. It is inside of a box. The box represents an octant. The curve begins in the upper right corner of the box and bends through the box to the other side.

r (t) = 〈 t^{2} + 1, 4 t^{3} + 3 〉, - 1 \leq t \leq 0

\frac{1}{54} (37^{3 / 2} - 1)

r (t) = 〈 e^{− t} cos t, e^{− t} sin t 〉

over the interval $[0, \frac{π}{2}] .$

Here is the portion of the graph on the indicated interval:

This figure is the graph of a curve in the first quadrant. It begins approximately at 0.20 on the y axis and increases to approximately where x = 0.3. Then the curve decreases, meeting the x-axis at 1.0.

Find the length of one turn of the helix given by $r (t) = \frac{1}{2} cos t i + \frac{1}{2} sin t j + \sqrt{\frac{3}{4}} t k .$

Length $= 2 π$

Find the arc length of the vector-valued function $r (t) = - t i + 4 t j + 3 t k$

over $[0, 1] .$

A particle travels in a circle with the equation of motion $r (t) = 3 cos t i + 3 sin t j + 0 k .$

Find the distance traveled around the circle by the particle.

6 π

Set up an integral to find the circumference of the ellipse with the equation $r (t) = cos t i + 2 sin t j + 0 k .$

Find the length of the curve $r (t) = 〈 \sqrt{2} t, e^{t}, e^{− t} 〉$

over the interval $0 \leq t \leq 1.$

The graph is shown here:

This figure is the graph of a curve in 3 dimensions. It is inside of a box. The box represents an octant. The curve begins in the upper left corner of the box and bends through the box to the bottom of the other side.

e - \frac{1}{e}

Find the length of the curve $r (t) = 〈 2 sin t, 5 t, 2 cos t 〉$

for $t \in [−10, 10] .$

The position function for a particle is $r (t) = a cos (ω t) i + b sin (ω t) j .$

Find the unit tangent vector and the unit normal vector at $t = 0.$

T (0) = j,

N (0) = - i

Given $r (t) = a cos (ω t) i + b sin (ω t) j,$

find the binormal vector $B (0) .$

Given $r (t) = 〈 2 e^{t}, e^{t} cos t, e^{t} sin t 〉,$

determine the tangent vector $T (t) .$

T (t) = 〈 2 e^{t}, e^{t} cos t - e^{t} sin t, e^{t} cos t + e^{t} sin t 〉

Given $r (t) = 〈 2 e^{t}, e^{t} cos t, e^{t} sin t 〉,$

determine the unit tangent vector $T (t)$

evaluated at $t = 0.$

Given $r (t) = 〈 2 e^{t}, e^{t} cos t, e^{t} sin t 〉,$

find the unit normal vector $N (t)$

evaluated at $t = 0,$

N (0) .

N (0) = 〈 \frac{\sqrt{2}}{2}, 0, \frac{\sqrt{2}}{2} 〉

Given $r (t) = 〈 2 e^{t}, e^{t} cos t, e^{t} sin t 〉,$

find the unit normal vector evaluated at $t = 0.$

Given $r (t) = t i + t^{2} j + t k,$

find the unit tangent vector $T (t) .$

The graph is shown here:

This figure is the graph of a curve in 3 dimensions. It is inside of a box. The box represents an octant. The curve begins in the bottom left corner of the box and bends through the box to the upper left side.

T (t) = \frac{1}{\sqrt{4 t^{2} + 2}} < 1, 2 t, 1 >

Find the unit tangent vector $T (t)$

and unit normal vector $N (t)$

at $t = 0$

for the plane curve $r (t) = 〈 t^{3} - 4 t, 5 t^{2} - 2 〉 .$

The graph is shown here:

This figure is the graph of a curve above the x-axis. The curve decreases in the second quadrant, passes through the y-axis at y=20. Then it intersects the origin. The curve loops at the origin, increasing back through y=20 into the first quadrant.

Find the unit tangent vector $T (t)$

for $r (t) = 3 t i + 5 t^{2} j + 2 t k$

T (t) = \frac{1}{\sqrt{100 t^{2} + 13}} (3 i + 10 t j + 2 k)

Find the principal normal vector to the curve $r (t) = 〈 6 cos t, 6 sin t 〉$

at the point determined by $t = π / 3 .$

Find $T (t)$

for the curve $r (t) = (t^{3} - 4 t) i + (5 t^{2} - 2) j .$

T (t) = \frac{1}{\sqrt{9 t^{4} + 76 t^{2} + 16}} ([3 t^{2} - 4] i + 10 t j)

Find $N (t)$

for the curve $r (t) = (t^{3} - 4 t) i + (5 t^{2} - 2) j .$

Find the unit normal vector $N (t)$

for $r (t) = 〈 2 sin t, 5 t, 2 cos t 〉 .$

N (t) = 〈 − sin t, 0, - cos t 〉

Find the unit tangent vector $T (t)$

for $r (t) = 〈 2 sin t, 5 t, 2 cos t 〉 .$

Find the arc-length function $s (t)$

for the line segment given by $r (t) = 〈 3 - 3 t, 4 t 〉 .$

Write r as a parameter of s.

Arc-length function: $s (t) = 5 t;$

r as a parameter of s: $r (s) = (3 - \frac{3 s}{5}) i + \frac{4 s}{5} j$

Parameterize the helix $r (t) = cos t i + sin t j + t k$

using the arc-length parameter s, from $t = 0.$

Parameterize the curve using the arc-length parameter s, at the point at which $t = 0$

for $r (t) = e^{t} sin t i + e^{t} cos t j .$

r (s) = (1 + \frac{s}{\sqrt{2}}) sin (ln (1 + \frac{s}{\sqrt{2}})) i + (1 + \frac{s}{\sqrt{2}}) cos [ln (1 + \frac{s}{\sqrt{2}})] j

Find the curvature of the curve $r (t) = 5 cos t i + 4 sin t j$

at $t = π / 3 .$

(Note: The graph is an ellipse.)

This figure is the graph of an ellipse. The ellipse is oval along the x-axis. It is centered at the origin and intersects the y-axis at -4 and 4.

Find the x-coordinate at which the curvature of the curve $y = 1 / x$

is a maximum value.

The maximum value of the curvature occurs at $x = \sqrt[4]{5} .$

Find the curvature of the curve $r (t) = 5 cos t i + 5 sin t j .$

Does the curvature depend upon the parameter t?

Find the curvature $κ$

for the curve $y = x - \frac{1}{4} x^{2}$

at the point $x = 2.$

\frac{1}{2}

Find the curvature $κ$

for the curve $y = \frac{1}{3} x^{3}$

at the point $x = 1.$

Find the curvature $κ$

of the curve $r (t) = t i + 6 t^{2} j + 4 t k .$

The graph is shown here:

This figure is the graph of a curve in 3 dimensions. It is inside of a box. The box represents an octant. The curve has a parabolic shape in the middle of the box.

κ \approx \frac{49.477}{{(17 + 144 t^{2})}^{3 / 2}}

Find the curvature of $r (t) = 〈 2 sin t, 5 t, 2 cos t 〉 .$

Find the curvature of $r (t) = \sqrt{2} t i + e^{t} j + e^{− t} k$

at point $P (0, 1, 1) .$

\frac{1}{2 \sqrt{2}}

At what point does the curve $y = e^{x}$

have maximum curvature?

What happens to the curvature as $x \to \infty$

for the curve $y = e^{x} ?$

The curvature approaches zero.

Find the point of maximum curvature on the curve $y = ln x .$

Find the equations of the normal plane and the osculating plane of the curve $r (t) = 〈 2 sin (3 t), t, 2 cos (3 t) 〉$

at point $(0, π, −2) .$

y = 6 x + π

and $x + 6 = 6 π$

Find equations of the osculating circles of the ellipse $4 y^{2} + 9 x^{2} = 36$

at the points $(2, 0)$

and $(0, 3) .$

Find the equation for the osculating plane at point $t = π / 4$

on the curve $r (t) = cos (2 t) i + sin (2 t) j + t .$

x + 2 z = \frac{π}{2}

Find the radius of curvature of $6 y = x^{3}$

at the point $(2, \frac{4}{3}) .$

Find the curvature at each point $(x, y)$

on the hyperbola $r (t) = 〈 a cosh (t), b sinh (t) 〉 .$

\frac{a^{4} b^{4}}{{(b^{4} x^{2} + a^{4} y^{2})}^{3 / 2}}

Calculate the curvature of the circular helix $r (t) = r sin (t) i + r cos (t) j + t k .$

Find the radius of curvature of $y = ln (x + 1)$

at point $(2, ln 3) .$

\frac{10 \sqrt{10}}{3}

Find the radius of curvature of the hyperbola $x y = 1$

at point $(1, 1) .$

A particle moves along the plane curve C described by $r (t) = t i + t^{2} j .$

Solve the following problems.

Find the length of the curve over the interval $[0, 2] .$

\frac{38}{3}

Find the curvature of the plane curve at $t = 0, 1, 2.$

Describe the curvature as t increases from $t = 0$

to $t = 2.$

The curvature is decreasing over this interval.

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).

[T] Use technology to graph the surface.

Find the curvature $κ$

of the generating curve as a function of x.

κ = \frac{6}{x^{2 / 5} (25 + 4 x^{6 / 5})}

[T] Use technology to graph the curvature function.

Arc Length and Curvature

Arc Length for Vector Functions

Arc-Length Parameterization

Curvature

Proof

The Normal and Binormal Vectors

Key Concepts

Key Equations

Glossary