How To Find Curvature If Given A Formula

Learning Objectives

• 3.three.1 Make up one's mind the length of a particle'due south path in infinite by using the arc-length function.
• iii.3.ii Explicate the meaning of the curvature of a curve in space and state its formula.
• 3.3.iii Describe the meaning of the normal and binormal vectors of a curve in space.

In this section, nosotros written report formulas related to curves in both two and three dimensions, and encounter how they are related to various properties of the aforementioned curve. For example, suppose a vector-valued function describes the motion of a particle in infinite. 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 information technology follows. Or, suppose that the vector-valued part describes a route we are edifice and we want to make up one's mind how sharply the road curves at a given signal. This is described past 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 part describes a bend in either two or three dimensions. Recall Alternative Formulas for Curvature, which states that the formula for the arc length of a curve defined by the parametric functions $x=x\left(t\right),y=y\left(t\right),t_{\mathrm{i}}\le t\le t_2$ is given by

$$s={\displaystyle {\int }_{t_1}^{t_{\mathrm{ii}}}\sqrt{{\left(x^{\prime }\left(t\right)\right)}^{\mathrm{ii}}+{\left(y^{\prime }\left(t\right)\right)}^{\mathrm{two}}}dt.}$$

In a similar fashion, if we ascertain a smooth curve using a vector-valued function $\text{r}\left(t\right)=f\left(t\right)\text{i}+\mathrm{1000}\left(t\right)\text{j},$ where $a\le t\le b,$ the arc length is given by the formula

$$s={\displaystyle {\int }_a^b\sqrt{{\left(f^{\prime }\left(t\right)\right)}^2+{\left({\mathrm{one\; thousand}}^{\prime }\left(t\right)\right)}^2}}dt.$$

In iii dimensions, if the vector-valued role is described by $\text{r}\left(t\right)=f\left(t\right)\text{i}+g\left(t\right)\text{j}+h\left(t\right)\text{1000}$ over the same interval $a\le t\le b,$ the arc length is given by

$$\mathrm{due\; south}={\displaystyle {\int }_a^b\sqrt{{\left(f^{\prime }\left(t\right)\right)}^2+{\left(m^{\prime }\left(t\right)\right)}^{\mathrm{two}}+{\left(h^{\prime }\left(t\right)\right)}^2}}dt.$$

Theorem 3.four

Arc-Length Formulas

1. Plane curve: Given a smoothen curve C defined past the function $\text{r}\left(t\right)=f\left(t\right)\text{i}+g\left(t\right)\text{j},$ where t lies within the interval $\left[a,b\right],$ the arc length of C over the interval is
   
   $$\mathrm{southward}={\displaystyle {\int }_a^b\sqrt{{\left[f^{\prime }\left(t\right)\right]}^2+{\left[g^{\prime }\left(t\right)\right]}^2}}dt={\displaystyle {\int }_a^b\Vert r^{\prime }\left(t\right)\Vert }dt.$$

   (3.11)

2. Space curve: Given a polish curve C defined by the function $\text{r}\left(t\right)=f\left(t\right)\text{i}+g\left(t\right)\text{j}+h\left(t\right)\text{1000},$ where t lies within the interval $\left[a,b\right],$ the arc length of C over the interval is
   
   $$s={\displaystyle {\int }_a^b\sqrt{{\left[f^{\prime }\left(t\right)\right]}^2+{\left[g^{\prime }\left(t\right)\right]}^{\mathrm{two}}+{\left[h^{\prime }\left(t\right)\right]}^2}dt}={\displaystyle {\int }_a^b\Vert r^{\prime }\left(t\right)\Vert }dt.$$

   (three.12)

The ii formulas are very like; they differ only in the fact that a space curve has 3 component functions instead of two. Note that the formulas are defined for smooth curves: curves where the vector-valued role $\text{r}(t)$ is differentiable with a non-null derivative. The smoothness condition guarantees that the curve has no cusps (or corners) that could make the formula problematic.

Example 3.9

Finding the Arc Length

Summate the arc length for each of the following vector-valued functions:

1. $\text{r}\left(t\right)=\left(\mathrm{iii}t-2\right)\text{i}+\left(4t+5\right)\text{j},1\le t\le \mathrm{v}$
2. $\text{r}\left(t\right)=\langle t\text{cos}t,t\text{sin}t,2t\rangle ,0\le t\le 2\pi$

Checkpoint iii.nine

Summate the arc length of the parameterized curve

$$\text{r}\left(t\right)=\langle 2t^2+1,2t^2-\mathrm{ane},t^{\mathrm{iii}}\rangle ,0\le t\le 3.$$

We at present return to the helix introduced before in this affiliate. A vector-valued function that describes a helix can be written in the form

$$\text{r}\left(t\right)=R\text{cos}\left(\frac{2\pi \mathrm{Due\; north}t}{h}\right)\text{i}+R\text{sin}\left(\frac{2\pi Nt}{h}\right)\text{j}+t\text{one thousand},0\le t\le h,$$

where R represents the radius of the helix, h represents the pinnacle (altitude between two consecutive turns), and the helix completes N turns. Let's derive a formula for the arc length of this helix using Equation three.12. Start of all,

$$r^{\prime }\left(t\right)=-\frac{\mathrm{ii}\pi NR}{h}\text{sin}\left(\frac{2\pi Nt}{h}\right)\text{i}+\frac{2\pi NR}{h}\text{cos}\left(\frac{2\pi Nt}{h}\right)\text{j}+\text{k}.$$

Therefore,

$$\begin{array}{cc}\hfill s& ={\displaystyle {\int }_a^b\Vert r^{\prime }\left(t\right)\Vert }dt\hfill \\ & ={\displaystyle {\int }_0^h\sqrt{{\left(-\frac{2\pi \mathrm{Due\; north}R}{h}\text{sin}\left(\frac{2\pi Nt}{h}\right)\right)}^2+{\left(\frac{\mathrm{two}\pi NR}{h}\text{cos}\left(\frac{2\pi Nt}{h}\right)\right)}^{\mathrm{ii}}+{\mathrm{ane}}^2}dt}\hfill \\ & ={\displaystyle {\int }_0^h\sqrt{\frac{4{\pi }^{\mathrm{ii}}{N}^{\mathrm{ii}}{R}^{\mathrm{two}}}{h^2}\left({\text{sin}}^{\mathrm{two}}\left(\frac{\mathrm{two}\pi \mathrm{Due\; north}t}{h}\right)+{\text{cos}}^2\left(\frac{\mathrm{two}\pi Nt}{h}\right)\right)+1}}dt\hfill \\ & ={\displaystyle {\int }_0^h\sqrt{\frac{\mathrm{four}{\pi }^2{\mathrm{Due\; north}}^2{R}^{\mathrm{two}}}{h^2}+1}}dt\hfill \\ & ={\left[t\sqrt{\frac{\mathrm{four}{\pi }^2{N}^{\mathrm{two}}{R}^{\mathrm{ii}}}{h^2}+1}\right]}_0^h\hfill \\ & =h\sqrt{\frac{4{\pi }^{\mathrm{two}}{\mathrm{Due\; north}}^2{R}^2+h^2}{h^{\mathrm{two}}}}\hfill \\ & =\sqrt{4{\pi }^2{\mathrm{Northward}}^2{R}^{\mathrm{two}}+h^{\mathrm{two}}}.\hfill \end{array}$$

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 at present have a formula for the arc length of a curve defined by a vector-valued office. Allow'southward 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 office 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 straight from the formula for arc length:

$$s\left(t\right)={\displaystyle {\int }_a^t\sqrt{{\left(f^{\prime }\left(u\right)\right)}^2+{\left(g^{\prime }\left(u\right)\right)}^2+{\left(h^{\prime }\left(u\right)\right)}^{\mathrm{two}}}}du.$$

(3.xiii)

If the bend is in ii dimensions, so only two terms appear nether the foursquare root within the integral. The reason for using the independent variable u is to distinguish between fourth dimension and the variable of integration. Since $\mathrm{southward}\left(t\right)$ measures distance traveled as a function of time, $s^{\prime }\left(t\right)$ measures the speed of the particle at any given fourth dimension. Since we have a formula for $s\left(t\right)$ in Equation 3.13, we tin differentiate both sides of the equation:

$$\begin{array}{cc}\hfill s^{\prime }\left(t\right)& =\frac{d}{dt}\left[{\displaystyle {\int }_a^t\sqrt{{\left(f^{\prime }\left(u\right)\right)}^{\mathrm{ii}}+{\left(m^{\prime }\left(u\right)\right)}^{\mathrm{two}}+{\left(h^{\prime }\left(u\right)\right)}^2}}du\right]\hfill \\ & =\frac{d}{dt}\left[{\displaystyle {\int }_a^t\Vert r^{\prime }\left(u\right)\Vert du}\right]\hfill \\ & =\Vert r^{\prime }\left(t\right)\Vert .\hfill \end{array}$$

If we assume that $\text{r}\left(t\right)$ defines a smooth bend, and so the arc length is always increasing, so $s^{\prime }\left(t\right)>0$ for $t>a.$ Last, if $\text{r}\left(t\right)$ is a curve on which $\Vert r^{\prime }\left(t\right)\Vert =1$ for all t, then

$$s\left(t\right)={\displaystyle {\int }_a^t\Vert r^{\prime }\left(u\right)\Vert du}={\displaystyle {\int }_a^{t}1du}=t-a,$$

which means that t represents the arc length every bit long equally $a=0.$

Theorem three.5

Arc-Length Function

Let $\text{r}\left(t\right)$ depict a shine curve for $t\ge a.$ And then the arc-length function is given by

$$\mathrm{south}\left(t\right)={\displaystyle {\int }_a^t\Vert r^{\prime }\left(u\right)\Vert du}.$$

(3.fourteen)

Furthermore, $\frac{d\mathrm{southward}}{dt}=\Vert r^{\prime }\left(t\right)\Vert >0.$ If $\Vert r^{\prime }\left(t\right)\Vert =1$ for all $t\ge a,$ then the parameter t represents the arc length from the starting point at $t=a.$

A useful application of this theorem is to discover an alternative parameterization of a given curve, called an arc-length parameterization. Recall that any vector-valued role can be reparameterized via a change of variables. For example, if we accept a role $\text{r}\left(t\right)=\langle 3\text{cos}t,3\text{sin}t\rangle ,0\le t\le \mathrm{ii}\pi$ that parameterizes a circle of radius iii, we can alter the parameter from t to $4t,$ obtaining a new parameterization $\text{r}\left(t\right)=\langle 3\text{cos}4t,3\text{sin}4t\rangle .$ The new parameterization still defines a circle of radius 3, but at present we demand merely utilize the values $0\le t\le \pi \text{/}2$ to traverse the circumvolve once.

Suppose that nosotros observe the arc-length function $\mathrm{southward}\left(t\right)$ and are able to solve this function for t every bit a function of south. We tin and so reparameterize the original function $\text{r}\left(t\right)$ by substituting the expression for t back into $\text{r}\left(t\right).$ The vector-valued function is now written in terms of the parameter south. Since the variable s represents the arc length, we call this an arc-length parameterization of the original function $\text{r}\left(t\right).$ One advantage of finding the arc-length parameterization is that the distance traveled along the curve starting from $s=0$ is at present equal to the parameter south. The arc-length parameterization also appears in the context of curvature (which we examine later in this section) and line integrals, which we report in the Introduction to Vector Calculus.

Case 3.ten

Finding an Arc-Length Parameterization

Detect the arc-length parameterization for each of the post-obit curves:

1. $\text{r}\left(t\right)=4\text{cos}t\text{i}+4\text{sin}t\text{j},t\ge 0$
2. $\text{r}\left(t\right)=\langle t+\mathrm{iii},2t-4,\mathrm{two}t\rangle ,t\ge 3$

Checkpoint 3.10

Find the arc-length function for the helix

$$\text{r}\left(t\right)=\langle 3\text{cos}t,3\text{sin}t,4t\rangle ,t\ge 0.$$

Then, use the human relationship between the arc length and the parameter t to detect an arc-length parameterization of $\text{r}\left(t\right).$

Curvature

An of import 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 bicycle to stay on the route. Now suppose the radius is smaller. In this case you lot would demand to turn more sharply to stay on the road. In the example 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 information technology is tangent to the bend at that bespeak and "hugs" the curve as closely as possible in a neighborhood of the point (Figure 3.6). The curvature of the graph at that signal is so defined to exist the same as the curvature of the inscribed circumvolve.

Figure 3.six The graph represents the curvature of a function $y=f\left(x\right).$ The sharper the turn in the graph, the greater the curvature, and the smaller the radius of the inscribed circumvolve.

Definition

Let C be a smooth bend in the plane or in infinite given by $\text{r}\left(s\right),$ where $s$ is the arc-length parameter. The curvature $\kappa$ at southward is

$$\kappa =\Vert \frac{d\text{T}}{ds}\Vert =\Vert T^{\prime }\left(s\right)\Vert .$$

The formula in the definition of curvature is not very useful in terms of adding. In detail, recall that $\text{T}\left(t\right)$ represents the unit tangent vector to a given vector-valued function $\text{r}\left(t\right),$ and the formula for $\text{T}\left(t\right)$ is $\text{T}\left(t\right)=\frac{r^{\prime }\left(t\right)}{\Vert r^{\prime }\left(t\right)\Vert }.$ To use the formula for curvature, it is first necessary to limited $\text{r}\left(t\right)$ in terms of the arc-length parameter due south, and so notice the unit of measurement tangent vector $\text{T}\left(\mathrm{southward}\right)$ for the function $\text{r}\left(s\right),$ then take the derivative of $\text{T}\left(\mathrm{due\; south}\right)$ with respect to s. This is a tedious process. Fortunately, there are equivalent formulas for curvature.

Theorem 3.six

Alternative Formulas for Curvature

If C is a smooth curve given by $\text{r}\left(t\right),$ then the curvature $\kappa$ of C at t is given by

$$\kappa =\frac{\Vert T^{\prime }\left(t\right)\Vert }{\Vert r^{\prime }\left(t\right)\Vert }.$$

(3.15)

If C is a three-dimensional curve, then the curvature can exist given by the formula

$$\kappa =\frac{\Vert r^{\prime }\left(t\right)\times \text{r″}\left(t\right)\Vert }{{\Vert r^{\prime }\left(t\right)\Vert }^{\mathrm{iii}}}.$$

(iii.sixteen)

If C is the graph of a function $y=f\left(\mathrm{10}\right)$ and both $y^{\prime }$ and $y\text{″}$ exist, and so the curvature $\kappa$ at indicate $\left(\mathrm{10},y\right)$ is given past

$$\kappa =\frac{\left|y\text{″}\right|}{{\left[1+{\left(y^{\prime }\right)}^2\right]}^{3\text{/}2}}.$$

(iii.17)

Proof

The offset formula follows directly from the chain dominion:

$$\frac{d\text{T}}{dt}=\frac{d\text{T}}{ds}\frac{ds}{dt},$$

where s is the arc length forth the bend C. Dividing both sides by $ds\text{/}dt,$ and taking the magnitude of both sides gives

$$\Vert \frac{d\text{T}}{ds}\Vert =\Vert \frac{T^{\prime }\left(t\right)}{\frac{d\mathrm{south}}{dt}}\Vert .$$

Since $ds\text{/}dt=\Vert r^{\prime }\left(t\right)\Vert ,$ this gives the formula for the curvature $\kappa$ of a curve C in terms of any parameterization of C:

$$\kappa =\frac{\Vert T^{\prime }\left(t\right)\Vert }{\Vert r^{\prime }\left(t\right)\Vert }.$$

In the case of a 3-dimensional curve, nosotros start with the formulas $\text{T}\left(t\right)=\left(r^{\prime }\left(t\right)\right)\text{/}\Vert r^{\prime }\left(t\right)\Vert$ and $ds\text{/}dt=\Vert r^{\prime }\left(t\right)\Vert .$ Therefore, $r^{\prime }\left(t\right)=\left(ds\text{/}dt\right)\text{T}\left(t\right).$ We can take the derivative of this function using the scalar product formula:

$$\text{r″}\left(t\right)=\frac{d^{2}s}{d{t}^2}\text{T}\left(t\right)+\frac{d\mathrm{due\; south}}{dt}{T}^{\prime }\left(t\right).$$

Using these final 2 equations we get

$$\begin{array}{cc}\hfill r^{\prime }\left(t\right)\times \text{r″}\left(t\right)& =\frac{d\mathrm{southward}}{dt}\text{T}\left(t\right)\times \left(\frac{d^{\mathrm{two}}\mathrm{due\; south}}{d{t}^2}\text{T}\left(t\right)+\frac{ds}{dt}{T}^{\prime }\left(t\right)\right)\hfill \\ & =\frac{ds}{dt}\frac{d^2\mathrm{due\; south}}{d{t}^{\mathrm{two}}}\text{T}\left(t\right)\times \text{T}\left(t\right)+{\left(\frac{d\mathrm{southward}}{dt}\right)}^2\text{T}\left(t\right)\times T^{\prime }\left(t\right).\hfill \end{array}$$

Since $\text{T}\left(t\right)\times \text{T}\left(t\right)=0,$ this reduces to

$$r^{\prime }\left(t\right)\times \text{r″}\left(t\right)={\left(\frac{ds}{dt}\right)}^{\mathrm{ii}}\text{T}\left(t\right)\times T^{\prime }\left(t\right).$$

Since $T^{\prime }$ is parallel to $\text{N},$ and $\text{T}$ is orthogonal to $\text{N},$ it follows that $\text{T}$ and $T^{\prime }$ are orthogonal. This means that $\Vert \text{T}\times T^{\prime }\Vert =\Vert \text{T}\Vert \Vert T^{\prime }\Vert \text{sin}\left(\pi \text{/}\mathrm{ii}\right)=\Vert T^{\prime }\Vert ,$ and so

$$\Vert {\mathbf{r}}^{\mathbf{\text{'}}}\left(t\right)\times \mathbf{\text{r″}}\left(t\right)\Vert ={\left(\frac{ds}{dt}\right)}^2\Vert {\mathbf{T}}^{\mathbf{\text{'}}}\left(t\right)\Vert .$$

Now we solve this equation for $\Vert T^{\prime }\left(t\right)\Vert$ and use the fact that $ds\text{/}dt=\Vert r^{\prime }\left(t\right)\Vert \text{:}$

$$\Vert T^{\prime }\left(t\right)\Vert =\frac{\Vert r^{\prime }\left(t\right)\times \text{r″}\left(t\right)\Vert }{{\Vert r^{\prime }\left(t\right)\Vert }^2}.$$

Then, nosotros divide both sides past $\Vert r^{\prime }\left(t\right)\Vert .$ This gives

$$\kappa =\frac{\Vert T^{\prime }\left(t\right)\Vert }{\Vert r^{\prime }\left(t\right)\Vert }=\frac{\Vert r^{\prime }\left(t\right)\times \text{r″}\left(t\right)\Vert }{{\Vert r^{\prime }\left(t\right)\Vert }^3}.$$

This proves Equation 3.xvi. To prove Equation iii.17, we start with the assumption that curve C is defined by the office $y=f\left(x\right).$ Then, we can ascertain $\text{r}\left(t\right)=\mathrm{10}\text{i}+f\left(x\right)\text{j}+0\text{k}.$ Using the previous formula for curvature:

$$\begin{array}{cc}\hfill r^{\prime }\left(t\right)& =\text{i}+f^{\prime }\left(x\right)\text{j}\hfill \\ \hfill \text{r″}\left(t\right)& =f\text{″}\left(\mathrm{ten}\right)\text{j}\hfill \\ \hfill r^{\prime }\left(t\right)\times \text{r″}\left(t\right)& =\left|\begin{array}{ccc}\text{i}& \text{j}& \text{k}\\ 1& f^{\prime }\left(x\right)& 0\\ 0& f\text{″}\left(\mathrm{ten}\right)& 0\end{array}\right|=f\text{″}\left(\mathrm{10}\right)\text{grand}.\hfill \end{array}$$

Therefore,

$$\kappa =\frac{\Vert r^{\prime }\left(t\right)\times \text{r″}\left(t\right)\Vert }{{\Vert r^{\prime }\left(t\right)\Vert }^3}=\frac{\left|f\text{″}\left(\mathrm{10}\right)\right|}{{\left(\mathrm{ane}+[{(f^{\prime }(x))}^{\mathrm{ii}}]\right)}^{3\text{/}2}}.$$

□

Example three.11

Finding Curvature

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

1. $\text{r}\left(t\right)=4\text{cos}t\text{i}+\mathrm{four}\text{sin}t\text{j}+3t\text{yard},t=\frac{4\pi }{3}$
2. $f\left(\mathrm{ten}\right)=\sqrt{\mathrm{iv}\mathrm{ten}-{\mathrm{ten}}^2},\mathrm{ten}=\mathrm{two}$

Checkpoint 3.11

Notice the curvature of the curve divers by the function

$$y=3x^2-2\mathrm{ten}+4$$

at the point $\mathrm{ten}=2.$

The Normal and Binormal Vectors

We accept seen that the derivative $r^{\prime }\left(t\right)$ of a vector-valued function is a tangent vector to the curve defined by $\text{r}\left(t\right),$ and the unit of measurement tangent vector $\text{T}\left(t\right)$ tin be calculated by dividing $r^{\prime }\left(t\right)$ past its magnitude. When studying move in iii dimensions, 2 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.

Definition

Permit C be a three-dimensional polish curve represented by r over an open interval I. If $T^{\prime }\left(t\right)\ne 0,$ then the principal unit of measurement normal vector at t is defined to exist

$$\text{Northward}\left(t\right)=\frac{T^{\prime }\left(t\right)}{\Vert T^{\prime }\left(t\right)\Vert }.$$

(3.18)

The binormal vector at t is divers as

$$\text{B}\left(t\right)=\text{T}\left(t\right)\times \text{North}\left(t\right),$$

(3.xix)

where $\text{T}\left(t\right)$ is the unit tangent vector.

Note that, past definition, the binormal vector is orthogonal to both the unit tangent vector and the normal vector. Furthermore, $\text{B}\left(t\right)$ is e'er a unit vector. This can be shown using the formula for the magnitude of a cross product

$$\Vert \text{B}\left(t\right)\Vert =\Vert \text{T}\left(t\right)\times \text{N}\left(t\right)\Vert =\Vert \text{T}\left(t\right)\Vert \Vert \text{N}\left(t\right)\Vert \text{sin}\theta ,$$

where $\theta$ is the angle between $\text{T}\left(t\right)$ and $\text{Northward}\left(t\right).$ Since $\text{Due north}\left(t\right)$ is the derivative of a unit of measurement vector, property (vii) of the derivative of a vector-valued function tells us that $\text{T}\left(t\right)$ and $\text{N}\left(t\right)$ are orthogonal to each other, so $\theta =\pi \text{/}\mathrm{ii}.$ Furthermore, they are both unit vectors, and then their magnitude is i. Therefore, $\Vert \text{T}\left(t\right)\Vert \Vert \text{N}\left(t\right)\Vert \text{sin}\theta =\left(\mathrm{i}\right)\left(1\right)\text{sin}\left(\pi \text{/}\mathrm{ii}\right)=1$ and $\text{B}\left(t\right)$ is a unit vector.

The main unit of measurement normal vector can be challenging to summate considering the unit tangent vector involves a quotient, and this quotient ofttimes has a square root in the denominator. In the three-dimensional case, finding the cantankerous product of the unit of measurement tangent vector and the unit normal vector can be even more cumbersome. Fortunately, nosotros have alternative formulas for finding these ii vectors, and they are presented in Motion in Infinite.

Instance three.12

Finding the Main Unit Normal Vector and Binormal Vector

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

1. $\text{r}\left(t\right)=\mathrm{four}\text{cos}t\text{i}-4\text{sin}t\text{j}$
2. $\text{r}\left(t\right)=\left(6t+2\right)\text{i}+5t^2\text{j}-8t\text{1000}$

Checkpoint 3.12

Observe the unit normal vector for the vector-valued function $\text{r}\left(t\right)=\left(t^{\mathrm{ii}}-3t\right)\text{i}+\left(4t+1\right)\text{j}$ and evaluate it at $t=\mathrm{ii.}$

For any smooth bend in 3 dimensions that is defined by a vector-valued office, we now take 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 bend at any point on the curve, chosen the normal airplane. In add-on, these three vectors form a frame of reference in three-dimensional infinite called the Frenet frame of reference (also called the TNB frame) (Figure 3.7). Lat, the plane determined by the vectors T and Northward forms the osculating plane of C at any betoken P on the curve.

Figure three.7 This effigy depicts a Frenet frame of reference. At every point P on a three-dimensional curve, the unit tangent, unit normal, and binormal vectors form a iii-dimensional frame of reference.

Suppose we form a circle in the osculating plane of C at point P on the bend. Assume that the circle has the same curvature as the bend does at bespeak P and allow the circle accept radius r. Then, the curvature of the circumvolve is given by $1\text{/}r.$ Nosotros 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 bend and is tangent to the curve at point P, so this circle is chosen the osculating circle of C at P, as shown in the following effigy.

Effigy 3.eight In this osculating circle, the circle is tangent to curve C at indicate P and shares the aforementioned curvature.

To detect the equation of an osculating circle in 2 dimensions, nosotros need find only the center and radius of the circle.

Case iii.13

Finding the Equation of an Osculating Circle

Find the equation of the osculating circle of the helix defined by the part $y=x^3-\mathrm{three}\mathrm{10}+1$ at $\mathrm{10}=\mathrm{one.}$

Checkpoint 3.13

Detect the equation of the osculating circumvolve of the curve defined by the vector-valued function $y=2{\mathrm{10}}^2-4x+5$ at $\mathrm{10}=1.$

Section 3.3 Exercises

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

102 .

$\text{r}(t)=t^2\text{i}+14t\text{j},0\le t\le 7.$ This portion of the graph is shown here:

103.

$\text{r}(t)=t^{\mathrm{two}}\text{i}+(2t^2+1)\text{j},\mathrm{one}\le t\le 3$

104 .

$\text{r}(t)=\langle 2\text{sin}t,5t,2\text{cos}t\rangle ,0\le t\le \pi .$ This portion of the graph is shown here:

105.

$\text{r}(t)=\langle t^2+1,4t^{\mathrm{iii}}+3\rangle ,-\mathrm{one}\le t\le 0$

106 .

$\text{r}(t)=\langle e^{\text{−}t}\text{cos}t,{\mathrm{eastward}}^{\text{−}t}\text{sin}t\rangle$ over the interval $\left[0,\frac{\pi }{2}\right].$ Here is the portion of the graph on the indicated interval:

107.

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

108 .

Observe the arc length of the vector-valued office $\text{r}(t)=-t\text{i}+4t\text{j}+3t\text{k}$ over $[0,\mathrm{i}].$

109.

A particle travels in one case around a circle with the equation of motility $\text{r}(t)=\mathrm{iii}\text{cos}t\text{i}+3\text{sin}t\text{j}+0\text{chiliad}.$ Notice the distance traveled around the circle by the particle.

110 .

Set an integral to observe the circumference of the ellipse with the equation $\text{r}(t)=\text{cos}t\text{i}+\mathrm{ii}\text{sin}t\text{j}+0\text{one thousand}.$

111.

Find the length of the curve $\text{r}(t)=\langle \sqrt{2}t,{\mathrm{due\; east}}^t,e^{\text{−}t}\rangle$ over the interval $0\le t\le 1.$ The graph is shown here:

112 .

Find the length of the curve $\text{r}(t)=\langle 2\text{sin}t,5t,\mathrm{ii}\text{cos}t\rangle$ for $t\in \left[\mathrm{-10},10\right].$

113.

The position part 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 $t=0.$

114 .

Given $\text{r}(t)=a\text{cos}(\omega t)\text{i}+b\text{sin}(\omega t)\text{j},$ observe the binormal vector $\text{B}(0).$

115.

Given $\text{r}(t)=\langle 2{\mathrm{east}}^t,{\mathrm{east}}^t\text{cos}t,e^t\text{sin}t\rangle ,$ determine the tangent vector $\text{T}(t).$

116 .

Given $\text{r}(t)=\langle 2e^t,e^t\text{cos}t,{\mathrm{east}}^t\text{sin}t\rangle ,$ determine the unit tangent vector $\text{T}(t)$ evaluated at $t=0.$

117.

Given $\text{r}(t)=\langle 2e^t,{\mathrm{due\; east}}^t\text{cos}t,{\mathrm{due\; east}}^t\text{sin}t\rangle ,$ observe the unit of measurement normal vector $\text{N}(t)$ evaluated at $t=0,$ $\text{Northward}(0).$

118 .

Given $\text{r}(t)=\langle 2e^t,{\mathrm{eastward}}^t\text{cos}t,e^t\text{sin}t\rangle ,$ find the unit binormal vector evaluated at $t=0.$

119.

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 hither:

120 .

Find the unit tangent vector $\text{T}(t)$ and unit normal vector $\text{N}(t)$ at $t=0$ for the plane curve $\text{r}(t)=\langle t^3-\mathrm{iv}t,\mathrm{five}{t}^{\mathrm{two}}-\mathrm{ii}\rangle .$ The graph is shown here:

121.

Find the unit of measurement tangent vector $\text{T}(t)$ for $\text{r}(t)=3t\text{i}+5t^2\text{j}+2t\text{chiliad}$

122 .

Find the principal normal vector to the curve $\text{r}(t)=\langle \mathrm{half-dozen}\text{cos}t,6\text{sin}t\rangle$ at the betoken determined past $t=\pi \text{/}3.$

123.

Find $\text{T}(t)$ for the curve $\text{r}(t)=\left(t^{\mathrm{three}}-4t\right)\text{i}+\left(5t^2-2\right)\text{j}.$

124 .

Find $\text{N}(t)$ for the bend $\text{r}(t)=\left(t^3-4t\right)\text{i}+\left(5t^2-2\right)\text{j}.$

125.

Notice the unit normal vector $\text{Northward}(t)$ for $\text{r}(t)=\langle \mathrm{ii}\text{sin}t,5t,\mathrm{ii}\text{cos}t\rangle .$

126 .

Find the unit tangent vector $\text{T}(t)$ for $\text{r}(t)=\langle \mathrm{two}\text{sin}t,\mathrm{v}t,\mathrm{two}\text{cos}t\rangle .$

127.

Notice the arc-length function $\mathrm{south}(t)$ for the line segment given by $\text{r}(t)=\langle 3-3t,4t\rangle .$ Write r as a parameter of s.

128 .

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 $t=0.$

129.

Parameterize the curve using the arc-length parameter s, at the point at which $t=0$ for $\text{r}(t)=e^t\text{sin}t\text{i}+{\mathrm{eastward}}^t\text{cos}t\text{j}.$

130 .

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.)

131.

Notice the x-coordinate at which the curvature of the curve $y=1\text{/}x$ is a maximum value.

132 .

Find the curvature of the curve $\text{r}(t)=\mathrm{v}\text{cos}t\text{i}+5\text{sin}t\text{j}.$ Does the curvature depend upon the parameter t?

133.

Find the curvature $\kappa$ for the bend $y=\mathrm{10}-\frac{1}{4}{\mathrm{10}}^{\mathrm{ii}}$ at the point $\mathrm{10}=2.$

134 .

Find the curvature $\kappa$ for the curve $y=\frac{1}{\mathrm{iii}}{x}^{\mathrm{three}}$ at the indicate $x=\mathrm{i.}$

135.

Find the curvature $\kappa$ of the curve $\text{r}(t)=t\text{i}+\mathrm{six}{t}^{\mathrm{two}}\text{j}+4t\text{1000}.$ The graph is shown hither:

136 .

Find the curvature of $\text{r}(t)=\langle 2\text{sin}t,5t,\mathrm{two}\text{cos}t\rangle .$

137.

Find the curvature of $\text{r}(t)=\sqrt{2}t\text{i}+e^t\text{j}+e^{\text{−}t}\text{k}$ at signal $P\left(0,1,1\right).$

138 .

At what point does the curve $y=e^x$ have maximum curvature?

139.

What happens to the curvature as $x\to \infty$ for the curve $y={\mathrm{eastward}}^{\mathrm{10}}?$

140 .

Notice the point of maximum curvature on the bend $y=\text{ln}x.$

141.

Find the equations of the normal plane and the osculating aeroplane of the curve $\text{r}(t)=\langle 2\text{sin}(\mathrm{three}t),t,2\text{cos}(3t)\rangle$ at bespeak $\left(0,\pi ,\mathrm{-2}\right).$

142 .

Find equations of the osculating circles of the ellipse $4y^2+9{\mathrm{ten}}^2=36$ at the points $(2,0)$ and $(0,3).$

143.

Detect the equation for the osculating plane at point $t=\pi \text{/}4$ on the curve $\text{r}(t)=\text{cos}(2t)\text{i}+\text{sin}(\mathrm{two}t)\text{j}+t\text{chiliad}.$

144 .

Observe the radius of curvature of $6y={\mathrm{10}}^3$ at the indicate $\left(2,\frac{4}{3}\right).$

145.

Find the curvature at each point $\left(\mathrm{10},y\right)$ on the hyperbola $\text{r}(t)=\langle a\text{cosh}(t),b\text{sinh}(t)\rangle .$

146 .

Calculate the curvature of the circular helix $\text{r}(t)=r\text{sin}(t)\text{i}+r\text{cos}(t)\text{j}+t\text{m}.$

147.

Find the radius of curvature of $y=\text{ln}(\mathrm{10}+1)$ at bespeak $\left(2,\text{ln}3\right).$

148 .

Notice the radius of curvature of the hyperbola $\mathrm{10}y=\mathrm{ane}$ at betoken $(1,\mathrm{one}).$

A particle moves along the airplane bend C described by $\text{r}(t)=t\text{i}+t^{\mathrm{ii}}\text{j}.$ Solve the following problems.

149.

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

150 .

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

151.

Depict the curvature as t increases from $t=0$ to $t=2.$

The surface of a large cup is formed past revolving the graph of the function $y=0.25{\mathrm{ten}}^{\mathrm{ane.half-dozen}}$ from $\mathrm{ten}=0$ to $\mathrm{ten}=5$ about the y-axis (measured in centimeters).

152 .

[T] Employ technology to graph the surface.

153.

Detect the curvature $\kappa$ of the generating curve as a function of x.

154 .

[T] Apply technology to graph the curvature function.

Source: https://openstax.org/books/calculus-volume-3/pages/3-3-arc-length-and-curvature

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