Vector Functions Arc Length and Curvature Objectives n
Vector Functions
Arc Length and Curvature
Objectives n Find the arc length of a space curve. n Use the arc length parameter to describe a plane curve or space curve. n Find the curvature of a curve at a point on the curve. n Use a vector-valued function to find frictional force. 3
Arc Length 4
Arc Length and Curvature We have defined the length of a plane curve with parametric equations x = f (t), y = g(t), a t b, as the limit of lengths of inscribed polygons and, for the case where f ' and g' are continuous, we arrived at the formula The length of a space curve is defined in exactly the same way (see Figure 1). Figure 1 The length of a space curve is the limit of lengths of inscribed polygons. 5
Arc Length and Curvature Suppose that the curve has the vector equation, r(t) = f (t), g(t), h(t) , a t b, or, equivalently, the parametric equations x = f (t), y = g(t), z = h(t), where f ', g', and h' are continuous. If the curve is traversed exactly once as t increases from a to b, then it can be shown that its length is 6
Arc Length and Curvature Notice that both of the arc length formulas (1) and (2) can be put into the more compact form because, for plane curves r(t) = f (t) i + g(t) j, and for space curves r(t) = f (t) i + g(t) j + h(t) k, 7
Example 1 Find the length of the arc of the circular helix with vector equation r(t) = cos t i + sin t j + t k from the point (1, 0, 0) to the point (1, 0, 2 ). Solution: Since r'(t) = –sin t i + cos t j + k, we have The arc from (1, 0, 0) to (1, 0, 2 ) is described by the parameter interval 0 t 2 and so, from Formula 3, we have 8
Example 2 – Solution integration with steps 11
Example 2 – Solution integration with steps 12
Example 2 – Solution integration with steps 13
Example 2 – Solution integration with steps 14
Example 2 – Solution integration with steps Plug in solved integrals: 15
Example 2 – Solution integration with steps The problem is solved. Apply the absolute value function to arguments of logarithm functions in order to extend the antiderivative's domain: 16
Example 2 – Solution integration with steps 17
Example 18
continue 19
Continu 20
Example: the Arc Length of a Curve Metal posts have been installed 6 m apart across a gorge. Find the length for the hanging bridge that follows the curve: f(x) = 5 cosh(x/5) Solution Here is the actual curve: Let us solve the general case first! A hanging cable forms a curve called a catenary: f(x) = a cosh(x/a) 21
Solution Larger values of a have less sag in the middle And "cosh" is the hyperbolic cosine function. The derivative is: f’(x) = sinh(x/a) The curve is symmetrical, so it is easier to work on just half of the catenary, from the center to an end at "b": Start with: Put in f’(x) = sinh(x/a): Use the identity 1 + sinh 2(x/a) = cosh 2(x/a): Simplify: 22
Solution alculate the Integral: S = a sinh(b/a) Now, remembering the symmetry, let's go from −b to +b: S = 2 a sinh(b/a) In our specific case a=5 and the 6 m span goes from − 3 to +3 S = 2× 5 sinh(3/5) = 6. 367 m (to nearest mm) This is important to know! If we build it exactly 6 m in length there is no way we could pull it hard enough for it to meet the posts. But at 6. 367 m it will work nicely. 23
Arc Length Parameter 24
Arc Length and Curvature A single curve C can be represented by more than one vector function. For instance, the twisted cubic r 1(t) = t, t 2, t 3 1 t 2 could also be represented by the function r 2(u) = eu, e 2 u, e 3 u 0 u ln 2 where the connection between the parameters t and u is given by t = eu. We say that Equations 4 and 5 are parametrizations of the curve C. 25
Arc Length and Curvature If we were to use Equation 3 to compute the length of C using Equations 4 and 5, we would get the same answer. In general, it can be shown that when Equation 3 is used to compute arc length, the answer is independent of the parametrization that is used. Now we suppose that C is a curve given by a vector function: r(t) = f (t) i + g(t) j + h(t) k a t b where r' is continuous and C is traversed exactly once as t increases from a to b. 26
Arc Length and Curvature We define its arc length function s by Thus s(t) is the length of the part of C between r(a) and r(t). (See Figure 3. ) Figure 3 27
Arc Length and Curvature If we differentiate both sides of Equation 6 using Part 1 of the Fundamental Theorem of Calculus, we obtain It is often useful to parametrize a curve with respect to arc length because arc length arises naturally from the shape of the curve and does not depend on a particular coordinate system. 28
Arc Length and Curvature If a curve r(t) is already given in terms of a parameter t and s(t) is the arc length function given by Equation 6, then we may be able to solve for t as a function of s: t = t(s). Then the curve can be reparametrized in terms of s by substituting for t: r = r(t(s)). Thus, if s = 3 for instance, r(t(3)) is the position vector of the point 3 units of length along the curve from its starting point. 29
Example 3 – Finding the Arc Length Function for a Line Find the arc length function s(t) for the line segment given by: r(t) = (3 – 3 t)i + 4 t j, 0≤t≤ 1 and write r as a function of the parameter s. (See Figure 12. 31. ) Figure 12. 31 30
Example 3 – Solution Because r′(t) = – 3 i + 4 j and you have 31
Example 3 – Solution cont’d Using s = 5 t (or t = s/5), you can rewrite r using the arc length parameter as follows. r(t) = (3 – 3 t)i + 4 t j, 0 ≤ t ≤ 1 32
Determine the length of the curve Solution We will first need the tangent vector and its magnitude. The length is then, 33
Curvature 34
Curvature You can calculate curvature by calculating the magnitude of the rate of change of the unit tangent vector T with respect to the arc length s, as shown in Figure 12. 33 35
Curvatures A parametrization r(t) is called smooth on an interval I if r' is continuous and r'(t) 0 on I. A curve is called smooth if it has a smooth parametrization. A smooth curve has no sharp corners or cusps; when the tangent vector turns, it does so continuously. If C is a smooth curve defined by the vector function r, recall that the unit tangent vector T(t) is given by and indicates the direction of the curve. 36
Curvatures From Figure 4 you can see that T(t) changes direction very slowly when C is fairly straight, but it changes direction more quickly when C bends or twists more sharply. Figure 4 Unit tangent vectors at equally spaced points on C 37
Curvatures The curvature of C at a given point is a measure of how quickly the curve changes direction at that point. Specifically, we define it to be the magnitude of the rate of change of the unit tangent vector with respect to arc length. (We use arc length so that the curvature will be independent of the parametrization. ) 38
Curvatures The curvature is easier to compute if it is expressed in terms of the parameter t instead of s, so we use the Chain Rule to write But ds/dt = | r'(t) | from Equation 7, so 39
Example 3 Show that the curvature of a circle of radius a is 1/a. Solution: We can take the circle to have center the origin, and then a parametrization is: r(t) = a cos t i + a sin t j Therefore r'(t) = –a sin t i + a cos t j so and | r'(t) | = a and T'(t) = –cos t i – sin t j 40
Example 3 – Solution cont’d This gives | T'(t)| = 1, so using Equation 9, we have 41
Solution: Back in the section when we introduced the tangent vector we computed the tangent and unit tangent vectors for this function. These were, The derivative of the unit tangent is, The magnitudes of the two vectors are, The curvature is then, In this case the curvature is constant. This means that the curve is changing direction at the same rate at every point along it. Recalling that this curve is a helix this result makes sense. 42
Curvature The result of Example 3 shows that small circles have large curvature and large circles have small curvature, in accordance with our intuition. We can see directly from the definition of curvature that the curvature of a straight line is always 0 because the tangent vector is constant. Although Formula 9 can be used in all cases to compute the curvature, the formula given by the following theorem is often more convenient to apply. 43
Curvature For the special case of a plane curve with equation y = f (x), we choose x as the parameter and write r(x) = x i + f (x) j. Then r'(x) = i + f '(x) j and r''(x) = f ''(x) j. Since i j = k and j j = 0, it follows that r'(x) r''(x) = f ''(x) k. We also have 10, and so, by Theorem 44
Example 4 – Finding the Curvature of a Circle Show that the curvature of a circle of radius r is K = 1/r. Solution: Without loss of generality you can consider the circle to be centered at the origin. Let (x, y) be any point on the circle and let s be the length of the arc from (r, 0) to (x, y) as shown in Figure 12. 34 45
Example 4 – Solution cont’d By letting be the central angle of the circle, you can represent the circle by r( ) = r cos i + r sin j. is the parameter. Using the formula for the length of a circular arc s = r , you can rewrite r( ) in terms of the arc length parameter as follows: 46
Example 4 – Solution So, , and it follows that which implies that the unit tangent vector is cont’d , and the curvature is given by at every point on the circle. 47
Example 5 – Finding the Curvature of a Space Curve Find the curvature of the curve given by Solution: It is not apparent whether this parameter represents arc length, so you should use the formula: r′(t) = 2 i + 2 t j – t 2 k 48
Example 5 – Solution cont’d Therefore, 49
Example 6 – Finding Curvature in Rectangular Coordinates Find the curvature of the parabola given by at x = 2. Sketch the circle of curvature at (2, 1). Solution: The curvature at x = 2 is as follows. 50
Example 6 – Solution cont’d Because the curvature at P(2, 1) is , it follows that the radius of the circle of curvature at that point is 2. So, the center of curvature is (2, – 1) as shown in Figure 12. 37. [In the figure, note that the curve has the greatest curvature at P. ] Figure 12. 37 51
Curvature 52
Example 7 – Tangential and Normal Components of Acceleration Find a. T and a. N for the curve given by: Solution: From Example 5, you know that Therefore, and 53
The Normal and Binormal Vectors 54
The Normal and Binormal Vectors At a given point on a smooth space curve r(t), there are many vectors that are orthogonal to the unit tangent vector T(t). We single out one by observing that, because | T(t) | = 1 for all t, we have T(t) T'(t) = 0, so T'(t) is orthogonal to T(t). Note that T'(t) is itself not a unit vector. But at any point where 0 we can define the principal unit normal vector N(t) (or simply unit normal) as: 55
The Normal and Binormal Vectors The vector B(t) = T(t) N(t) is called the binormal vector. It is perpendicular to both T and N and is also a unit vector. (See Figure 6. ) Figure 6 56
Example 6 Find the unit normal and binormal vectors for the circular helix. r(t) = cos t i + sin t j + t k Solution: We first compute the ingredients needed for the unit normal vector: r'(t) = –sin t i + cos t j + k 57
Example 6 – Solution cont’d This shows that the normal vector at a point on the helix is horizontal and points toward the z-axis. The binormal vector is B(t) = T(t) N(t) 58
The Normal and Binormal Vectors The plane determined by the normal and binormal vectors N and B at a point P on a curve C is called the normal plane of C at P. It consists of all lines that are orthogonal to the tangent vector T. The plane determined by the vectors T and N is called the osculating plane of C at P. The name comes from the Latin osculum, meaning “kiss. ” It is the plane that comes closest to containing the part of the curve near P. (For a plane curve, the osculating plane is simply the plane that contains the curve. ) 59
The Normal and Binormal Vectors The circle that lies in the osculating plane of C at P, has the same tangent as C at P, lies on the concave side of C (toward which N points), and has radius = 1/ (the reciprocal of the curvature) is called the osculating circle (or the circle of curvature) of C at P. It is the circle that best describes how C behaves near P; it shares the same tangent, normal, and curvature at P. 60
The Normal and Binormal Vectors We summarize here the formulas for unit tangent, unit normal and binormal vectors, and curvature. Ex 2. Find the unit normal and unit binormal vectors for the vector equation r (t)=(6, t 2, −t). Ex 3. Find the binormal vector of v(t)=5 cos(t)i+5 sin(t)j+2 k. 61
The Normal and Binormal Vectors Ex 5. Find the binormal vector of v(t)=5 cos(t)i+5 sin(t)j+2 k. Possible Answers: Does not exist 62
Application 63
Application There are many applications in physics and engineering dynamics that involve the relationships among speed, arc length, curvature, and acceleration. One such application concerns frictional force. A moving object with mass m is in contact with a stationary object. The total force required to produce an acceleration a along a given path is The portion of this total force that is supplied by the stationary object is called the force of friction. 64
Application For example, if a car moving with constant speed is rounding a turn, the roadway exerts a frictional force that keeps the car from sliding off the road. If the car is not sliding, the frictional force is perpendicular to the direction of motion and has magnitude equal to the normal component of acceleration, as shown in Figure 12. 39. The potential frictional force of a road around a turn can be increased by banking the roadway. Figure 12. 39 65
Example 8 – Frictional Force A 360 -kilogram go-cart is driven at a speed of 60 kilometers per hour around a circular racetrack of radius 12 meters, as shown in Figure 12. 40. To keep the cart from skidding off course, what frictional force must the track surface exert on the tires? Figure 12. 40 66
Example 8 – Solution The frictional force must equal the mass times the normal component of acceleration. For this circular path, you know that the curvature is Therefore, the frictional force is 67
Application 68
Application cont’d 69
Application cont’d 70
- Slides: 68