CURVES: VELOCITY, ACCELERATION, AND LENGTH


 Esmond Barton
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1 CURVES: VELOCITY, ACCELERATION, AND LENGTH As examples of curves, consier the situation where the amounts of ncommoities varies with time t, qt = q 1 t,..., q n t. Thus, the amount of the commoities are functions of time. We can also consier the prices of these commoities as functions of time, pt = p 1 t,..., p n t. Another way that curves can arise in economics is the following. Assume there is a fixe prouction of a single goo Q using inputs of labor L an capital K, Q 0 = FL, K = L K. Let w be the price of labor an r the price of capital. The quantities L, K that minimize cost of prouction epens on the parameters w an r, Lw, r, K w, r = r Q0 /w, wq 0 /r. As the price of labor is varie keeping r = r 0 fixe, a curve of optimal choices Lw, r 0, K w, r 0 = r0 Q 0 /w, wq 0 /r 0 is etermine as a function of the single variable, which is the price of labor. 3.1 Derivatives Definition. Let r : R R n be a ifferentiable function. The position vector at time t is rt. The velocity vector is given by the erivatives of the position vector with respect to time, vt = r t. The spee is the length of the velocity vector, an is a scalar quantity. So, the velocity inclues both the spee an the irection of current motion. The acceleration is the erivative of the velocity an the secon erivative of the position, at = v t = r t. Example 1.2 Circle. As an example, we consier the point on a circle of raius 7 about the origin as a function of time, i.e., the point changes with time. Using polar coorinates, let rt = 7 an θt = 3t, or xt = 7 cos3t an yt = 7 sin3t. We can put the two components together to get the position vector, rt = xt, yt = 7 cos3t, 7 sin3t. The position vector is given as a function of time t, so this way of presenting this circle is calle the parametric form of the circle. The point moves aroun the circle with increasing angle in polar coorinates, so the point moves counterclockwise: i when t = 0 then r0 = 7, 0; ii when 3t = π /2 or t = π /6 then r π /6 = 0, 7; iii when 3t = π or t = π /3 then r π /3 = 7, 0; ii when 3t = 3π /2 or t = π /2 then r π /2 = 0, 7; iii when 3t = 2π or t = 2π /3 then r 2π /3 = 7, 0 an the point has move once aroun the circle. The velocity vector is vt = 21 sin3t, 21 cos3t. Notice that the velocity is a vector. The spee vt = 21 is a scalar. Finally, the acceleration at = 63 cost, 63 sint is a vector. Notice that for this example rt vt = 0. This says that the velocity vector is perpenicular to the position vector, an is tangent to the circle. We see later that this is a consequences of the fact that rt 2 is a constant. Example Ellipse. By making the coefficients of sine an cosine ifferent, we obtain an ellipse rather than a circle. Consier rt = xt, yt = 4 cost, 3 sint. 1
2 2 CURVES: VELOCITY, ACCELERATION, AND LENGTH Then xt 2 yt 2 + = cos 2 t + sin 2 t = 1, 4 3 so the curve lies on the ellipse with semiaxes of length 4 an 3. Its velocity vector is vt = 4 sint, 3 cost, an its spee is vt = 16 sin 2 t + 9 cos 2 t an is not a constant. In this example the velocity vector is not perpenicular to the position vector, rt vt = 7 sint cost = 0 except at certain times. Example Circular Helix in R 3. For this example let xt = 7 cos3t, yt = 7 sin3t, an zt = 5t. Compare with Examples 1.3 an 1.4 on pages The position vector is rt = 7 cos3t, 7 sin3t, 5t. The x an y coorinate are the same functions as the first example, so the point lies on a cyliner of raius 7 about the z axis. The time to go once aroun the circle is still 2π /3. However, in this time, z 2π /3 = 10π /3 > 0 = z0, so the point moves up in the zirection. The motion is on what is calle a circular helix. The velocity is vt = r t = 21 sin3t, 21 cos3t, 5. The spee is vt = [ ] = 466, which is a constant. The velocity vector has constant length, but it is not a constant vector since is changes irection. The acceleration is at = v t = r t = 63 cos3t, 63 sin3t, 0. Notice that at = 0 even though the spee is a constant. The acceleration measure both the turning change of irection of the velocity an the change of spee. In this example, at vt = 0, i.e., the acceleration is perpenicular to the velocity. This is a result of the fact that the spee is a constant as we shall see. The tangent line to the curve at the point r π /9 = 7 /2, 73 /2, /9 5π is parallel to the vector v π /9 = /2, 21 1 /2, 5, an is the line x, y, z = 7 /2, 73 /2, /9 5π + t π / /2, 21 /2, 5. Note that we have use the parameter t π /9 so that at t = π /9 the point on the line is the point r π /9. See Proposition 1.3. There are a few rule for the ifferentiations of proucts. If we consier the vectors as single objections, then the prouct rules look very similar to the prouct rule for real value functions. Proposition 1.4. Assume that pt an qt are C 1 curves of vectors in R n an gt is a scalar C 1 function. a. Dot prouct: pt qt = p q + p q. b. Multiplication by a scalar function: gtqt = g tqt + gtq t. c. Cross prouct: If n = 3 an the curves are in R 3, then v w = v w + v w.
3 CURVES: VELOCITY, ACCELERATION, AND LENGTH 3 Proof. a pt qt = p 1tq 1 t + p 2 tq 2 t + p 3 tq 3 t = p 1 q 1 + p 1 q 1 + p 2 q 2 + p 2 q 2 + p 3 q 3 + p 3 q 3 = p q + p q. Thus, the erivative of the ot prouct is the erivative of the first term ot prouct with the secon plus the first term ot prouct with the erivative of the secon. The proof of cases b an c an we leave the proof of c to Exercise 3.1:28. Remark 1. If we consier p as the bunle vector of prices an q as the bunle vector of quantities, then the ot prouct gives the value of the commoities. The erivative pt qt = p q + p q contains two terms: the first gives the change of value ue to the change of prices with the amount of commoities fixe, an the secon gives the change of value ue to the change in the amount of commoities with the prices fixe. We use the rule for the erivative of the ot prouct to prove two of the facts which we illustrate above for examples. Theorem 1.7. Let rt be a C 1 curve of vectors. Then rt = constant if an only if rt an r t are perpenicular for all t. Proof. Using the rule for the erivative of a prouct, [rt rt] = r t rt + rt r t = 2 rt r t. If rt = c for all t, then 0 = c2 = [rt rt] = 2rt r t an rt r t = 0. On the other han, if rt r t = 0, then 0 = 2rt r t = [rt rt] an rt = rt rt must be a constant. Theorem. Let rt be a C 2 curve of vectors an vt = r t. Then, the spee vt = constant if an only if the velocity vt an the acceleration at = v t are perpenicular for all t. Proof. Using again the rule for the erivative of a prouct, [vt vt] = v t vt + vt v t = 2 vt at. If vt = c for all t, then 0 = c2 = [vt vt] = 2vt at, vt at = 0, an these two vectors are perpenicular for all t. On the other han, if vt at = 0, then 0 = 2vt at = [vt vt], an the spee vt = vt vt must be a constant. Remark 2. For the circular helix consiere earlier, the spee is a constant an the acceleration is perpenicular to the velocity as we note.
4 4 CURVES: VELOCITY, ACCELERATION, AND LENGTH Decomposition of the Acceleration cf. 3.2 We give a treatment that avois using the parameterization by arc length an oes not efine curvature. cf. Section 3.2. We are given the two vector quantities of velocity an acceleration. It is natural to breakup the acceleration into the component along vt an the normal component. We shall show that these two components measures the change in spee an the change in irection respectively. Definition. The unit tangent vector is the vector of length one in the irection of the velocity vector, Tt = vt / vt. This gives the irection of motion. The scalar component of acceleration along the velocity is vt at a T t = comp v a =. v The vector component of acceleration along the velocity is vt at a T t = proj v a = a T t Tt = v. v v The vector normal component of the acceleration is a N t = at a T t. The scalar normal component of the acceleration is a N t = at 2 a T t 2 = a N t. Theorem. Let rt be a ifferentiable curve. Then the following hol. a. The change in spee, erivative of the spee, equals the tangential component of the acceleration: vt = a T t. b. The change in irection, erivative of the unit tangent vector, equals the normal component of the acceleration: T t = 1 vt a N t. Proof. The following two calculation using the prouct rule proves the claims: vt = vt vt 1 2 = 1 2 vt vt vt at vt at = vt = a T t. vt T t = vt at at t = vt vt 2 vt 1 = at a T t vt 1 = a N t. vt
5 CURVES: VELOCITY, ACCELERATION, AND LENGTH Antierivatives Example 1.6 in Colley, iscusses the way in which the antierivative can be applie to vector quantities. We illustrate this process for another example. Example. Assume that r t = 3t 2, t, 2t is known an also that r0 = 1, 2, 3. Since x t = 3t 2, xt = t 3 + C x. Since 1 = x0 = C x, we get that xt = t In a similar way, yt = t2 /2 + 2, an zt = t Combining, we get the vector rt = t 3 + 1, t2 /2 + 2, t We coul o this all in one process by consiering vectors. First, taking the antierivative of r t, rt = t 3 + C x, t2 /2 + C y, t 2 + C z. Evaluating at t = 0, Therefore, 1, 2, 3 = r0 = C x, C y, C z. rt = t 3 + 1, t2 /2 + 2, t This example illustrates the fact that if we know the initial position an the velocity at each time then we can etermine the position at each time. Example 1.6. Assume that at = g j, where g is a constant, the gravitational constant. Assume also that r0 = 0 an v0 = v 0 = v 0 cosθi + v 0 sinθj is known. Integrating the vector at once, we get vt = gt j + C. Then, v0 = v 0 = C, so vt = gt j + v 0. Integrating a secon time, we get rt = g 2 t2 j + v 0 t + C 2. Again, taking the values at t = 0, C 2 = r0 = 0, so rt = g 2 t2 j + v 0 t. Roger Ramjet hits the groun again when yt 1 = 0, 0 = yt 1 = g 2 t2 1 + v 0 sinθt 1 t 1 = 2v 0 sinθ. g The horizontal istance travele is xt 1 x0 = v 0 cosθt 1 = 2v2 0 sinθ cosθ g This is maximize for sin2θ = 1, 2θ = π /2, or θ = π /4. = v2 0 sin2θ. g
6 6 CURVES: VELOCITY, ACCELERATION, AND LENGTH 3.2 Length of a Curve We cover the part of Section 3.2 ealing with the length of a curve. Since it is in our textbook, we merely sketch the results. Definition. Let rt be a continuous curve in R 3. The efinitions in R n are similar. The portion of the curve from t = a to t = b is split up into pieces using the points rt i = x i, y i, z i with a = t 0 < t 1 < < t n = b. Letting x i = x i x i 1, y i = y i y i 1, an z i = z i z i 1, the istance between the points rt i 1 an rt i is xi 2 + yi 2 + zi 2. The istance along the curve is probably a bit longer. The length of the curve from t = a to t = b is Lr = lim max t i 0 n i=1 x 2 i + y 2 i + z 2 i, provie the limit exists. When the length exists, the curve is calle rectifiable, an when the limit oes not exist it is calle nonrectifiable. The following is a theorem an not a efinition. Theorem Definition 2.1. Assume that r : [a, b] R 3 is a C 1 curve. Then the curve is rectifiable an Iea of the proof. Lr = = = Lr = lim max t i 0 x b a b a b a r t. [ ] n x 2 i + yi 2 + zi t i i=1 2 + r t, t 2 i y 2 + z 2 Example 2.2. For the helix rt = a cost, a sint, bt, the spee r t = a 2 + b 2, so the length of the curve from t = 0 to t = 2π is 2π 0 a 2 + b 2 = 2π a 2 + b 2. Example. The curve rt = t, t sin 1 /t is nonrectifiable. The velocity is r t = 1, sin 1 /t 1 t cos1 /t an the spee is 1 + sin 2 1 /t 2 t sin1 /t cos 1 /t + 1 cos 2 1 /t t not integrable. 2 that is It is possible to calculate the istance travele up to a given time an etermine the time t in terms of this istance or arc length. The following two examples illustrate this proceure.
7 CURVES: VELOCITY, ACCELERATION, AND LENGTH 7 Example 2.3. If the spee is a constant, as for rt = a cost, a sint, bt, then it is possible to solve for the time as a function of the istance travele: r t = a 2 + b 2 t st = a 2 + b 2 τ = a 2 + b 2 t, t = 0 s a 2 + b 2, Then, the position as a function of the istance travele is s xs = a cos, a sin a 2 + b 2 s, a 2 + b 2 bs. a 2 + b 2 Example 2.4. As given in Example 2.4 in our textbook, for rt = t, 2 2 t2, 1 3 t3, r t = 1, 2 t, t 2, r t = t 2 + t 4 = 1 + t 2, an the length from 0 to t is st = t τ 2 τ = t + t3 3. It is not easy to write t as an explicit function of s, but s = 1 + t2 > 0, so the istance is strictly increasing as a function of t. Therefore, the istance travele etermines the time an the position on the curve. 3.2 The Differential Geometry of Curves Our textbook uses the length of a curve to give a new parameterization using arc length. Using this parametrization, it is possible to efine two geometric quantities, curvature an torsion, that etermine the shape of the curve. I will not ask you to calculate the curvature an torsion. However, you will be hel responsible to using the tangential an normal components of the acceleration. As long as s = v = 0, v = r = r s r s = v v = Tt. s = r s v Thus, the erivative of the position with respect to arc length gives the unit tangent vector. I fin the equation r t = r s s in the book confusing. The book an the usual treatment in ifferential geometry procees to efine some quantities that epen on the shape of the curve an not the spee in which it is traverse. The curvature κ = T s = T v. an
8 8 CURVES: VELOCITY, ACCELERATION, AND LENGTH Because Tt 1, T t is perpenicular to Tt. If T t = 0, then the unit vector in the irection of T t, N = T T T = s T s. is calle the principal normal vector. Completing T an N to a basis of R 3, B = T N. is calle the binormal vector. At each point with T t = 0, the three vectors T, N, B form a basis of R 3. The erivative of B efines the torsion τ by the equation Some proof is require to show that B s B s = τ N. is a scalar multiple of N.
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