OHSx XM521 Multivariable Differential Calculus: Homework Solutions 14.1

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1 OHSx XM5 Multivariable Differential Calculus: Homework Solutions 4. (8) Describe the graph of the equation. r = i + tj + (t )k. Solution: Let y(t) = t, so that z(t) = t = y. In the yz-plane, this is just a parabola. Since the x-component of r is the constant, the parabola is shifted from the yz-plane by in the negative x-direction. (40) Show that the graph of lies in the plane x y + z + = 0. r = ti + + t t j + t k, t > 0 t Solution: r is represented by the parametric equations x(t) = t, y(t) = ( + t)/t, z(t) = ( t )/t. Letting x, y, and z denote the components of r, we have: x y + z + = t + t t = t + + t t t + t t = t t + = 0. Therefore the components of r satisfy the equation of the given plane, for t > 0. So the graph of r must lie in the plane x y + z + =

2 (4) Show that the graph of r = cos ti + sin tj + sin tk is an ellipse, and find the lengths of the major and minor axes. [Hint: Show that the graph lies on both a circular cylinder and a plane and use the result in Exercise 60 of Sections.4.] Solution: The result of Exercise 60 in Section.4 states that if a plane is not parallel to the axis of a right circular cylinder, then the intersection of the plane and cylinder is an ellipse. Observe that if x(t), y(t), and z(t) denote the components of r(t), then (x(t)) + (y(t)) = 9, and y(t) = z(t). Therefore, r lies on the cylinder x + y = 9 and on the plane y = z. Consequently, by Exercise 60, Section.4, the graph of r is an ellipse. To find the lengths of the major and minor axes, consider r = cos ti + sin tj + sin tk = cos ti + sin t( j+k ). Let u = j+k so that r = cos ti + sin tu. From the form of this equation, it is clear this graph is an ellipse in the xu-plane where the u-axis is defined by the direction of u. The minor axis is b = () = 6 and the major axis is a = ( ) = 6. (4) For the helix r = a cos ti + a sin tj + ctk, find c (c > 0) so that the helix will make one complete turn in a distance of units measured along the z-axis. Solution: The above helix will make one complete turn when t = π since this is the period for the functions cos t and sin t. Therefore, when t = π we want z =, or equivalently c π =. So c = /π. (45) Show that the curve r = t cos ti + t sin tj + tk, t 0, lies on the cone z = x + y. Describe the curve. Solution: Letting x, y, and z denote the components of r, we have: x + y = (t cos t) + (t sin t) = t cos t + sin t = t

3 = z. Since the components satisfy the cone equation, r lies on that cone. Qualitatively, the curve is helical in nature with radius always increasing due to the multiplicative t factor in the first two components of r(t). The curve starts at the origin and spirals upward in the positive z-direction around the z-axis but with increasing distance away from this axis as it moves farther in the z-direction. The curve will wrap once around the z-axis every time t increases by π. (47) In each part, match the vector equation with one of the accompanying graphs, and explain your reasoning. (a) r = ti tj + t k. Solution: Let x(t) = t, y(t) = t, and z(t) = t. The graph must have x = y, and at t = 0 it has the value (0, 0, ). The only graph that has these characteristics is III. (b) r = sin πti tj + tk. Solution: Let x(t) = sin πt, y(t) = t, and z(t) = t. The graph must have y = z, and at t = 0 it has the value (0, 0, 0). The only graph that has these characteristics is IV. (c) r = sin ti + cos tj + sin tk. Solution: Let x(t) = sin t, y(t) = cos t, and z(t) = sin t. Observe that (x(t)) +(y(t)) =, so r lies on the cylinder x + y =. Graph II is the only one with this property. (d) r = ti + cos tj + sin tk. Solution: The form of the components of r are of a helix with rotation in planes parallel to the yz-plane and distance from the yz-plane (given by x ) increasing as t increases. This corresponds to graph I.

4 OHSx XM5 Multivariable Differential Calculus: Homework Solutions 4. (8) Find lim t t, ln t t, sin t. Solution: To find the limit of a vector function, take the limit of each individual component: lim t t, ln t t, sin t = lim t t, lim ln t t t, lim sin t =, t, sin. The y-component was evaluated using L Hopital s rule: lim t ln t t = lim /t t t =. (0) Determine whether r(t) is continuous at t = 0. Explain your reasoning. (a) r(t) = e t i + j + csc tk. Solution: r(t) is not continuous at t = 0 because csc t is undefined for t = 0. (b) r(t) = 5i t + j + e t k. Solution: r(t) is continuous at t = 0 because lim t 0 r(t) = lim t 0 5i lim t 0 t + j + lim t 0 e t k = 5i j + k = r(0). We ve used the facts that t + and e t are continuous at t = 0. (46) Solve the vector initial-value problem for y(t) by integrating and using the initial condition to find the constants of integration. Solution: Integrate y (t) to obtain y (t): y (t) = t i tj, y(0) = i 4j, y (0) = 0. y (t) = y (t)dt = (t i tj)dt = 4t i t j + C. To satisfy the initial condition y (0) = 0, we must have 0 = y (0) = 4 0 i 0 j + C. Therefore C = 0, and y (t) = 4t i t j. Now integrate y (t) to obtain y(t): y(t) = y (t)dt = (4t i t j)dt = t 4 i t j + C. To satisfy the initial condition y(0) = i 4j, we must have C = i 4j, implying that y(t) = (t 4 + )i ( t + 4)j.

5 (5) Find where the tangent line to the curve at the point (,, 0) intersects the yz-plane. r(t) = e t i + cos tj + sin tk Solution: The point (,, 0) corresponds to t = 0 for r(t). To find the tangent line to r(t) at this point, first calculate r (t) = e t i sin tj + cos tk, so that r (0) = i + k. The tangent line can therefore be represented by the vector equation L(t) = L 0 + vt where L 0 =,, 0 is a vector from the origin to a point on the line and v = r (0) =, 0, gives the direction of the tangent line. This line can be represented parametrically as x = t, y =, z = t. At the intersection of this line and the yz-plane, x = 0, which forces t = /. At this t value z = /. Therefore, the intersection point is (0,, /). (54) Show that the graphs of r (t) and r (t) intersect at the point P. Find, to the nearest degree, the acute angle between the tangent lines to the graphs of r (t) and r (t) at the point P (,, ). r (t) = e t i + cos tj + (t + )k, r (t) = ( t)i + t j + (t + 4)k. Solution: r passes through P when t = 0, and r passes through P when t =. The vectors in the directions of the respective tangent lines can be calculated by differentiating r (t) and r (t) and evaluating at t = 0 and t = respectively: r (t) = e t i sin tj + tk, r (0) = i. r (t) = i + tj + t k, r ( ) = i j + k. Using a b = a b cos θ, the angle between the above vectors is given by cos θ = [( )( )]/[()( 4)] = / 4, so that θ = arccos(/ 4) 74.5 o. Since this angle is acute, it must be the acute angle between the corresponding tangent lines. (57) Use Formula (9) to derive the differentiation formula d dt [r(t) r (t)] = r(t) r (t). Solution: Apply Formula (9) to d dt [r(t) r (t)] to get: d dt [r(t) r (t)] = r (t) r (t) + r(t) r (t). Since a a = 0 for all vectors a, the first term is 0, and we obtain the desired result. (58) Let u = u(t), v = v(t), w = w(t) be differentiable vector-valued functions. Use Formulas (8) and (9) to show that d du dw [u (v w)] = [v w] + u [dv w] + u [v dt dt dt dt ].

6 Solution: Let x = v w so that d dt [u (v w)] = d dt [u x]. Apply Formula (8) to get: d du [u x] = x + u dx dt dt dt. Now substitute v w for x and use Formula (9) to compute d[v w] dt : du [v w] + u [dv dt dt w + v dw dt ]. Finally, apply the distributive property of the dot product on the second term of this expression to get the desired formula. (59) Let u, u, u, v, v, v, w, w, and w be differentiable functions of t. Use Exercise 58 to show that d dt u u u v v v w w w = u u u v v v w w w + u u u v v v w w w + u u u v v v w w w Solution: For any vectors a = a, a, a, b = b, b, b, and c = c, c, c, a a a a b c = b b b c c c. So the left-hand side of the desired equation can be written as d dt (u v w), while the right-hand side is du dt v w + u dv dt w + u v dw dt. Therefore, the desired result follows immediately from Exercise 58.

7 OHSx XM5 Multivariable Differential Calculus: Homework Solutions 4. (7) Calculate dr/dτ by the chain rule, and then check your result by expressing r in terms of τ and differentiating. r = e t i + 4e t j; t = τ. Solution: dr/dt = e t i 4e t j = e τ i 4e τ j and dt/dτ = τ. Using the chain rule: dr dτ = dr dt dt dτ = (eτ i 4e τ j) τ = τe τ i 8τe t j. To check this, writing r as a function of τ gives r = e t i + 4e t j = e τ i + 4e τ j. differentiating and applying the chain rule to each component gives So dr dτ = eτ (τ)i 4e τ (τ)j = τe τ i 8τe τ j, which agrees with the earlier computation of dr/dτ. () (a) Find the arc length parametrization of the line x = 5 + t, y = t, z = 5 + t that has the same direction as the given line and has reference point ( 5, 0, 5). Solution: Since dx/dt =, dy/dt =, dz/dt =, and the point ( 5, 0, 5) corresponds to t = 0, the arc length parameter for the line is: s = t 0 t (dx/dt) + (dy/dt) + (dz/dt) dt = + + dt = 4t. So t = s/ 4, and the arc length parametrization of the line is x = 5 + (s/ 4), y = s/ 4, z = 5 + s/ 4. 0 (b) Use the parametric equations obtained in part (a) to find the point on the line that is 0 units from the reference point in the direction of increasing parameter. Solution: Substituting s = 0 into the result of (a) gives ( ,, ) (8) Find an arc length parametrization of the curve that has the same orientation as the given curve and has t = 0 as the reference point. r(t) = sin e t i + cos e t j + e t k; t 0.

8 Solution: dr(t)/dt = e t cos e t i e t sin e t j + e t k, so that dr(t)/dt = e t cos e t + e t sin e t + e t = e t ( + ) = e t. The arc length parameter s is given by: s = t 0 dr/du du = t 0 e u du = (e t ), implying that t = ln((s/) + ) = ln((s + )/). Using this to find an expression for e t as a function of s: e t = e ln((s+)/) = s +. So the desired arc length parametrization is: ( s + ) r(s) = sin i + cos ( s + ) (s + ) j + k. () Show that in cylindrical coordinates a curve given by the parametric equations r = r(t), θ = θ(t), and z = z(t) for a t b has arc length L = b a (dr/dt) + r (dθ/dt) + (dz/dt) dt [Hint: Use the relationships x = r cos θ and y = r sin θ.] Solution: Let x(t), y(t), z(t), a t b, be a parametrization of the curve in rectangular coordinates. We know that for all t, x(t) = r(t) cos θ(t) and y(t) = r(t) sin θ(t). Differentiating with respect to t, and using dot notation (popular with physicists and engineers) for the derivatives with respect to t (e.g. dx/dt = ẋ) gives This implies: ẋ = ṙ cos θ r θ sin θ, ẏ = ṙ sin θ + r θ cos θ. ẋ = ṙ cos θ rṙ θ cos θ sin θ + r θ sin θ, ẏ = ṙ sin θ + rṙ θ cos θ sin θ + r θ cos θ. Adding these two expressions, grouping like terms, and using sin θ + cos θ = gives ẋ + ẏ = ṙ + r θ. Using this result in the formula for arc length L in -space, we obtain L = b a ẋ + ẏ + ż dt = b which is the desired result in dot notation. a ṙ + r θ + ż dt, (8) What change of parameter g(τ) would you make if you wanted to trace the graph of r(t) (0 t ) in the opposite direction with τ varying from 0 to. Solution: The substitution t = g(τ) = τ will suffice. Simply note that as τ varies from 0 to, t varies from to 0. In fact g can be any (smooth) strictly decreasing function with domain 0 τ and range 0 t.

9 (40) Let x = cos t, y = sin t, z = t /. Find (a) r (t). Solution: r (t) = x (t)i + y (t)j + z (t)k = sin ti + cos tj + t/ k. So: r (t) = sin t + cos t t = 4 + 9t. (b) ds/dt. Solution: If s is an arc length parameter, then s = t t 0 r (u) du. Differentiating with respect to t gives ds/dt = r (t). So from part (a), ds/dt = 4 + 9t. (c) 0 r (t) dt. Solution: Substituting the result from (a) gives 0 r (t) dt = t dt = (4 + 9t)/ 7 0 = 7 (()/ 8). (4) Prove: If r(t) is a smoothly parameterized function, then the angles between r (t) and the vectors i, j, and k are continuous functions of t. Solution: Let r(t) = x(t)i + y(t)j + z(t)k. Since r(t) is a smoothly parameterized function, r is continuous and r 0. The continuity of r implies that x, y, and z must be continuous. The angle α between i and r (t) can be found by using the dot product: or α = cos i r (t) = i r (t) cos(α) i r (t) i r (t) = x (t) cos r (t). (Note: In general, the equation cos(θ) = c does not necessarily imply that θ = cos (c) because θ may not lie between 0 and π. However, the angle between two vectors is always between 0 and π, so in this case cos(θ) = c does imply θ = cos (c).) Since x and r are continuous and r 0, x / r is a continuous function of t. Finally, since cos is continuous, α is a continuous function of t. Similar arguments will hold for β and γ, the angles that r (t) makes with j and k respectively.

10 OHSx XM5 Multivariable Differential Calculus: Homework Solutions 4.4 (8) Find T(t) and N(t) at the given point. r(t) = ti + t j + t k; t = 0. Solution: r (t) = i + tj + t k, and So T(0) = i. T(t) = To find N(t), first calculate T (t): r (t) r (t) = + t + t 4 (i + tj + t k). T (t) = (t + 4t )( + t + t 4 ) (i + tj + t k) + (j + tk). + t + t4 So T (0) = j (which makes sense since it must be orthogonal to T(0)), and N(0) = T (0)/ T (0) = j. (7) Use the formula B(t) = T(t) N(t) to find B(t), and then check your answer by using Formula () to find B(t) directly from r(t). r(t) = (sin t t cos t)i + (cos t + t sin t)j + k. Solution: r (t) = t sin ti + t cos tj, r (t) = t, and T(t) = r (t) r = sin ti + cos tj. (t) So T (t) = cos ti sin tj, T (t) =, and therefore N(t) = T (t)/ T (t) = cos ti sin tj. This implies that B(t) = T(t) N(t) = (sin ti + cos tj) (cos ti sin tj) = sin tk cos tk = k. Formula () states that B(t) = (r (t) r (t))/ r (t) r (t), so first calculate r (t) = (sin t + t cos t)i + (cos t t sin t)j. Then r (t) r (t) = (t sin ti + t cos tj) ((sin t + t cos t)i + (cos t t sin t)j) = t (sin t + cos t)k = t k, and r (t) r (t) = t. So B(t) = (r (t) r (t))/ r (t) r (t) = k, which is the same as the earlier result.

11 (0) Find T(t), N(t), and B(t) for the given value of t. Then find equations for the osculating, normal, and rectifying planes at the point that corresponds to that value of t. r(t) = e t i + e t cos tj + e t sin tk; t = 0. Solution: The following calculations are easily checked: r (t) = e t i + (e t cos t e t sin t)j + (e t sin t + e t cos t)k; r (t) = e t + (cos t sin t) + (sin t + cos t) ) = e t ; T(t) = r (t) r (t) = (i + (cos t sin t)j + (sin t + cos t)k); T(0) = (i + j + k); T (t) = ( (sin t + cos t)j + (cos t sin t)k); T (t) = (sin t + cos t) + (cos t sin t) = N(t) = ; T (t) T (t) = ( (sin t + cos t)j + (cos t sin t)k); N(0) = ( j + k); B(t) = T(t) N(t) = 6 (i (cos t sin t)j (sin t + cos t)k); B(0) = 6 (i j k). With these calculations and r(0) = r 0 = i + j, the various planes can be calculated using the point-normal form plane equation n (r r 0 ) = 0. The osculating plane has B(0) as a normal vector, so an equation for that plane is which simplifies to x y z = 0. 6 (i j k) ((x )i + (y )j + zk) = 0, The normal plane has T(0) as a normal vector, so an equation for that plane is which simplifies to x + y + z = 0. (i + j + k) ((x )i + (y )j + zk) = 0, The rectifying plane has N(0) as a normal vector, so an equation for that plane is which simplifies to y + z + = 0. ( j + k) ((x )i + (y )j + zk) = 0, () (a) Use the formula N(t) = B(t) T(t) and Formulas () and () to show that N(t) can be expressed in terms of r(t) as N(t) = r (t) r (t) r (t) r (t) r (t) r (t).

12 Solution: To derive this result, just substitute T(t) = r (t)/ r (t) (which is Formula ()) and B(t) = r (t) r (t)/ r (t) r (t) (which is Formula ()) into N(t) = B(t) T(t). (b) Use properties of cross products to show that the formula in part (a) can be expressed as N(t) = r (t) r (t) r (t) r (t) r (t) r (t). Solution: From part (a), N(t) = r (t) r (t) r (t) r (t) r (t) r (t) = (r (t) r (t)) r (t) r (t) r (t) r (t). The numerator of this expression is of the form (a b) a. But note that (a b) a = a (a b) = a ( (b a)) = a (b a). Therefore, the numerator in the expression for N(t) can be written as r (t) r (t) r (t) without any confusion regarding the order in which the cross products are performed. In the denominator, r (t) must be orthogonal to r (t) r (t), so the angle between these two vectors is π/, implying that r (t) r (t) r (t) = r (t) r (t) r (t) sin(π/) = r (t) r (t) r (t). Using this relation in the denominator of our expresssion for N(t) gives the desired result. (c) Use the result in part (b) and Exercise 9 of Section.4 to show that N(t) can be expressed directly in terms of r(t) as N(t) = u(t) u(t) where u(t) = r (t) r (t) (r (t) r (t))r (t). Solution: Starting with the result of (b), let u(t) = r (t) r (t) r (t), so that N(t) =, and apply the identity (a b) c = (c a)b (c b)a to obtain: u(t) u(t) as desired. u(t) = (r (t) r (t))r (t) (r (t) r (t))r (t) = r (t) r (t) (r (t) r (t))r (t),

13 OHSx XM5 Multivariable Differential Calculus: Homework Solutions 4.5 (7) (a) Use Formula () to show that in -space the curvature of a smooth parametric curve x(t), y(t) is κ(t) = x y y x (x + y ) /, where primes denote differentiation with respect to t. Solution: Formula () states: κ(t) = r (t) r (t) r (t). Let r = xi + yj, r = x i + y j and r = x i + y j, (with the functional dependence on t omitted for notational convenience). This implies r r = (x i + y j) (x i + y j) = (x y y x )k, so that r r = x y y x, which gives the desired form for the numerator. Since r = x + y, substituting this in the denominator yields the desired result. (b) Use the result in part (a) to show that in -space the curvature of the plane curve given by y = f(x) is d y/dx κ(x) = [ + (dy/dx) ]. / [Hint: Express y = f(x) parametrically with x = t as the parameter.] Solution: Expressing y = f(x) parametrically with x = t as the parameter gives y = f(t), y = dy/dt = (dy/dx)dx/dt = dy/dx (since dx/dt = ), y = d y/dt = (d y/dx )dx/dt = d y/dx, x =, and x = 0. Substituting these expressions into the result from (a) yields the desired formula. (4) The graphs of f(x) and the associated curvature function κ(x) are shown. Determine which is which, and explain your reasoning. Solution (a): The curve II flattens out as x goes to either or, so curvature must go to 0 as x goes to either or. Also curve II appears to have its maximum curvature at x = 0, corresponding to where curve I reaches a maximum. Hence, curve II is the function and I is the curvature function. Solution (b): A circle has a constant curvature, so II is the function and I is the curvature function. (47) Find the radius of curvature of the parabola y = 4px at (0, 0). Solution: Use the result of Exercise 7(a) with the given parabola expressed parametrically as y = t, x = t /4p. It s easy to check that x = t/p, x = /p, y =, and y = 0, which implies that /p κ(t) = ((t /4p ) + ). / The point (0, 0) corresponds to t = 0, so the curvature at this point is κ(0) = /p, and the radius of curvature is ρ = /κ = p = p.

14 (48) At what point(s) does y = e x have maximum curvature? Solution: It should be clear from the shape of the graph of y = e x that there is at least one point of maximum curvature. (In other words the curvature does not get arbitrarily large.) Using d dx (ex ) = e x and the result of Exercise 7(b) gives κ(x) = d y/dx ( + (dy/dx) ) = e x / ( + e x ). / This function is differentiable everywhere, so any maximum must occur where the derivative is zero: dκ(x) dx = ex ( + e x ) / ( + ex ) / (e x )e x ( + e x ) = 0. Using simple algebra, the numerator of the above quotient can be written e x ( + e x ) / ( e x ). Therefore, dκ/dx = 0 when e x = 0, or x = ln and y =. Moreover, dκ/dx > 0 (κ is increasing) for x < ln, while dκ/dx < 0 (κ is decreasing) for x > ln. Thus, (x, y) = ( ln, ) must be the point of maximal curvature. (59) In Exercises 59, we will be concerned with the problem of creating a single smooth curve by piecing together two separate smooth curves. If two smooth curves C and C are joined at a point P to form a curve C, then we will say that C and C make a smooth transition at P if the curvature of C is continuous at P. The accompanying figure shows the arc of a circle of radius r with center at (0, r). Find the value of a so that there is a smooth transition from the circle to the parabola y = ax at the point where x = 0. Solution: The curvature for a circle of radius r is /r. For the parabola y = ax, dy/dx = ax and d y/dx = a. So, using the result of Exercise 7(b), κ(x) = d y/dx [ + (dy/dx) ] / = a [ + 4a x ] /, and κ(0) = a = a (since a > 0 for the parabola in the figure). Equating the curvature for the circle and parabola results in /r = a or a = /r. (6) In Exercises 6-64, we assume that s is an arc length parameter for a smooth vector-valued function r(s) in -space and that dt/ds and dn/ds exist at each point on the curve. This implies that db/ds exists as well, since B = T N. Show that dt ds = κ(s)n(s) and use this result to obtain the formula in (0). Solution: Therefore, We know that T(s) = r (s), N(s) = r (s)/ r (s), and κ(s) = r (s). dt ds = T (s) = r (s) = r r (s) (s) r (s) = κ(s)n(s).

15 (6) (a) Show that db/ds is perpendicular to B(s). Solution: B is unit vector, so B(s) B(s) =. Differentiating both sides with respect to s gives db B + B db ds ds = 0, or Thus, B db ds B db ds = 0. = 0, implying that db/ds is perpendicular to B(s). (b) Show that db/ds is perpendicular to T(s).[Hint: Use the fact that B(s) is perpendicular to both T(s) and N(s), and differentiate B T with respect to s.] Solution: Since B and T are orthogonal, B T = 0. Differentiating this equation with respect to s gives db T + B dt ds ds = 0. Since dt/ds = κn (from Exercise 6) and B is orthogonal to N, the second term on the left is 0, implying that db T = 0. ds Hence, db/ds is perpendicular to T(s). (c) Use the results in (a) and (b) to show that db/ds is a scalar multiple of N(s). The negative of this scalar is called the torsion of r(s) and is denoted by τ(s). Thus, db ds = τ(s)n(s). Solution: Since N = B T, N is orthogonal to both B and T, as is db/ds from the results in (a) and (b). Since B and T are orthogonal unit vectors that define a plane, both N and db/ds must be orthogonal to this plane. Hence, db/ds must be a scalar multiple of N. (d) Show that τ(s) = 0 for all s if the graph of r(s) lies in a plane. [Note: For reasons that we cannot discuss here, the torsion is related to the twisting properties of the curve, and τ(s) is regarded as a numerical measure of this tendency for the curve to twist out of the osculating plane.] Solution: If the graph of r(s) lies in a plane, then both T and N must lie in this plane since they are the unit tangent and unit inward normal vectors respectively. (Note that if T lies in a plane with normal vector n then T n = 0. Differentiating gives T n = 0, which implies that N n = 0. Therefore N also lies in the same plane.) Since T and N lie in the plane containing r, B = T N must always be normal to this plane. Since the unit vector B has constant magnitude and the direction of the vector is always perpendicular to the plane of r, B must be constant, implying that db/ds = 0. Since db ds = τ(s)n(s) and N(s) is never zero, τ(s) must be 0 for all s.

16 (6) Let κ be the curvature of c and τ the torsion (defined in Exercise 6). By differentiating N = B T with respect to s, show that dn/ds = κt + τb. Solution: Consider dn/ds = (db/ds) T + B (dt/ds). Since db/ds = τn and dt/ds = κn, we obtain dn/ds = τn T + B κn. Since N T = B, B N = T, and a scalar can be factored out of a cross product, dn/ds = τb κt = κt + τb. (64) The following derivatives, known as the Frenet-Serret formulas, are fundamental in the theory of curves in -space: dt ds dn ds db ds = κn [Exercise 6] = κt + τb [Exercise 6] = τn [Exercise 6(c)] Use the first two Frenet-Serret formulas and the fact that r (s) = T if r = r(s) to show that τ = [r (s) r (s)] r (s) r (s) and B = r (s) r (s) r (s) Solution: The equation for B is easy, and we leave it to you. To find τ, dot both sides of the second Frenet-Serret formula with B: B dn ds = B ( κt + τb) = κb T + τb B. Since B is a unit vector orthogonal to T, B B = and B T = 0, which implies that the above equation simplifies to B dn ds = τ. Since N = r (s)/ r (s) = r (s)/κ(s), we can differentiate with respect to s to obtain dn ds = κ(s)r (s) r (s)κ (s) (κ(s)) = r (s) κ(s) r (s) κ(s) κ (s) κ(s) Therefore, B dn ds = r (s) κ(s) N κ (s) κ(s). = B r (s) κ(s) = B r (s) r (s) B N κ (s) κ(s) because B N = 0 and κ(s) = r (s). Finally, since B = r (s) r (s) r (s), we can write as desired. τ = B dn ds = B r (s) r = (s) r (s) r (s) r (s) r (s) r = [r (s) r (s)] r (s) (s) r (s), 4

17 OHSx XM5 Multivariable Differential Calculus: Homework Solutions 4.6 (9) As illustrated in the accompanying figure, suppose that the equations of motion of a particle moving along an elliptic path are x = a cos ωt, y = b sin ωt. (a) Show that the acceleration is directed toward the origin. Solution: Let the parametric equations of the particle motion be represented by r(t) = x(t)i + y(t)j = a cos ωti + b sin ωtj, so that v(t) = r (t) = ω( a sin ωti + b cos ωtj) and a(t) = r (t) = ω (a cos ωti + b sin ωtj) = ω r(t). Since the acceleration vector is proportional to the negative of the position vector, the acceleration vector always points toward the origin. (b) Show that the magnitude of the acceleration is proportional to the distance from the particle to the origin. Solution: The magnitude of the acceleration is given by a(t) = ω r(t) = ω r(t), which shows that it is proportional to the distance from the particle to the origin. () Suppose that the motion of a particle is described by the position vector r = (t t )i t j. Find the minimum speed of the particle and the location when it has this speed. Solution: r (t) = ( t)i tj and speed of the particle is v(t) = r (t) = ( t) + 4t = 8t 4t +. The expression 8t 4t + is minimal at t = /4. So the minimum speed occurs when t = /4, which corresponds to the point (x, y) = ( 6, 6 ). The speed at this point v(/4) = /. (9) What can you say about the trajectory of a particle that moves in -space or -space with zero acceleration? Justify your answer. Solution: If a(t) = 0 for all t, then v(t) = a(t) dt = C for some constant vector C. So the velocity is constant. Integrating again gives r(t) = v(t) dt = C t + C for some constant C. Therefore the trajectory of the particle must lie on a line. (9) The position vector of two particles are given. Show that the particles move along the same path but the speed of the first is constant and the speed of the second is not. r = cos ti + sin tj r = cos(t )i + sin(t )j (t 0) Solution: If x(t) and y(t) denote the components of the vector functions, then for both r and r we have (x(t)) + (y(t)) = 4. So the path of both particles is the circle of radius centered at the origin. The speed of r is given by r (t) = 6 sin ti + 6 cos tj = 6, so it is constant. The speed of r is given by r (t) = (t) sin t i + (t) cos t j = (t) = 4t, so it is not constant.

18 (48) Suppose that a particle moves with nonzero acceleration along the curve y = f(x). Use part (b) of Exercise 7 in Section 4.5 to show that the acceleration vector is tangent to the curve at each point where f (x) = 0. Solution: According to Theorem 4.6., the acceleration of a particle can be expressed as a = (d s/dt )T + κ(ds/dt) N, where s is an arc length parameter for the curve, κ is the curvature, T is the unit tangent vector, and N is the unit normal vector. The result of Exercise 7(b) in Section 4.5 states that the curvature for a curve y = f(x) can be expressed as: d y/dx κ(x) = [ + (dy/dx) ]. / So κ = 0 when f (x) = d y/dx = 0. Thus, a = (d s/dt )T, implying that the (nonzero) acceleration vector must be tangent to the curve. (6) A ball rolls off a table 4 ft high while moving at a constant speed of 5 ft/s. (a) How long does it take for the ball to hit the floor after it leaves the table? Solution: Choose a coordinate system such that the origin lies at the edge of the table, the positive x-axis points horizontally in the initial direction of the ball, and the positive y-axis points upward. The parametric equations for the motion of the ball are x(t) = (v 0 cos α)t and y(t) = s 0 + (v 0 sin α)t gt, where s 0 = 0 is the initial height, v 0 = v 0 = 5 ft/s is the initial speed, α = 0 is initial angle of release, g is the acceleration due to gravity (ft/s ), and t is the time in seconds. So y(t) = s 0 + (v 0 sin α)t gt = 6t, and the ball hits the floor when y(t) = 6t = 4, or t = /. (b) At what speed does the ball hit the floor? Solution: The components of the velocity of the ball are given by v x = v 0 cos α and v y = v 0 sin α gt. So at t = /, v x = 5 ft/s and v y = 6 ft/s. The final speed is v f = v x + v y = 8 ft/s. (c) If a ball were dropped from rest at table height just as the rolling ball leaves the table, which ball would hit the ground first? Justify your answer. Solution: If the ball were dropped from rest, v 0 = 0 and α = π/, so the parametric equations of motion would be x = 0 and y = 6t. At the ground, y = 4 = 6t, implying that the ball would hit the ground at t = / s, which is the same as before. Therefore, both balls would strike simultaneously. This makes sense: the time it takes for the ball to hit the floor is completely dependent on the vertical component y(t), which itself does not depend on the initial horizontal velocity. (67) A shell is fired from ground level at an elevation angle of α and a muzzle speed of v 0. (a) Show that the maximum height reached by the shell is maximum height = (v 0 sin α). g Solution: Choose the (obvious) coordinate system such that the origin is the point from which the shell is fired, the ground is y = 0, and the positive x-axis corresponds to zero angle of elevation. The parametric equations for the motion of the shell are x = (v 0 cos α)t

19 and y = s 0 + (v 0 sin α)t gt, where s 0 = 0 is the initial height, and g is the acceleration due to gravity. The shell is at its maximum height when dy/dt = v 0 sin α gt = 0, which occurs when t = v 0 sin α/g. So the maximum height is: y(v 0 sin α/g) = (v 0 sin α)(v 0 sin α/g) g(v 0 sin α/g) = (v 0 sin α). g (b) The horizontal range R of the shell is the horizontal distance traveled when the shell returns to ground level. Show that R = (v 0 sin α)/g. For what elevation angle will the range be maximum? What is the maximum range? Solution: When the shell returns back to the ground, y = (v 0 sin α)t gt = 0, implying that either t = 0 or t = v 0 sin α/g. Since t = 0 corresponds to the initial release of the shell, t = v 0 sin α/g is the desired time for the shell to return to the ground. Therefore, the horizontal range is: R = x(v 0 sin α/g) = (v 0 cos α)t = (v 0 cos α)(v 0 sin α/g) = (v 0 sin α)/g. (We ve made use of the identity sin α = sin α cos α.) The maximum range occurs when sin α =, or α = π/4, and this maximum range is v 0 /g.