Problem Set 2 Solution

Winter 2012 Problem Set 2 1) Prove the chain rule ( − δ proofs are welcome) dF (f (t)) dF df = dt df dt Solution Method 1: F (f (t + δt)) − F (f (t)) δt→0 δt F (f (t + δt)) − F (f (t)) f (t + δt) − f (t) = lim δt→0 f (t + δt) − f (t) δt f (t + δt) − f (t) F (f (t + δt)) − F (f (t)) lim = lim δt→0 δt→0 f (t + δt) − f (t) δt dF df = df dt lim where we have used the continuity of f (t) and the fact that the limit of a product of functions is equal to the product of the limits of two functions. Method 2: We recall that if limx→a y(x) = b then for every  > 0 such that |x − z| <  there exists a δ() such that |y(x) − b| < δ(). Since F (f ) and f (t) are differentiable, we know that F (f + ∆f ) − F (f ) dF = ∆f →0 ∆f df lim and f (t + ∆t) − f (t) df = . ∆t→0 ∆t dt lim 1 Winter 2012 Thus for every F > 0 there exists δF > 0 such that if |∆f | < F , then F (f + ∆f ) − F (f ) dF < δF − ∆f df and f > 0 there exists δf > 0 such that if |∆t| < f , then f (t + ∆t) − f (t) df < δf . − ∆t dt We therefore need to show that for every F f > 0 there exists δF f (F f ) such that if |∆t| < F f then F (f (t + ∆t)) − F (f (t)) dF df < δF f . − ∆t df dt Now F (f (t + ∆t)) − F (f (t)) dF df − ∆t df dt F (f (t + ∆t)) − F (f (t)) f (t + ∆t) − f (t) dF df = − f (t + ∆t) − f (t) ∆t df dt   F (f (t + ∆t)) − F (f (t)) dF f (t + ∆t) − f (t) = − f (t + ∆t) − f (t) df ∆t   dF f (t + ∆t) − f (t) df + − df ∆t dt F (f (t + ∆t)) − F (f (t)) dF f (t + ∆t) − f (t) − ≤ f (t + ∆t) − f (t) df ∆t dF f (t + ∆t) − f (t) df + − df ∆t dt   df dF ≤ δF + δf + δf dt df 2 Winter 2012 In the last line, we have used the fact that f is continuous, so for ∆t < F f there is a ∆f such that f (t + ∆t) − f (t) = ∆f and as ∆t → 0 then ∆f → 0. dF We assume that df dt and df are bounded for the values of interest, and taking their maxima, we deduce that we have found our δF f . Section 11.3, Pages 705–715: 3) The parametric equations of the cycloid are given by x = r(θ − sin θ), y = r(1 − cos θ). Eliminate θ from these equations to get a cartesian equation in x and y. Solution We get that θ = cos−1 (1 − y/r) and so 
 x = r cos−1 (1 − y/r) − sin cos−1 (1 − y/r) . 4) Find the asymptotes to the tractrix, which is given by x = a sin t and y = a (cos t + log(tan t/2)) Solution We observe that y → ∞ when t → 0, π, 2π..., for which x → 0. Hence the curve has an asymptote along the y-axis. The curve is sketched in Fig. 1. 5) (72) Use a graphing device to graph the polar curve r = sin2 (4θ) + cos(4θ). Choose the parameter interval to make sure that you produce the entire curve. Solution The curve is sketched in Fig. 2. Section 11.4, Pages 715–720: 6) (14) Sketch the curve r = 2 + cos(2θ) and find the area it encloses. 3 Winter 2012 2.5 -2.5 0 2.5 5 Figure 1: The Tractrix 1.5 0.5π 1 0.75π 0.25π 0.5 -2.4 -2 -1.6 π -1.2 -0.8 -0.4 0 0.4 0.8 1.2 1.6 2 2.4 -0.5 1.25π 1.75π -1 1.5π -1.5 Figure 2: The curve r = sin2 (4θ) + cos(4θ) 4 Winter 2012 Solution The curve is sketched R 2 in Fig. 3. We recall that the area is given by r dθ. In this case, we can use the symmetry of the curve and find Z π/2 9π 4 . (2 + cos(2θ))2 dθ = 2 0 2.4 1.6 0.5π 0.8 π -4.8 -4 -3.2 -2.4 -1.6 -0.8 0 0.8 1.6 2.4 3.2 4 4.8 -0.8 1.5π -1.6 -2.4 Figure 3: The curve r = 2 + cos(2θ) 7) (42) Find all the points of intersection of r2 = sin(2θ) and r2 = cos(2θ). Solution We need to find the points where tan(2θ) = 1, we find that θ = π/8, 5π/8, 9π/8, and so r2 = ±2−1/2 . We require r to be positive and real, hence r = 2−1/2 and θ = π/8, 9π/8. We check our results by plotting a graph. The graph shows that the curves also cross at the origin, but this occurs for different values of θ, thus we also need to include the point r = 0. 8) (48) Find the exact length of the polar curve r = θ for 0 ≤ θ ≤ 2π. 5 Winter 2012 1.5 0.5π 1 0.75π 0.25π 0.5 -2.4 -2 -1.6 π -1.2 -0.8 -0.4 0 0.4 0.8 1.2 1.6 2 2.4 -0.5 1.25π 1.75π -1 1.5π -1.5 Figure 4: The curves r2 = cos(2θ) and r2 = sin(2θ) Solution We use the formula for arclength in polar coordinates s  2 Z b dr S= r2 + dθ dθ a to find that Z 2π p S= θ2 + 1dθ 0 p p = 0.5θ 1 + θ2 + 0.5 ln(θ + 1 + θ2 )|2π 0 p p = π 1 + 4π 2 + 0.5 ln(2π + 1 + 4π 2 ) where we have used a table to find the integral. We could also use trigonometric substitution tan α = θ and then integrate sec3 α. 9) (56) a) Find a formula for the area of the surface generated by rotating the polar curve r = f (θ), a ≤ θ ≤ b (where f’ is continuous and 0 ≤ a < b ≤ π), 6 Winter 2012 about the line θ = π/2. Solution We recall that the area generated by rotating a curve around the x axis is given by s   2 Z 2 dx dy S = 2π y + dt. dt dt For a rotation around the y axis we get s   2 Z 2 dx dy S = 2π x + dt. dt dt We choose θ instead of t to be our parameter and recall that x = r cos θ, y = r sin θ, and letting r = f (θ), after differentiating and substituting we get that s  Z 2 df S = 2π f (θ) cos θ + (f )2 dt. dθ b) Find the surface area generated by rotating the lemniscate r2 = cos 2θ about the line θ = π/2. Solution The lemniscate is defined for −π/4 ≤ θ ≤ π/4 and 3π/4 ≤ θ ≤ 5π/4. By symmetry, we can use one of these curves to integrate with. The area 7 Winter 2012 is given by Z s √ π/4 2π cos 2θ cos θ −π/4 Z sin2 (2θ) dθ cos(2θ) + cos(2θ) π/4 = 2π cos(θ)dθ −π/4 π  = 4π sin 4 √ = 2 2π. Read sections 11.5 and 11.6 on conic sections and conic sections in polar coordinates because similar material occurs in Chapter 13.5, 13.6 and 13.7. Sections 11.5 and 11.6 are not directly examinable, but 13.5, 13.6 and 13.7 are examinable. Section 13.1, Pages 829–834: 10) (4) What are the projections of the point (2, 3, 5) on the xy−, yz−, and xz−planes? Draw a rectangular box with the origin and (2, 3, 5) as opposite vertices and with its faces parallel to the coordinate planes. Label all vertices of the box. Find the length of the diagonal of the box. Solution On the xy plane we have (2, 3, 0), on the yz plane we have (0, 3, 5) and on the xz plane we have (2, 0, 5). The length of the diagonal is p √ √ 22 + 32 + 52 = 4 + 9 + 25 = 38. A sketch of the box is in fig. 5. 11) (42) Find the volume of the solid that lies inside both 8 Winter 2012 Figure 5: A picture of a box. of the spheres x2 + y 2 + z 2 + 4x − 2y + 4z + 5 = 0 and x2 + y 2 + z 2 = 4. Solution We first rewrite these equations in standard form to get (x + 2)2 + (y − 1)2 + (z + 2)2 = 4 and x2 + y 2 + z 2 = 4. Thus the √ spheres both have radius of 2 and centers that are 22 + 12 + 22 = 3 units apart. To make the calculations easier, we can therefore consider the volume between the spheres (x − 1.5)2 + y 2 + z 2 = 4 and (x + 1.5)2 + y 2 + z 2 = 4. 9 Winter 2012 This is a volume of revolution (see pg. 387 or Stewart) so we need to calculate Z 0.5 2π y 2 dx 0 Z 0.5 = 2π 4 − (x + 1.5)2 dx Z0 0.5 = 2π 4 − 1.52 − x2 − 3xdx  0 x3 3x2 7 = 2π x − − 4 3 2 11π = 12 This agrees with problem 49 on pg. 392. Section 13.2, Pages 834–842: 12) Find the sum of the vectors < −1, 3, 4 > and < 1, 3, 2 >. Illustrate this geometrically. Solution < −1, 3, 4 > + < 1, 3, 2 >=< 0, 6, 6 > This is illustrated in fig. 6. 13) a) Find |a|, a + b, a − b, 2a and 3a + 4b for a =< 1, 2, 3, 4 > b =< 5, 6, 0, 1 > Solution |a| = p 1+ 22 + 32 + 42 = a + b =< 6, 8, 3, 5 > 10 √ 30 Winter 2012 Figure 6: A picture of demonstrating the sum of two vectors. a − b =< −4, −4, 3, 3 > 2a =< 2, 4, 6, 8 > 3a + 4b =< 23, 30, 9, 16 > b) Describe a situation where you may want to use vectors with more than 3 entries. Solution Many possible choices, for example, a shopping list, exam scores, coordinates in space and time. 14) A clothesline is tied between two poles 7m apart. The line is quite taut and has negligible sag. When a wet shirt with a mass of 0.5Kg is hung at the middle of the line, the midpoint is pulled down 5cm. Find the tension in each half of the clothesline. Solution We decompose the tension into vertical and horizontal components. The sum of the vertical components of the tensions in the two halves of the string must be equal to 0.5Kg×10m/s2 = 5N. Thus 2.5N= T sin θ, where T is the tension in the string and θ is the angle 11 Winter 2012 between √ the horizontal and the string. √ Hence sin θ = 0.05/ 3.52 + 0.052 and so T = 2.5 3.52 + 0.052 /0.05N≈ 175.0179N. Bonus problem 15) (pg. 735 no. 4) Four bugs are placed at the four corners of a square with side length a. The bugs crawl counterclockwise at the same speed and each bug crawls directly toward the next bug at all times. They approach the center of the square along spiral paths. a) Find the polar equation of a bug’s path assuming the pole is the center of the square (Use the fact that the line joining one bug to the next is tangent to the bug’s path). Solution Let1 us label the bugs 1, 2, 3 and 4. By symmetry all the bugs each have the same distance from the origin as a function of time and also the angles between radii from the origin to the bugs also remain at π/2 or 90◦ . We let ri denote the position of bug i, so that the equations of motion are given by ṙi = α(rmod(i+1,4) − ri ). We let ri = r(t) [cos θi (t), sin θi (t)] and convert to polar coordinates. We then using the fact that θ4 = θ3 + π/2 = θ2 + π = θ1 + 3π/2 and choose α = − (rK√2 , where K is a constant, so that |ṙ| = 1 We can, following Gardner (1957), assume 1 and 3 are female and 2 and 4 are male. 12 Winter 2012 K. Now K ṙ1 = √ (sin θ1 − cos θ1 , − cos θ1 − sin θ1 ) 2 and also ṙ1 = (ṙ cos θ1 − θ̇r sin θ, ṙ sin θ + rθ̇ cos θ). After equating components, we get that θ˙1 = − √K2 and so since r(t = 0) = √a2 , we get that K √ . We also get that r θ̇ = − √ and so r(t) = a−Kt 2 2 choosing θ(t = 0) = 0, we get that √ Kt 2r = log . θ = log 1 − a a It is also possible to derive the equation for this curve without calculus. b) Find the distance traveled by a bug by the time it meets the other bugs at the center. Solution We find that r = 0 when t = Ka . We recall that the distance traversed is given by s s    2  2 Z Z 2 dr dθ dr r2 + dθ = r2 + dt. dθ dt dt In our parameterization this becomes Z ar 2 K K K2 + dt = a. 2 2 0 c) Download the extension by Chapman, Lottes and Trefethen Four bugs on a rectangle, Proc. Roy. 13 Winter 2012 Soc. A, vol. 467 pg. 881-8962 . Who originally came up with the problem of Four Bugs on a Square? Write a one paragraph summary of what you consider to be the most important point in Four bugs on a rectangle. Solution According to the paper cyclic pursuit first appeared in a Cambridge University tripos exam of 1871. It is also possible to get points for stating; in a 1735 paper by Pierre Bouguer on the problem of a pirate ship chasing a merchant vessel. Anything reasonable for the one paragraph summary will receive credit. 2 Bing, Google, Yahoo or any other search engine should locate this for you. 14