MATHEMATICS 121, Worked Problems

Transcription

1 MATHEMATICS, Worked Problems ) Evaluate a) b) n Solution. a) = i i = d i = d i= j= j j i= = ( )+( ) ( ) = + ( ) The right hand side is not defined for =, but this case ma be handled b using l Hôpital s rule to take the limit : = = lim = lim + ( ) = lim 99 + ( ) = lim = 55 You can also use n = n(n+) to handle this case. b) Let =. n j= j j = n j j = j= n j j = d j= = (n+)n ( )+( n+ ) ( ) n j = d j= n+ Subbing = n j= j = (n+)(/) n+ +( (/) n+ ) = (n + ) ( ) n+ ] j (/) = n+ n

2 ) Find the area of a circle as a limit of the areas of inscribed polgons. Solution. Inscribe a polgon with n equal sides in a circle of radius r. The figure at the right illustrates such a polgon with n = 6. The polgon is built up out of n equal triangles. Each triangle has one verte, with angle nπ, at the centre of the circle. Drop a π/n perpendicular from that verte to the centre of the opposite side of the triangle. This perpendicular bisects the angle we were just talking about. The length of the perpendicular is r cos π. We can n also view this as the height of the triangle. The base of the triangle (the side opposite the centre of the circle) has length r sin π n. So one triangle has area (base)(height) = ( r cos π n )( ) r sin π n = r( sin π n cos ) π n = r sin π n The area of the polgon is the area of n triangles or n r sin π n. The area of the circle is n lim n r sin π n = lim π r sin = lim πr sin = πr sin lim where we have substituted = π n, or equivalentl n = π. sin lim = lim cos =, so the area of the circle is πr. B L Hôpital s rule, ) Evaluate the following integrals b interpreting it as an area. a) 4 b) 5 c) 4 Solution. Recall that the area between the ais and the curve = f() with running from = a to = b is b f(). a a) The integral 4 represents the area between the ais and the curve = 4 with running from to. As the curve = 4 is the top half of the curve = 4 or + = 4, and = ± are the two values of for which =, the integral 4 represents the area of the top half of the interior of the circle + = 4. This circle has radius and hence area π = 4π. The integral is half of this, or π. b) Observe that 5 takes the value zero for = 5, is positive for > 5 and negative for < 5. Hence { 5 if > 5 5 = ( 5) if < 5 and 5 = 5/ 5/ = ( 5) + ( 5) 5/ 5/

3 The first integral represents the area between the ais and the straight line = 5 with running from (where = 5) to 5 (where = ). This is a triangle with height 5, base 5 and area 5 5 = 5 6. The second integral represents the area between the ais and the straight line = 5 with running from 5 (where = ) to = (where = 4). This is a triangle with height 4, base 5 = 4 and area 4 4 = 6 6. All together 5 = 5/ ( 5) + 5/ ( 5) = = 4 6 c) The integral 4 represents the area between the ais and the top half of the circle + = 4 with. This regions consists of the circular sector OAB, which is the fraction π/6 π = of a full disk of radius, and the triangle OBC, which has base and height. The total area is thus π + = π +. A = (, ) O π B = (, ) C = (, ) 4) Let h() = sin t cos ( t ) dt. Evaluate h (). 5 Solution. Define f(u) = u 5 t cos ( t ) dt and g() = sin Then df du = u cos ( u ) and h() = f ( g() ) so that, b the chain rule, and dg = cos h () = f ( g() ) g () = u cos ( u ) u=g() cos = sin cos ( sin ) cos 5) Find the area of the region bounded b the parabola =, the tangent line to this parabola at (, ) and the ais. Solution. The slope of the parabola at = is m = d = = =. This is = also the slope of the tangent line to = at =. So the equation of the tangent line is = m( ) = ( ) or =. We are to determine the area of the region in the figure (, ) = =

4 Use horizontal slices. On each slice, is constant (with some fied value between and ) and runs from to ( + )/. So, the area is ( + )/ ] d = + ] / / ] = + ] / / = 4 = 6) Find the number b such that the line = b divides the region bounded b the curves = and = 4 into two regions with equal area. Solution. The area of the part of region with b is A = b d = / / b = b/ The area of the part of region with b 4 is (, 4) = b = A = 4 b d = / / 4 b = 4/ b/ The two areas are equal when b/ = 4/ b/ b/ = 4/ b / = 4/ b = 4 = = 4/ / / 7) Set up, but do not evaluate, an integral for the volume of the solid obtained b rotating the region bounded b the specified curves about the given line. a) = ln, =, =, about the ais. b) = cos, =, =, = π about the line =. Solution. a) When the vertical strip shown in the figure is rotated about the ais it forms a washer with inner radius ln and outer radius. The area of this washer is π π(ln ). As runs from to e (which is where ln = ) the volume of the solid is π e (ln ) ]. = P = ln = ln P P P side view of slice b) When the vertical strip shown in the figure below is rotated about the line = it forms a washer with inner radius cos and outer radius. The area of this washer is π π( cos ) 4

5 As runs from to π, the volume of the solid is π π/ ( cos ) ]. = cos = 8) Each of the following integrals represents the volume of a solid. Describe the solid. a) π 6 d b) π 4 ] 6 ( ) 4 Solution. a) When the horizontal strip shown in the figure on the left below is rotated about the ais it forms a disk with radius f() and hence of area πf(). The volume of the region obtained when one rotates the curve = f(), a b about the ais is π b a f() d. This matches the given integral when a =, b = and f() =. So the integral represents the volume of the region obtained when one rotates the curve =, about the ais. = f() = f() = g() b) The integrand π 6 ( ) 4] = π4 π ( ) ] is the area of a washer of outer radius 4 and inner radius ( ). Suppose in general that the region g() f() is rotated about the ais. When the vertical strip shown in the figure on the right above is rotated about the ais it forms a washer with outer radius f() and inner radius g() and hence of area πf() g() ]. The volume of the region obtained when one rotates g() f(), a b about the ais is π b a f() g() ]. This matches the given integral when a =, b = 4, f() = 4 and g() = ( ). So the integral represents the volume of the region obtained when one rotates ( ) 4, 4 about the ais. 5

6 9) Find the volume of a solid torus (donut) obtained b rotating the circle ( R) + = r about the ais. Solution. ( R) + = r r R ( R r, ) P P ( R + r, ) R r P P R + r top view of slice The torus ma be formed b rotating the circle in the figure on the left above about the ais. Call this circle the generating circle. It is centred on (R, ) and has radius r and so has equation ( R) + = r. Slice the donut horizontall. The figure on the right shows a top view of a tpical slice. The points on one slice all have the same value of (with some ). The slice looks like a washer. Our main problem is to determine the inner and outer radii of the washer. The portion of the washer that lies in the generating circle is marked in both figures as a horizontal line. The end points of the line are denoted P and P. Both P and P lie on the washer and hence have coordinate. The also both lie on the generating circle. We ma determine their -coordinates b solving ( R) + = r for. The coordinate of P is = R r that of P is = R + r. So the washer has inner radius R r and outer radius R + r. The area of the washer is π ( R + r ) π ( R r ) = π 4R r ] The volume of the donut is the integral of the cross sectional area of the slice from = r to = r. That is, r 4πR r d = 4πR(area of right half of a circle of radius r) r = 4πR ( πr) = π Rr ) Evaluate π π sin + 6. Solution. The critical observations here are that the integrand is odd and the domain of integration is smmetric about =. Recall that f() is odd if f( ) = f(). I claim that, for an odd function f() and an positive number a, a f() =. a Intuitivel, here is the reason. The part of the graph = f() from = to = a and part of the graph = f() from = to = a are just upside down images of 6

7 = f() a a each other. So the magnitude of the area between the ais and the part of the graph = f() from = a to = is the same as the magnitude of the area between the ais and the part of the graph = f() from = to = a. The signed areas are negatives of each other. That is a f() = a f(), so that a f() = a a f() + a f() =. To see algebraicall that a f() = a f(), just make the substitution = t, = dt: a f() = a a a f( t) ( dt) = f( t) dt = f( t) dt = f(t) dt a The oddness of f was used in the last step. Choosing f() = sin + 6 and a = π gives π π sin + 6 =. ) Evaluate a) e +e b) π/ π/ + cos c) sin Solution. a) Make the substitution = e, d = e =. = e +e +/ d = b) Using the trig identit cos θ = cos θ with θ = π/ + d = tan + C = tan e + C + cos = π/ cos Since cos for all π π/ cos = π/ b) (Second solution.) π/ cos = sin π/ + cos = (+cos )( cos ) cos = Now substitute = cos, d = sin π/ sin cos = 7 π/ π/ cos d = = = sin π 4 = cos = π/ sin cos

8 c) Make the substitution = π, d =. π/ sin = π/ cos ( d) = cos d π/ B the trig identit cos θ = sin θ with θ =, π/ cos d = π/ sin d = π/ = ( cos π 4 ) = ( ) sin d = cos π/ ) Find the volume of the solid obtained b rotating the region bounded b = 4 and = 8 about =. Solution. The region bounded b = 4 and = 8 is sketched on the right. Note that the = 8 two parabolas meet at (, ) and (4, ). Consider the thin slice in the figure on the right. It runs verticall from (, 4 ) to (, 8 ) and has width. = 4 When this slice is rotated about =, it sweeps out a clindrical shell. A radius for the shell is shown in the 4 figure on the right. It is the horizontal line half wa up the thin slice. The coordinate of the right hand end of the radius is and the coordinate of the left hand end is. So the radius has length ( ) = +. The height of the shell is the difference between the coordinates at the top and bottom of the thin slice. So the height of the shell is (8 ) (4 ) = 4. The thickness of the shell is. So its volume is π( + )(4 ). The total volume of the solid is 4 π( + )(4 ) = π 4 (8 + ) = π ] 4 = π ] = 56 π ) Find the volume of the solid obtained b rotating the region bounded b = 4 and = 8 about = 5. Solution. The region bounded b = 4 and = 8 is sketched below. Note that the two parabolas meet when 4 = 8 or = 4 or = ±. The corresponding = 4 (±) =. So, the two parabolas meet at (, ±). Consider the thin slice in the figure on the right. It runs horizontall from (4, ) to (8, ) and has width d. When this slice is rotated about = 5, it sweeps out a clindrical shell, as illustrated in the figure on the left. A radius for the shell is shown in the figure on the right. It is the 8

9 5 5 = 8 = 4 vertical line half wa along the thin slice. The coordinate of the top end of the radius is 5 and the coordinate of the bottom end is. So the radius has length 5. The height of the shell is the difference between the coordinates at the right and left hand ends of the thin slice. So the height of the shell is (8 ) (4 ) = 4. The thickness of the shell is d and its volume is π(5 )(4 )d. The total volume of the solid is π(5 )(4 )d = π ( ) d For an odd power n of, and an a, the integral a a n d =. This is because the area with a has the same magnitude but opposite sign as the area with a. See the figure on the left below. Thus the integrals d = d =. = = For an even power n of, and an a, the integral a a n d = a n d. This is because the area with a has the same magnitude and same sign as the area with a. See the figure on the right above. The volume of the solid is π ( ) d = π ( 5 ) d = 4π ( 5 ) d = π (4 ) d = π 4 ] = π ] 8 8 = π 4) The region below the curve =, above the ais and between the ais and the line = is rotated about the line =. a) Epress the volume of the solid of revolution so obtained as a definite integral. 9

10 b) Evaluate the integral of part a. Solution. a) When a point (, ), between the ais and the line =, is rotated about the line =, it sweeps out a circle of radius and hence of circumference π( ). So, b the method of clindrical shells, b) Vol = / Vol = / π( ) π( ) = 4π / For the second integral, sub in t =, dt =. Vol = 4π arcsin / /4 + π t dt = 4π π 6 + π t / π / ] /4 = π + π π 5) Let R be the plane region bounded b =, =, = and = c +, where c is a positive constant. (a) Find the volume V of the solid obtained b revolving R about the ais. (b) Find the volume V of the solid obtained b revolving R about the ais. (c) If V = V, what is the value of c? Solution. a) Let V be the solid obtained b revolving R about the ais. The portion of V with coordinate between and + is a disk of radius c + and thickness. The volume of this disk is π(c + ) = πc ( + ). The total volume of V is V = πc ( + ) = πc ] + = 4 πc b) Let V be the solid obtained b revolving R about the ais. Then V can be thought of as a union of clindrical shells. The shell whose intersection with the ais is the interval from to + has radius, height c + and thickness. The volume of this shell is π(c + ) = πc +. The total volume of V is (making the substitution = +, d = ) V = c) If V = V, πc + = πc d = πc / / = πc/ ] 4 πc = πc/ ] = c = or c = / ] 6) A 45 notch is cut to the centre of a clindrical log having radius cm. One plane face of the notch is perpendicular to the ais of the log. What volume of wood was removed?

11 Solution. Slice the notch into rectangles as in the figure on the right below. Suppose that the base of the notch is in the plane. Then the circular part of the boundar of the base of the notch has equation + =. If our coordinate sstem is such that is constant on each slice, then the slice has width = and height (since the upper face of the notch is at 45 to the base). So the slice has cross sectional area and the volume is V = Make the change of variables t =, dt =. V = t ( dt) = t / / = = 6, Solution Suppose that the base of the notch is in the plane with the skinn edge along the ais. Slice the notch into triangles parallel to the ais as in the figure below. Then the circular part of the boundar of the base of the notch has equation + =. Our coordinate sstem is such that is constant on each slice, so that the slice has both base and height = (since the upper face of the notch is at 45 to the base). So the slice has cross sectional area ( ) and the volume is V = ( ) d = ( ) d = ] = = 6, 7) Evaluate the integrals a) tan b) sin(ln ) c) 4 e

12 Solution. a) Integrate b parts with u = tan and dv =. Then du = + and v =, so tan = tan + The integral on the right hand side is + = ] + + = + = + = tan + C All together tan = tan + tan + C with C = C. b) First, substitute = ln, d =, in order to simplif the argument of sin. Because = e d, sin(ln ) = e sin d Integrate b parts using u = e and dv = sin d so that du = e d and v = cos. e sin d = e cos ( cos )e d = e cos + e cos d Integrate b parts a second time using u = e and dv = cos d so that du = e d and v = sin. e sin d = e cos + e cos d = e cos + e sin (sin )e d Moving the (sin )e d from the right hand side to the left hand side, e sin d = e cos + e sin = e sin d = e cos + e sin ] All together sin(ln ) = e sin d = e sin e cos ] + C = sin(ln ) cos(ln ) ] + C c) First, substitute =, d =, in order to simplif the argument of the eponential. Because = d = d and () =, (4) =, 4 e = e d = e d

13 Integrate b parts using u = and dv = e d so that du = d and v = e. 4 e = e d = e = ] e d = e e ] ] = e (e e ) (e e ) 8) Evaluate sin. Solution. sin = +sin sin = = tan +sin cos = d(cos ) cos = tan + cos + C sec + sin cos Solution Using the trig identities sin = cos ( π ) and cos() = sin, sin = cos( π ) = = cot( π 4 ) + C sin ( π 4 )] = sin ( π 4 ) The answers in the two solutions are reall the same. Using the trig identities sin(a b) = sin a cos b cos a sin b and cos(a b) = cos a cos b + sin a sin b, cot( π 4 ) = cos( π 4 ) sin( π 4 ) = cos π 4 cos +sin π 4 sin sin π 4 cos cos π 4 sin Since sin π 4 = cos π 4 cot( π 4 ) = cos +sin cos sin = cos +sin ] cos sin = + sin cos cos = +sin cos = sec + tan 9) Household electricit is supplied in the form of alternating current that varies from 55 V to 55 V with a frequenc of 6 ccles per second. The voltage is given b the equation V (t) = 55 sin(πt) where t is the time in seconds. Voltmeters read the RMS (root-mean-square) voltage, which is the square root of the average value of V (t)] over one ccle. Calculate the RMS voltage of household current.

14 Solution. The voltage goes through 6 ccles in one second. So one ccle takes /6 sec and /6 /6 RMS voltage = V (t) dt = 6 55 /6 sin (πt) dt /6 = 6(55) = cos( πt) dt 6(55) t sin( πt) 4π = V ] /6 = 6(55) 6 ) If a freel falling bod starts from rest, then its displacement is given b s = gt. Let the velocit after a time T be v T. Show that if we compute the average of the velocities with respect to t we get v ave = v T, but if we compute the average of the velocities with respect to s we get v ave = v T. Solution. The speed at time t is ds dt = gt. In particular v T = gt. The average velocit over the time interval t T is T T gt dt = T gt] T = T gt = gt = v T The bod reaches displacement s at the time t that obes s = gt. That is, at t = s/g. At displacement s, the velocit of the bod is g s/g = sg. The average velocit over the displacement interval s gt (the final displacement is the value of s at time T ) is gt / gt / sg ds = gt / g s / / = gt = v T ] gt / = gt / g (gt /) / = g T / gt / ) Show that if f(a) = f(b) = and f is twice differentiable, then b a ( a)(b )f () = b a f() Solution. Integrating b parts with u = ( a)(b ), du = (b ) ( a)], dv = f () and v = f () b a ( a)(b )f () = ( a)(b )f () b a = b a b (a + b )f () 4 a (a + b )f ()

15 Integrating b parts with u = a + b, du =, dv = f () and v = f() b a ( a)(b )f () = b a (a + b )f () = (a + b )f() b a b = f() a b a f() ) Find the volume enclosed b the ellipsoid a + b + z c = Solution. Suppose that we compute the volume b slicing it into horizontal disks. z c h a The height z of a disk varies from z = c to z = c. The disk at height z has equation a + b = z c. This is an ellipse with semi aes a z c and b z c. Hence the cross sectional area of the disk is πa z c b z c = πab ( ) z c. The volume of the the ellipsoid is V = c c πab ( z c ) dz = πab c b ( z c ) dz = πab z z c ] c = 4 πabc ) Evaluate the following integrals. a) 4 b) +4+8 c) (5 4 ) 5/ Solution. a) Make the substitution u = 4. Then du = so that 4 = 4 = (4 u) u du. When =, u = 4 and when =, u =. Hence 4 = (4 u) u du = ( ] u + u/) du = 4 u/ + 5 u5/ = = ( ) 5 =

16 b) = +4+8 = (+) +4 Sub in + = tan θ. Then = sec θ dθ and ( + ) + 4 = 4 tan θ + 4 = 4 sec θ = sec θ. So = +4+8 sec θ dθ sec θ = sec θ dθ = ln sec θ + tan θ + C Subbing in θ in terms of c) = ln = (5 4 ) 5/ C = (9 4 4 ) 5/ ( + ) + 4 (9 (+) ) 5/ θ + Sub in + = sin θ. Then = cos θ dθ and ( 9 ( + ) ) 5/ = ( 9 9 sin θ ) 5/ = ( 9 cos θ ) 5/ = 5 cos 5 θ. So = (5 4 ) 5/ = 8 cos θ dθ 5 cos 5 θ = 8 + tan θ] sec θ dθ dθ cos 4 θ = 8 sec θ sec θ dθ Now sub in = tan θ. Then d = sec θ dθ so Subbing back in = tan θ, = (5 4 ) 5/ 8 = (5 4 ) 5/ 8 + ] d = 8 + ] tan θ + tan θ + C ] + C Finall we have sub back in θ in terms of, using + = sin θ. From the triangle of + the right, tan θ = = + so 9 (+) 5 4 = (5 4 ) 5/ (+) 5 4 (5 4 ) / ] + C θ + 9 ( + ) 6

17 4) Find the area of the crescent shaped region bounded b arcs of circles with radii r and R, as in the figure to the right. Solution. The large circle has equation + = R. The small one is centred at =, = R r and has radius r, so its equation is + ( R r ) = r. The area is r R r r R r + r R ] = r r R r + r r r r r R The first integral represents the area of a rectangle of width r and height R r. So the first integral equals r R r. The second integral represents the area of the top half of a circle of radius r. So the second integral equals πr. For the third integral substitute = R sin θ. Then = R cos θ dθ. When = ±r, θ = ± sin r R. So r r R = = sin r R sin r R sin = R θ + r R R R sin θ R cos θ dθ sin R cos θ dθ = R ] sin r sin θ R When θ = sin r R, sin θ = r R and cos θ = r R and the area is r r r R r + πr R sin r R R r R r R = R θ + sin θ cos θ ] R = R sin r R + r R r R r R so +cos θ dθ ] sin r R = r R r + πr R sin r R 5) Let R be the region in the plane bounded b = sin, =, = and = π. a) Find the volume of the solid obtained b rotating R about the ais. b) Find the volume of the solid obtained b rotating R about the ais. Solution. a) B clindrical shells, Vol = π π ( sin ) π = π sin 7

18 B integration b parts with u =, dv = sin, du =, v = cos Vol = π ( cos ) π π + 4π π cos = π + 4π cos B integration b parts, a second time, with u =, dv = cos, du =, v = sin Vol = π + 4π sin π 4π π sin = π 4π( cos ) π = π 8π b) Vol = = π π = π π ( sin ) = π π cos ] π sin cos + 4 sin ] π = π π π ] 6) Let R be the region bounded b the line = and = ( 4). Calculate the position of the centroid of R. Solution. The region is smmetric about = 4, so = 4. The line and parabola intersect when ( 4) = or = ( 4) ( 4) = 4 or 4 = ± or =, 6. The area of R is A = 6 ( 4) ] ] = 6 = + 4 ] 6 = ] = + 8 ] = We use vertical approimating rectangle to compute. A tpical rectangle has width and running from ( 4) to. Thus the average value of on the rectangle is + ( 4) ] ] = ( 4) and the area of the rectangle is ( 4) ] ] = + 8 ]. ȳ = A 6 Sub in = t + 4, = dt. ȳ = A = ( 4) + 8 ] 6 = A ( 4) (6 )( ) t ( t)(t + ) dt = 64 ] 4 t t5 5 ] = 5 = 5 4t t 4 ] dt = 4t t 4 ] dt 8

19 7) Find the coordinate of the centroid (centre of gravit) of the plane region R that lies in the first quadrant, and inside the ellipse = 6. (The area bounded b the ellipse a + b = is πab square units.) Solution. In standard form = 6 is 9 + =. The ellipse has a = and 4 b =. So, on R, runs from to a = and R has area A = 4 π = π. For each fied, between and, runs from to () = 9. = A () = A Sub in t = 9, dt = 9. = 9 4 π t dt = 9 9 = 4 π 9 4 π ] t / / = 9 ] 4 π = 4 π 8) a) Find the centroid of the quarter circular disk,, + a. b) Find the centroid of the region R in the diagram. (, ) R quarter circle Solution. a) B smmetr, = ȳ. The area of the quarter disk is A = 4 πa. = A a a To evaluate the integral, sub in t = a, dt =. a a = a t dt = ] t / / a = a ( ) So ] = 4 a πa = 4a π 9

20 b) Again, b smmetr, = ȳ. The area of the region is A = 4 π = 6 π 4. = ] A ] + ] = 4 6 π + ] ] = 4 6 π 4 b ( ) with a = = π Solution When the quarter disk is rotated about the ais a hemisphere is formed. The hemisphere has radius a and hence volume 4 πa = πa. The quarter disk has area 4 πa = 4 πa. B Pappus s Theorem πa = V = π A = π a, so that = 4a. To get ȳ = either invoke smmetr or rotate about the ais. π When R is rotated about the ais a clinder with a hemispherical hole is formed. The clinder has radius and height and hence volume π and the hole has radius and hence volume 4 π. So the volume of the solid formed b rotating R is π. R has area π. B Pappus s Theorem = V 4 πa = π/ π(6 π)/4 = 44. Again, to 48 π get ȳ = either invoke smmetr or rotate about the ais. 9) Prove that the centroid of an triangle is located at the point of intersection of the medians. Solution. Choose a coordinate sstem so that the vertices of the triangle are located at (a, ), (, b) and (c, ). (In the figure, a is negative.) (, b) (a, ) (c, ) The line joining (a, ) and (, b) has equation b + a = ab. (Check that (a, ) and (, b) both reall are on this line.) The line joining (c, ) and (, b) has equation b + c = bc. (Check that (c, ) and (, b) both reall are on this line.) Hence for each fied between and b, runs from a a b to c c b. A thin strip at height has length l() = (c c b ) (a a b )] = c a (b ). On this strip has average value and b has average value (a a b ) + (c c b )] = a+c b (b ). As the area of the triangle is A = (c a)b ȳ = A b = b b 6 = b b l() d = (c a)b c a b (b ) d = b b (b ( ) d = b b b ) b

21 = A = a+c b b b a+c (b )l() d = b (c a)b ( b)] b = a+c b b = a+c a+c b (b ) c a b (b ) d = a+c b b ( b) d The midpoint of the line joining (, b) and (c, ) is (c, b). The point two thirds of the wa from (a, ) to (c, b) is (a, ) + (c, b) = ( a+c, ) b = (, ȳ) as desired. ) A water storage tanker has the shape of a clinder with diameter ft. It is mounted so that the circular cross sections are vertical. If the depth of the water is 7 ft, what percentage of the total capacit is being used? Solution. Here is an end view of the tank. The shaded part of the circle is filled with water. The cross sectional area of the the filled part of the tank is the area of a circle θ 5 7 of radius 5, namel π5, minus the area of the unfilled cap. The area of the cap is ( ) ( Area = Area θ 5 ) ( Area θ 5 ) The first term is the fraction θ of the area of a circle of radius 5. The triangle in the π second has base 5 cos θ and height 5 sin θ. So the area of the cap is θ π π5 ( 5 cos θ)(5 sin θ) = 5θ 5 sin θ cos θ The tank is filled to a height of 7 ft, which is ft above the centre of the tank. So cos θ = 5 =.4 and sin θ = and θ = arccos The area of the cap is ] 9.8. The area of the full circle is A = 5π, so the fraction filled is =.74 = 74.8% A 9.8 A = ) Evaluate the following integrals. a) b) c) (+) 4 + d) +

22 Solution. a) First, we have to factor the denominator P () = This is a polnomial with integer coefficients. So an integer roots of P () must factor the constant term, 6, of P (). The onl possible candidates for integer roots of P () are ±, ±, ±, ±6. Tr =. Because P () =, = is a root of P () and must be a factor of P (). B long division, P () = ( )( ) = ( )( + )( + ) The degree of the numerator 8 4 is, which is smaller than the degree of the denominator, which is. So the integrand must be epressible in the form 8 4 ( )(+)(+) = a + b + + c + The eas wa to find a is to multipl both sides of this equation b a + ( ) b c + + ( ) + = 8 4 (+)(+) and then set = Similarl, a = 8 4 (+)(+) = = = b = 8 4 ( )(+) = ( ) = = c = 8 4 ( )(+) = ( 4) ( ) = = To check this, put everthing back over the common denominator ( )( + )( + ) = (+)(+) ( )(+) ( )(+) ( )(+)(+) = ( ) +(5 4 )+(6+6+6) ( )(+)(+) This is the same as the original integrand. So ] = = ln ln + ln + + C + b) Substitute t = +, dt =. (t ) = dt = (+) t t t +t ( dt = t t + t t ) dt with C = C. = t ln t t + t + C = ln (+) + C c) Trick question! The numerator is, aside from a factor of, the derivative of the denominator. So make the change of variables u = 4 +, du = = du/ u = ln u + c = ln c

23 d) Solution Sub in = tan θ. Then = sec θ dθ and + = tan θ + = sec θ = sec θ. When =, tan θ = so θ =. When =, tan θ =, so θ = π 4. + = = = π/4 = sec θ sec θ dθ = π/4 sec θ dθ sec θ tan θ + ln sec θ + tan θ ] π/4 + ln + ] + ln ( + ) ] + ln + ] d) Solution Sub in = ( e e ). Then = ( e + e ) d ( + = 4 e e ) ] / + = 4( e + e ) + ( = 4 e + + e )] / ( = 4 e + e ) ] / ( e + e ) = = + = 4( e + e )( e + e ) d = 4( e + + e ) d ] / = + = 4 (e + + e ) d = 8 e + 8 e + C When =, e e =, so e = e so e = so =. When =, e e =, so e = e so e e =. View this as a quadratic equation in z = e. The solutions to z z = are z = ( ± 4 + 4) = ±. Since e >, onl the solution z = + is allowed. So when =, e = + and = ln ( + ). + = To simplif this use + = 8 8 e + 8 e ] ln( + = +) ( +)( ) =. ) ( ) = 8( ln ( + ) ( ) ( ) ln ( + ) = + ln ( + ) ) Evaluate a) e + e b) e + c) + + d) e t 9 dt

24 Solution. a) Make the substitution = e, d = e =. e + e = ] + d = d = ln ln + C = ln e ln e + C = ln e / e / + C = ln e + e + C b) Make the substitution = e, d = e =. e + = ] + d = d = ln ln + + C + = ln e ln(e + ) + C = ln ( + e ) + C c) Let = u = +u +u 6u 5 du B long division u 8 +u 5 +u = u 6 u 4 + u + u u + u+ +u Hence + + = 6 u 6 u 4 + u + u u + u+ = 6 +u ] u 7 7 u5 5 + u4 4 + u u u + ln( + u ) + tan u du ] + C = 6 7 7/ /6 + / + / / 6 /6 + ln( + / ) + 6 tan /6 + C a) Make the substitution = e t, d = e t dt = dt. e t 9 dt = 9 d Now substitute = sec u, d = sec u tan u du. Since 9 sec u 9 = 9 tan u = tan u, e t 9 dt = 9 sec u tan u du sec u 9 = sec u tan u du = (sec u ) du = tan u u ] + C = sec u u ] + C = ( ) ] sec + C = e t et 9 sec + C ) The circle + ( ) = and the parabola = ( ) are tangent to each other. 4

25 a) Carefull plot these two curves and determine the verte of the parabola and the points of tangenc. b) Find the area of the crescent shaped region bounded b these curves. Solution. a) The parabola is smmetric about the ais. It hits the ais at = /. So the verte is (, /). An point of tangenc must also be a point of intersection. At a point of intersection, = ( ) and = + so + = ( ) or = or =. The corresponding values of are = ( ) = ±. We are told that there is at least one point of tangenc. B smmetr about the ais ±(, ) are both points of tangenc. b) The bottom half of the circle has equation. So the area is ( ) ] = = ] ] = 8 For the remaining integral, sub in = sin t, = cos t dt. When =, t = and when =, sin t =, so t = π. The area is 4 8 π/4 = 8 sin t π/4 cos t dt = 8 4 cos t dt π/4 = 8 + cos t] dt = t 8 + ] = 8 π 4 + = 5 π ] π/4 sin t 4) Let R be the region in the plane bounded b = +, =, = and =. a) Find the volume of the solid obtained b rotating R about the ais. b) Find the volume of the solid obtained b rotating R about the ais. Solution. a) Vol = b) B clindrical shells, π ( ) + ( ) ] = π + = π ] = π Vol = π ( ) + = π + 5

26 Sub in = tan t, = sec t dt, = tan π 4, = tan π. Vol = π π/ π/4 tan t + sec t dt = π π/ π/4 π/ sec cos t t dt = π cos 4 t dt π/4 Sub in u = sin t, du = cos t dt, cos 4 t = ( sin t) = ( u ), sin π 4 =, sin π = / Vol = π / / = π / / = π / = π = π = π = π u u u ( u ) / du = π / / u +u] + / du ( u) + ( u)(+u) + (+u) ] ( u) + u + +u + (+u) ] ln u + ln + u +u ] + ln +u / u / = π du ] / / du ] /4 + ln + / ln ln ln ] ] + ln( + ) ln( + ) 5) A swimming pool has the shape shown in the figure below. The vertical cross sections of the pool are squares. The distances in feet across the pool are given in the figure at foot intervals along the siteen foot length of the pool. a) Denote b f() the width of the pool a distance from the left hand end. Epress the volume of the pool as an integral in terms of f. b) Find an approimate value for the integral of part a

27 Solution. a) The volume of the part of the pool with coordinate running from to + is f(). So the total volume is V = 6 f() b) Divide the domain of integration into 8 intervals each of length =. Denote b i the right hand end point of interval number i. Thus =, = 4,, 8 = 6. The corresponding values of f are f( ) = f() =, f( ) = f(4) =, f( ) =, f( 4 ) = 8, f( 5 ) = 6, f( 6 ) = 8, f( 7 ) = and f( 8 ) =. Then 6 V = f() ] f( ) + f( ) + f( ) + f( 4) + f( 5) + f( 6) + f( 7) + f( 8) ] = = 6 ft 6) Consider the function F () = t e t dt t4 e t4 dt. Find the values of for which F () takes its maimum and minimum values on + a. The number a has the propert that F ( + a ) >. Solution. We first compute F (). To do so we write F () = G( ) + H( ) with G() = B the Fundamental Theorem of Calculus, t e t dt, H() = t 4 e t4 dt G () = e H () = 4 e 4 Hence, b the chain rule F () = G ( ) H ( ) = 5 e 4 4 e 4 = 4 ( )e 4 Observe that F () < for < and F () > for >. Hence F () is decreasing for < and increasing for > and F () must take its minimum value when =. It ma take its maimum value at = or = + a. But F () = and we are told that F ( + a ) >. Hence F () takes its maimum value when = + a. 7) Recall that, when n is odd, it is eas to integrate sin m cos n using the substitution = sin. The integral sec is of the form sin m cos n with m = and n =, which is odd. Integrate sec b substituting = sin. 7

28 Solution. Substitute = sin, d = cos. Then cos sec = cos = cos sin = d = ] = / / + = (sin +) sin d ( )(+) d = ln + ln + + C = ln sin + + C sin To make the connection to the indefinite integral sec = ln sec + tan + C, observe that ( ) ( ) sin + (sin +) sin + sin = cos = cos = tan + sec Hence sec = ln sin + + C = ln tan + sec + C = ln tan + sec + C sin 8) Evaluate, using the substitution = tan θ. a) dθ +sin θ b) π/ dθ +sin θ+cos θ c) dθ + cos θ Solution. If = tan θ, then, from the right angled triangle with sides and and hpotenuse +, sin θ = and cos θ + = + so that sin θ = sin θ cos θ = + cos θ = cos θ = + = + dθ = cos θ = + a) dθ +sin θ = + + = = ( + ) + 4 Now sub + =, = d ( + ) + 4 = = b) π/ dθ +sin θ+cos θ = = d = ( ) tan tan θ + + C = ln( + ) 8 + d = tan ( ( + ) ) + C + = = ln + + +

29 c) dθ + cos θ = + + = = 5 + = 5 tan 5 + C = 5 tan ( tan θ 5 ) + C 9) Use cos θ = ( e iθ + e iθ) and sin θ = i( e iθ e iθ) to find identities for sin(θ) and cos(θ) in terms of sin θ and cos θ. Solution. Note that sin ( θ = 8i e iθ e iθ) ( = 8i e iθ e iθ + e iθ e iθ) = 4 i( e iθ e iθ) ( 4 i e iθ e iθ) = 4 sin θ 4 sin(θ) and Hence ( cos θ = 8 e iθ + e iθ) ( = 8 e iθ + e iθ + e iθ + e iθ) = 4 ( e iθ + e iθ) ( + 4 e iθ + e iθ) = 4 cos θ + 4 cos(θ) sin(θ) = sin θ 4 sin θ and cos(θ) = 4 cos θ cos θ 4) For each natural number n and each nonzero comple number a, the equation z n = a has precisel n distinct solutions, called the n th roots of a. a) Find the three cube roots of. b) Show that the sum of the n n th roots of is zero. Solution. a) Note that ( e iπ/ ) = e iπ = ( e iπ/ ) = e iπ = e iπ = since e iπ = ( e i5π/ ) = e i5π = e iπ = since e i4π = Thus e iπ/ = + i, eiπ/ = and e i5π/ = e iπ/ = i are three distinct cube roots of. b) For all integers k, e kπi =. Thus for ever integer k ( ) e ikπ/n n = e ikπ = 9

30 and e ikπ/n is an n th root of. The n comple numbers = e iπ/n, e iπ/n, e i π/n, e i π/n,,, e i(n ) π/n are all different and so are the n n th roots of. Write z p = e iπ/n. The n n th roots of are, z p, zp, z p,, zn p. Using the formula for the partial sum of a geometric series with ratio z p + z p + zp + z p + + zn p = zn p z p The numerator vanishes because z p is an n th root of. 4) Evaluate the given integrals, using comple numbers. a) b) π/ sin 4 θ dθ ( +) Solution. a) We use partial fractions. ( +) = (+i) ( i) = a +i + b (+i) + c i + = a(+i)( i) +b( i) +c( i)(+i) +d(+i) (+i) ( i) d ( i) For this to be true for all, we need a(+i)( i) +b( i) +c( i)(+i) +d(+i) = for all. Setting = i gives (i) d = 4 or d =. Setting = i gives ( i) b = 4 or b =. Subbing b = d = back in and using b( i) + d( + i) = ( i) + ( + i) = gives a( + i)( i) + c( i)( + i) = or a = c =. Hence = ( +) = + b) Since sin θ = i( e iθ e iθ), + (+i) = ( i) ] = ] ] +i = i++i i (+i)( i) π/ π/ π/ sin 4 ( θ dθ = 6 e iθ e iθ) 4 ( dθ = 6 e 4iθ 4e iθ + 6 4e iθ + e 4iθ) 4 dθ 4i e4iθ i eiθ + 6θ + i e iθ 4i e 4iθ] π/ = 6 = 6 sin(4θ) 4 sin(θ) + 6θ] π/ = 6π 6 = 6 π 4) Solve a) dt = + t t d b) dt = t+t t +, () = d c) =

31 Solution. a) dt b) = + t t = ( + t)( ) = ( + t) dt ln = t + t + D = e D e t t / = ±e D e t t / = + Ce t t / d dt = t+t t + = ( + ) t t + d + = t t + ln + = ln(t + ) + D + = e D t + = + C t + where C = ±e D To satisf () =, we need = + C 5 or C = 5, so = + 5t + 5. c) When, d = = d = = = + c = = +c When =, this computation breaks down because d contains a division b. We can check if the function () = satisfies the differential equation b just subbing it in: () = = () =, () = = () = () So () = is a solution and the full solution is () = or () = +c, for an constant c 4) Find an equation for the curve that passes through the point (, ) and whose slope at (, ) is. Solution. We need to find a function () obeing d = and () =. d = = d = = = + c To satsif () =, we need = + c or c =. So = = + = +. or 44) In 986, the population of the world was 5 billion and was increasing at a rate of % per ear. Using the logistic growth model with an assumed maimum population of billion, predict the population of the world in the ears, and 5. Solution. Let (t) be the population of the world, in billions of people, at time t. The logistic growth model assumes = k(m )

32 We know that, if at time zero the population is below M, then as time increases the population increases, approaching the limit M as t tends to infinit. So in this problem M is the maimum population. That is, M =. We are also told that, at time zero, the percentage rate of change of population,, is, so that, at time zero, =.. But, from the differential equation, = k(m ). Hence at time zero,. = k( 5), so that k =. We now know k and M and can solve the differential equation 95 d dt = k(m ) d (M ) = kdt M ] M d = kt + C M ln ln M ] = kt + C ln M = kmt + CM with D = e CM. We know that remains between and M, so that M M = De kmt = M = De kmt. We are given that at t =, = 5. Subbing this, and the values of M and k, in 5 5 = De D = 5 95 = 5 95 et/95 = ( ) 5 95 et/95 95 = (5 5)e t/95 = 5et/95 et/95 = et/ e = t/95 + 9e t/95 In the ear, t = 4 and = +9e 8/ billion. In the ear, t = 4 and = +9e 8/ billion. In the ear, t = 54 and = billion. +9e 8/95 45) When a raindrop falls it increases in size so that its mass m(t), is a function of time t. The rate of growth of mass is km(t) for some positive constant k. According to Newton s law of motion, d dt (mv) = gm, where v is the speed of the raindrop and g is the acceleration due to gravit. Find the terminal velocit lim v(t) of a raindrop. t Solution. First we determine m(t). dm dt = km = dm m = k dt = ln m = kt + c = m = dekt where d = e c Define (t) = m(t)v(t). Then d dt = gm(t) = gdekt = d = gde kt dt = = gd ekt k + C Subbing back in (t) = m(t)v(t) = de kt v(t), = gd ekt k + C dekt v(t) = gd ekt k + C v(t) = g k + C de kt C Since k >, lim = and lim v(t) = g t e kt t k.

33 46) A population of rare South American monkes has been found to have a population growth rate which decreases with time. Specificall, the population size, (t), satisfies the differential equation (t) = (t+) a) Given that at time t = the population is, solve for the population size (t) after time t. b) In the limit t, what is the size of the population? Solution. a) (t) = (t+) d = (t+) dt d = dt (t+) = Ke (t+) ln = (t+) + C where K = e C. At t =, =. Subbing this in = Ke, so K = e and (t) = e (t+) +. b) As t, t+ approaches. So lim (t) = lim + = e + = e t t e (t+) 47) Suppose that (t) (measured in thousands of dollars kilobucks ) represents the balance in a savings account at time t ears after the account was opened with an initial balance of kilobucks. Interest is being paid into the account continuousl at a nominal rate of 5% 5 per annum; that is to sa, interest is flowing into the account at a rate of (t) kilobucks per ear at time t. At the same time, the account is being continuousl depleted b taes and other charges at a rate (t) kilobucks per ear. No other deposits or withdrawals are made on the account. a) What is balance in the account at time t = ears? b) How large can the balance in the account grow over time? Solution. a) The differential equation determining the time dependence of (t) is d dt (t) = 5 (t) (t) = d dt = 5 + = 5 d = dt The integral of the left hand side is ] 5 d = ( 5) d = 5 d = ln 5 ln + C = ln 5 + C Before continuing let s discuss the sign of 5. We start with =. The rate of change of is determined b d dt (t) = 5 (t) (t) = (5 ). Note that the rate of change is positive for 5 and negative otherwise. So, as we start with =,

34 increases at the beginning and will continue to do so as long as 5. But can never get above 5, because d is negative, that is is decreasing in time, for all > 5. So dt must remain between and 5 and 5 = 5. Now back to the solution. d dt (t) = 5 (t) (t) = 5 d = dt = ln + C = t 5 Dividing b and then taking the eponential of both sides 5 = Ke t/ where K = e C. At t =, =, so K = 5 = 4. 5 = 4e t/ = 5 = 4e t/ = ( + 4e t/) = 5 = = 5 + 4e t/ and () 5.87 b) As t increases, + 4e t/ decreases, approaching the limit as t tends to infinit. Hence increases over time, but remains below 5, approaching 5 as time tends to infinit. 48 a) Let T n and M n be the n step Trapezoidal and Midpoint Rule approimations, respectivel, for cos( ). Find T 4, T 8, M 4 and M 8. b) Estimate the errors involved in these approimations. Solution. a) T 4 = 4 cos + cos ( ( 4) + cos ( 4) + cos ) ] 4 + cos = T 8 = 8 cos + cos ( ( 8) + cos ( 8) + cos ( 8) + + cos 7 8 M 4 = 4 cos ( ( 8) + cos ( 8) + cos 5 ( 8) + cos 7 ) ] 8 =.9897 M 8 = 8 cos ( ) ( 6 + cos ) ( 6 + cos 5 ) ( cos 5 ) ] 6 =.956 b) ) + cos ] =.9 f() = cos( ) = f () = sin( ) = f () = sin( ) 4 cos( ) On the interval, sin( ) is never bigger than and cos( ) is never bigger than and is never bigger than, so f () is never bigger than + 4 = 6. Hence ET4 6( ). 4 ET8 6( ).78 8 EM4 6( ) EM8 6( )

35 49) How large should n be to ensure that the Simpson s Rule approimation to e is accurate to within.? Solution. f() = e = f () = e = f () = e + 4 e = ( + )e = f () () = (4)e + 4( + )e = 4( + )e = f (4) () = 4( + 6 )e + 8( + )e = 4( )e On the interval, e is never bigger than e and ( ) is never bigger than = 9, so f (4) () is never bigger than 4 9 e = 76e. Hence error in Simpson s rule is at most 76e( )5 8n. This error is smaller than 5 if 4 76e 5 n 4 76e 5 n 4 76e 5 = 8.4 8n For Simpson s rule, n must be even so take n. 5 a) State Simpson s Rule to approimate b f(). a b) Use Simpson s Rule to compute with n = 4. Solution. a) Let n be a strictl positive even integer and h = b a n. Define j = a + jh. Simpson s Rule is b a f() h f( )+4f( )+f( )+4f( )+f( 4 )+ +f( n )+4f( n )+f( n ) ] b) When a =, b = and n = 4 we have h =. 6 f() + 4f(.5) + f() + 4f(.5) + f() ] = 6 + 4(.5) + () + 4(.5) + ] = 4 5) Suppose that f() has a continuous second derivative that obes f () K for all a b. Let M n be the result of appling the midpoint rule to b f() with n a subintervals. Prove that b f() M n K(b a) 4n a Hint: use the error estimate for the tangent line approimation to show that f() f(m) f (m)( m) K ( m) 5

36 Solution. Set, for j =,,,, n, j = a + b a n j. This partitions a b into n equal subintervals, each of width b a n. Also, let m j = j + b a n be the midpoint of the jth interval. The midpoint rule approimates j+ j f() b j+ j f(m j ) = f(m j ) b a n. The error in this approimation is e j = j+ j f() f(m j ) b a n j+ = f() j j+ j f(m j ) = j+ j f() f(mj ) ] Talor epanding f() about m j to first order (with a second order error) gives f() = f(m j ) + f (m j )( m j ) + f (z)( m j ) for some z between and m j. B hpothesis f (z) K so that Furthermore, subbing = m j, f() f(mj ) f (m j )( m j ) K( m j) j+ j f (m j )( m j ) = f (m j ) b a n b a n d = So, the error arising in the j th interval ej = j+ j j+ j = K ( m j ) f() f(mj ) ] j+ = f() f(mj ) f (m j )( m j ) ] f() f(mj ) f (m j )( m j ) j j+ K( m j) j = K ( j+ m j ) = K( ) b a n = 4 K (b a) n j j+ The total error, from all n intervals is b a f() M n e + e + + e n n 4 K (b a) n K(b a) 4n 6

37 5) Suppose that f() has a continuous second derivative that obes f () K for all a b. Let T n be the result of appling the Trapezoidal Rule to b f() with n a subintervals. Use the generalization of the Mean Value Theorem given below to prove that b f() T n K(b a) n a Solution. Set, for j =,,,, n, j = a + b a n j. This partitions a b into n f() equal subintervals, each of width b a n. The Trapezoidal Rule approimates j+ j b f( j) + f( j+ )] b a n, which is the area between j and j+ under the straight line through ( j, f( j ) ) and ( j+, f( j+ ) ). The straight line through ( j, f( j ) ) and ( j+, f( j+ ) ) is = p() = f( j ) + f( j+) f( j ) j+ j ( j ). Note that j+ j p() = f( j ) j+ j ]+ f( j+) f( j) j+ j ( j+ j ) = f( j)+f( j+ )] j+ j ] as desired. The error in this approimation is e j = j+ f() f( j)+f( j+ )] b a n j = j+ j+ j+ ] f() p() = f() p() j j j B the generalization of the Mean Value Theorem given below, with n =, = j and = j+ f() p() = ( j )( j+) f (c ) for some c between and m j. B hpothesis f (c ) K so that f() p() K j j+ = K( j)( j+ ) for all j j+. So, the error arising in the j th interval j+ ] j+ ej = f() p() K( j)( j+ ) j Subbing t = j, dt =, j+ j = b a n, b a ej K n j ] b a t( b a n t) dt = K b a t n t n The total error, from all n intervals is b a = K (b a) n f() M n e + e + + e n n K (b a) n ] K(b a) n = K (b a) n 7

38 Theorem (Mean Value Theorem for Interpolation) Let a < < < < n b. Let f() and its first n derivatives be continuous on a b and let f (n+) () eist on a < < b. Suppose that p() is a polnomial of degree at most n that coincides with f() at,, n. That is, assume that f( ) = p( ), f( ) = p( ),, f( n ) = p( n ) Then, for each a b, there is a number a < c < b, depending on, such that f() p() = ( )( ) ( n ) f (n+) (c ) (n+)! Proof: If is one of,, n then both of f() p() and ( )( ) ( n ) f (n+) (c ) (n+)! are zero, for an c, and hence are equal. So it suffices to consider a that is different from all of,, n. Fi an such and define g(z) = f(z) p(z) (z ) (z n ) f() p()] ( ) ( n ) Note that g(z) is zero for at least n+ different values of z, namel, z = and z =,, n. The Mean Value Theorem implies that dg dz has a zero between an two successive zeroes of g(z). So dg dz (z) is zero for at least n+ different values of z. B the Mean Value Theorem, d g dz has a zero between an two successive zeroes of dg dz. So d g dz (z) is zero for at least n different values of z. Continuing in this wa, we see that dn+ g As p(z) is a polnomial in z of degree at most n, dn+ p dz n+ fied in this computation. So (z ) (z n ) ( ) ( n ) z whose degree n + term is n+ ( ) ( n )f() p()]. Hence dz has at least one zero. Call it c n+. =. Recall that is being held f() p()] is a polnomial in z of degree n + d n+ (z ) (z n ) dz n+ ( ) ( n ) f() p()] = (n+)! ( ) ( n )f() p()] and = dn+ g dz n+ (c ) = f (n+) (c ) (n+)! ( ) ( n )f() p()] Moving f (n+) (c ) to the other side of the equation and cross multipling gives as desired f() p() = ( ) ( n ) (n+)! f (n+) (c ) 8

39 5) Let I = π/ ln(sin ). π/6 a) Determine the maimum, K, of f (), for π 6 π, where f() = ln(sin ). b) How large should n be in order that the approimation I M n be accurate to within 4? Solution. a) f() = ln(sin ) = f () = cos sin = cot = f () = csc = sin For π 6 π, = sin π 6 sin sin π =. So the largest value of f () on the interval π 6 π occurs at = π 6 and is (/) = 4 = K. b) We wish to choose n so that In this case K = 4, a = π 6 and b = π so K(b a) 4n 4 4(π/) 4n 4 n 4(π/) 4 4 = π 4 4 n π/ = 4.75 So n 44 will do the job. 54) Find the values of p for which each of the following integrals converges and evaluate the integrals for those values of p. a) b) p p Solution. a) p = lim ɛ + ɛ p = lim In conclusion, b) if p < p = if p > if p = = p p ɛ + R = lim p R = lim p ] p p ɛ if p ln if p = ɛ = lim ɛ + if p < and diverges otherwise. R if p > p = if p < if p = ] R p p ln R 9 if p if p = = lim R { p ɛ p p ] if p ln ɛ if p = { ] R p p p if p ln R if p =

40 In conclusion, = p p if p > and diverges otherwise. 55) Evaluate (+). Solution. Trick question! The integrand has a singularit when + = or =. The domain of integration has to be split up into two parts: < < and < <. (+) = lim a + a (+) = = lim / a + a+ / (+) lim / a + ] = (+) a since a + approaches zero as a tends to. So the integral diverges. Similarl, (+) = a lim a = lim (+) = / a ( +) / a+ and the other part of the integral diverges too. lim / a ] = (+) a 56) Determine if the following integrals converge or diverge. Justif our answers. a) / ( + ) b) / ( + / ) Solution. With both of these integrals, there are two potential problems which could cause the integrals to diverge (i.e. become infinite). First, the integrands become infinite at = and secondl runs all the wa to =. To separate these two potential problems, split up the domain of integration into and. / ( + ) = / ( + ) / ( + / ) = + / ( + / ) + / ( + ) / ( + / ) First consider the domain of integration. On this domain / (+) / and / (+ / ). So both integrals / / ( + ), / ( + / ) 4 / / = / =

41 So both / (+) and / (+) / =. So 4/ converge. Now for the second domain. For, / (+ / ) / ( + ) 4/ = / / = and / (+) converges. On the other hand, for, we have / (+ / ) / ( / + / ) = 5/6 and hence / ( + / ) 5/6 = /6 /6 = and / (+ / ) diverges. 57) Find the length, L, of the curve = e,. Solution. L = + ( ) d = + e Make the change of variables u = + e, du = L = +e u u du = +e e = u +e u. + / ] = u + ln u ln u + +e u / u+ = + e + ln +e +e + ln + ] du 58) Find the length of the curve = from ( 7, ) to ( 67 4, ). Solution. On this curve = + 4. So d = 4 + ( ) d = = = ( ) Hence length = + ( ) d d = ( ) + 4 d = ] 4 =

42 59) Find the length of the curve / + / =. Solution. This equation does not change if ou substitute or. So the curve is invariant under reflection in the or aes. It suffices to find the length of the part of the curve in the first quadrant and multipl b 4. On this curve = ( /) /. So + ( d ( ) / /( /) = ( /) / / ) = + ( / ) / = / d = Hence length = 4 + ( ) d = 4 / = 4 / = 6 6) Set up two integrals, one of which is improper, representing the part of the curve = between (, ) and (, ). Evaluate both integrals. Solution. Since d = d = = L = = / + ( d d = ) L = / = = / To evaluate the integral over, substitute z = : L = /4 + ( d ) d d z 4 9 dz = 4 9 z/] /4 = 8 / 7 8 ] To evaluate the integral over, which is improper because of the singularit at =, substitute = /, = / d. This ields the integral that we just evaluated. 6) Find the area of the surface obtained b rotating the given curve about the ais. a) = 4 ln, 4 b) 8 = ( ), 4

43 Solution. a) ( d + ( d d = ) ( = + 4 ) = + 4 ) ( + ) = 4 ( + + ) = 4 ( ) + So Area = = π = π 4 4 π() + ( d ) = π 4 ( + ln ln ) ] ln + 4 (ln ) ( 4 ln ) ( ) + = π ln (ln )] b) Differentiating both sides of 8() = ( ) with respect to, 6 d = 4 = ( ) = d = 8 ( ) ( d ) = 64 ( ) = ( ) 8( ) + ( ) d = + ( ) 8( ) = ( ) = ( ) = ( ) 8( ) So Area = = π 4 π() + ( d ( ) = π 4 ) = π 4 ] 8 8( ) = π 4 6) Let L be the length of the part of the curve = e ling between = and = 4. Use the Trapezoidal Rule with four subintervals to estimate L. Solution. Let f() = e. Then f () = e so L = 4 + f () = + e. Denoting g() = + e, the Trapezoidal Rule with = 4 gives L g()+g()+g()+g()+ g(4)] = 4.8 6) At what point does the curve = cos t, = (tan t)( cos t) cross itself. Find the equations of both tangents at that point. 4