Eploration 8-a: Number-Line Graphs for Derivatives Objective: Relate the signs of the first and second derivatives to the graph of a function. The graph shows the quartic function f() H 3 D 52 3 C 3 2 D 672 C 9 f (). From the arrows ou drew in Problem 3, ou can tell whether the critical point is a maimum or a minimum. Write ma or min in the f() row. Do these agree with the graph of f at the beginning of this Eploration? 5 2 3 5 6 7 8 9. Find an equation for fq(). Factor the epression for fq() into three linear factors. Use the result to write all values of for which fq() H. 5. Find an equation for f P(), the second derivative. Find the two values of that make f P() H. Mark these on this number line the wa ou marked the critical points in Problem 2, putting in the f P() row. Put C or D in the f P() row for intervals where f P() is positive or negative. Put arcs in the f() row, opening up if ou put C for f P() or down if ou put D, like this: or f () f () 2. The values of in Problem determine critical points of the graph of f. On this number line, put a short vertical arrow pointing down to each of these -values. Write above each arrow, in the row corresponding to fq(). 2 3 5 6 7 8 9 6. Do the arcs ou drew in Problem 5 agree with the concavit of the graph of f? f() f () 2 3 5 6 7 8 9 3. The arrows in Problem 2 divide the number line into four intervals. In the fq() row for each interval, put C if fq() is positive in that interval or D if fq() is negative. In the f() row, put diagonal arrows pointing upward if there is a C in the fq() row or downward if there is a D in the fq() row, like this: 7. On the number line in Problem 5, write p.i. (for point of inflection) at each value of where the sign of f P() changes. Do these values agree with the points of inflection in the graph of f? or (Over) 6 / Eploration Masters Calculus: Concepts and Applications Instructor s Resource Book
Eploration 8-a: Number-Line Graphs for Derivatives continued 8. Demonstrate that ou understand the relationships among fq(), f P(), and the graph of f b sketching the graph of a continuous function that has the features indicated on these number lines. 9. What did ou learn as a result of doing this Eploration that ou did not know before? f() 2 3 5 6 7 8 9 f() f () + + und. + 2 3 5 6 7 8 9 f() f () + und. 2 3 5 6 7 8 9 Calculus: Concepts and Applications Instructor s Resource Book Eploration Masters / 7
Eploration 8-2a: Maima, Minima, and Points of Inflection Objective: Find the first and second derivatives for a function, and show how values of these correspond to the graph. The figure shows the graph of H 5 2/3 D 5/3 as it might appear on our grapher. In this Eploration ou will make connections between the graph and the first and second derivatives.. Find an equation for the second derivative, P. This is the derivative of the (first) derivative. Factor out an common factors. 5 5 5. If P is negative, the graph is concave down. Show algebraicall that the graph is concave down at H.. Find an equation for Q. Factor out an common factors, including D/3 (the lowest power of ). 6. Find a value of at which P equals zero. The corresponding point on the graph is a point of inflection. 2. What value of makes Q equal zero? What is true about the graph of at that value of? 7. B picking a value of on either side of the point of inflection in Problem 6, show that the graph is reall concave up on one side and concave down on the other, even though the graph appears to be straight in a neighborhood of this -value. 3. You realize that zero raised to a negative power is infinite because it involves dividing b zero. What value of makes Q infinite? What is true about the graph at this value of? 8. What did ou learn as a result of doing this Eploration that ou did not know before? 8 / Eploration Masters Calculus: Concepts and Applications Instructor s Resource Book
Eploration 8-2b: Derivatives and Integrals from Given Graphs Objective: Given the graph of a function, sketch the graph of its derivative function or its integral function.. The graph below shows a function h. On the same aes, sketch the graph of the derivative function, hq. At the cusp, hq() is undefined. Make sure our graph accounts for this fact. h() 5. The graph below shows a function. On the same aes, sketch the graph of the derivative, Q. 6 5 5 5 5 2. For h in Problem, is it also correct to sa that hq() is infinite? Eplain. 6. The graph shows the derivative of a function, Q. On the same aes, sketch the graph of the function if (, ) is on the graph of. 3. For Problem, is h continuous at H? Is h differentiable at H? 6. The graph below shows the derivative, gq, of function g. On the same aes, sketch the graph of g, using as an initial condition g() H 3. 5 3 g () g() 2 3 2 2 3 5 7. What did ou learn as a result of doing this Eploration that ou did not know before? 2 3 Calculus: Concepts and Applications Instructor s Resource Book Eploration Masters / 9
Eploration 8-3a: Minimum-Cost Motel Three Problem Objective: Find an equation for a function, and use the equation to find the minimum value of that function. Developers plan to build motels units with three rooms in a row. Each room must have square feet of floor space. To lower construction costs, the want each unit to have the minimum total length of walls.. The figure shows three possible configurations for a motel. One has 2-b-2 rooms, another has -b- rooms, and the third has -b- rooms. Calculate the total wall length (interior walls plus eterior walls) for each configuration.. Plot the graph of L as a function of. Sketch the result. Use MINIMUM on our grapher to find the minimum value of L and the value of at which this occurs. Show this point on our sketch. 5. Use derivatives to calculate algebraicall the value of that gives the minimum value of L. Justif that this is a minimum rather than a maimum. Does the answer agree with Problem? 2. Let be the breadth of each room and let be the depth of each room. Based on our answers to Problem, sketch the graph of the total wall length, L, as a function of. 3. Write an equation for L as a function of and. Use the fact that each room must have area to transform the equation for L in terms of alone. 6. Write the overall dimensions of the motel with the minimum value of L. 7. What did ou learn as a result of doing this Eploration that ou did not know before? 2 / Eploration Masters Calculus: Concepts and Applications Instructor s Resource Book
Eploration 8-3b: Maimal Clinder in a Cone Problem Objective: Find an equation for a function, and maimize the function. The figures show clinders inscribed in a cone. The cone has radius in. and altitude 2 in. If the clinder is tall and skinn as in the figure at left, the volume is small because the radius is small. If the clinder is short and fat as in the figure at right, the volume is small because the altitude is small. In this Eploration ou will learn how to find the dimensions that give the inscribed clinder its maimum volume.. As increases the volume increases, reaches a maimum, then decreases again. At the maimum point its rate of change will be zero. Set dv/d from Problem 3 equal to zero and solve for. 2 2 (, ) 5. Plot the graph of V. Sketch the graph here. Show on the graph that the maimum volume occurs at the value of ou found algebraicall in Problem. (, ). Calculate the volumes of clinders for which the radius,, equals,, 2, 3, and units. 6. Find the radius, altitude, and volume of the maimum-volume clinder. 2. Write the volume of the clinder in terms of the sample point (, ) on an element of the cone. Then transform the equation so that the volume is in terms of alone. You ma find it useful to determine the equation of the line in Quadrant I that the sample point lies on. 7. Summarize the steps ou went through in Problems 2 through 6 in order to find the maimum-volume clinder. 3. Differentiate both sides of the volume equation in Problem 2 with respect to. 8. What did ou learn as a result of doing this Eploration that ou did not know before? Calculus: Concepts and Applications Instructor s Resource Book Eploration Masters / 2
Eploration 8-a: Volume b Clindrical Shells Objective: Find the volume of a solid of revolution b slicing the rotated region parallel to the ais of rotation. The left-hand figure below shows the region under the graph of H D 2 from H to H 3. The right-hand figure shows the solid formed when this region is rotated about the -ais. (, ) (, ). Demonstrate that ou understand slicing a solid into clindrical shells b using the technique to find the volume of the solid shown below. 2 3 3. As the region rotates, the strip parallel to the -ais generates a clindrical shell as shown below. The volume, dv, of this shell equals the circumference at the sample point (, ), times the altitude of the shell, times the thickness of the shell, d. Write an equation for dv in terms of,, and d. In this solid, the region bounded b the -ais, the lines H and H 2, and the graph of H ln is rotated about the -ais to form a solid. (For clarit, onl the back side of the solid is shown.) (, ) 2. Substitute for to get dv in terms of and d alone. 3. Calculate the volume of the solid b adding up all the dv s and taking the limit (i.e., integrate). Tell wh the limits of integration are to 3, not D3 to 3. 5. What did ou learn as a result of doing this Eploration that ou did not know before? 22 / Eploration Masters Calculus: Concepts and Applications Instructor s Resource Book
Eploration 8-b: Volumes b Shells or Washers Objective: Find volumes of revolution b appropriate calculus. The figure shows the region R in Quadrant I bounded b the graphs of H e. and 2 H D 2.. In doing calculus with this region, tell wh it would be more convenient to slice the region verticall rather than horizontall.. On the figure above, sketch the solid formed b rotating R around the -ais. Write an integral equal to its volume. Evaluate the integral numericall. 2. Find numericall the value of H b in Quadrant I where the two graphs intersect. 3. On the figure above, sketch the solid generated b rotating R about the line H 3. Write an integral equal to its volume. Evaluate the integral numericall. 5. What is the main difference between finding the volume of a solid of revolution b clindrical shells and finding the volume b washers as ou have done earlier? 6. What did ou learn as a result of doing this Eploration that ou did not know before? Calculus: Concepts and Applications Instructor s Resource Book Eploration Masters / 23
Eploration 8-5a: Length of a Plane Curve (Arc Length) Objective: Calculate the arc length of a plane curve approimatel b geometr and eactl b calculus. The figure shows the graph of H 2 from HD to H 2. In this Eploration ou will find the arc length of this graph segment. 3. In Problems and 2 the arc length, L, is approimated b the sum L M 2 C 2 In Problem, H and in Problem 2, H.5. What would ou have to do to find the eact value of L? 2. Draw line segments connecting (consecutivel) the points on the graph where is D,,, and 2. Estimate the length of the graph b calculating the lengths of these segments. Does this overestimate or underestimate the arc length?. If = f(), and f is a differentiable function, then an term in the sum in Problem 3 can be written eactl as L H Q(c)] + [f 2 where c is some number in the respective subinterval. Consult the tet, then tell what theorem is the basis for this fact. 5. Eplain wh, in the limit, L can be written as dl H d d 2 C 2 6. Find dl for H 2. Then write L eactl as a definite integral. 2. Find an estimate of the length using segments connecting the points on the graph where is D, D.5,,.5,,.5, and 2. Eplain wh this estimate is better than the one in Problem. 7. Find a decimal approimation for the eact arc length b evaluating the integral in Problem 6 numericall. (You can consult Section 9-6 to see how to do the integrating eactl, b trigonometric substitution.) 8. What did ou learn as a result of doing this Eploration that ou did not know before? 2 / Eploration Masters Calculus: Concepts and Applications Instructor s Resource Book
Eploration 8-6a: Area of a Surface of Revolution Objective: Calculate eactl the area of a doubl-curved surface of revolution. If a line segment is rotated about an ais coplanar with it, a singl-curved surface is formed (a cone or clinder). Such a surface can be rolled out flat and its area found b calculus or geometr techniques ou have learned before. If a plane curve is rotated about an ais coplanar with it, a doubl-curved surface is formed. Spheres, ellipsoids, paraboloids, and so on, are eamples. Because such a surface cannot be flattened without distortion, its area must be found without first flattening. The figure below shows a doubl-curved surface formed b rotating a graph about the -ais. r l R L L l Doublcurved surface 3. The diagram above shows a frustum of a cone with base radii and slant heights R and r, and L and l. Prove that the lateral area of the frustum equals the circumference at the average radius multiplied b the slant height of the frustum. Rotate a curved graph, get a doubl-curved surface. In the problems below, ou will turn the new problem of finding the area of a doubl-curved surface into the old problem of finding areas of man singl-curved surfaces. Then ou will add the areas of these surfaces and take the limit (i.e., integrate).. Suppose that a curve in the figure above has been sliced into curved segments b partitioning the - ais into subintervals of equal length. Show on the diagram how rotating one of these segments gives a surface that resembles a frustum of a cone. 2. Prove that the lateral area of a cone is πrl, where R is the radius of the cone s base and L is the slant height of the cone.. The slice ou drew in Problem has a slant height approimatel equal to dl, the differential of arc length. Write an equation for ds, the differential of surface area. 5. The graph of H 3 from H to H is rotated about the -ais to form a surface. Find its area. 6. What did ou learn as a result of doing this Eploration that ou did not know before? Calculus: Concepts and Applications Instructor s Resource Book Eploration Masters / 25
Eploration 8-7a: Area of an Ellipse in Polar Coordinates Objective: Find the area of an ellipse from its polar equation, then compare with the area found b familiar geometr. The figure shows the ellipse in polar coordinates r H 3 D 2 cos θ. Show that in general, the area da of the sector is da H 2 r 2 dθ Ellipse Circle of radius r d (r, ) r 5. The eact area of the elliptical region is the limit of the sum of the sectors areas. That is, the area equals the definite integral of da. Write an integral representing the eact area. Evaluate the integral numericall.. Set our grapher to POLAR mode and plot the graph. Use equal scales on both - and -aes. Does our graph agree with the figure above? 2. The sample point (r, θ) shown in the figure is at θ H.3 radian. Calculate r,, and for this point. Show that all three agree with the graph. 6. The -radius, a, of the ellipse shown is 6 units. Measure or calculate the -radius, b. Then confirm that the answer ou got b integration in Problem 5 agrees with the answer ou get using the ellipse area formula A H πab. 3. The area of a wedge-shaped piece of the elliptical region bounded b the graph is approimatel equal to the area of the sector of a circle. Calculate the area of the sector shown in the figure above, if θ H.3 radian and dθ H. radian. 7. What did ou learn as a result of doing this Eploration that ou did not know before? 26 / Eploration Masters Calculus: Concepts and Applications Instructor s Resource Book
Eploration 8-7b: Polar Coordinate Area, Length, and Intersection Problem Objective: Find the area and arc length of a region bounded b graphs in polar coordinates, and find intersections of polar curves. This problem concerns the region in Quadrant I outside the circle r H and inside one leaf of the four-leaved rose r 2 H 6 sin 2θ. 5. Find the area of the entire leaf in Quadrant I. Is the answer an interesting multiple of π? 2 3 5 6. Plot the entire graph (one revolution) on our grapher. Use equal scales on the two aes. (Using SIMULTANEOUS mode will reveal some things ou need to know later in this problem.) Does our grapher confirm what is shown in the figure above? 2. Find the values of θ in Quadrant I at which the two graphs cross. You ma do this algebraicall or numericall. Store the answers, without round-off, as a and b, respectivel. 6. Find the length of the boundar of the leaf in Quadrant I. Is the answer an interesting multiple of π? 7. The two graphs seem to intersect at two points in Quadrant II. Eplain wh neither of these points is a true intersection point. (Tracing the graph ma help.) 3. Find the area of the region between the two graphs.. Confirm from the given figure that our answer to Problem 3 is reasonable. 8. What did ou learn as a result of doing this Eploration that ou did not know before? Calculus: Concepts and Applications Instructor s Resource Book Eploration Masters / 27
Eploration 8-7c: Comet Polar Coordinates Problem Objective: Find the area and arc length of a region bounded b graphs in polar coordinates. Asmptotes Comet. How far does the comet travel along its curved path from the time θ H.8 to the time θ H 5.5? Show our work. Earth =.8 radian Hperbola When a comet approaches Earth, it is drawn into a more and more curved path b gravit as it gets closer and closer. If the velocit is high enough, the comet follows a hperbolic path with Earth s center at one focus as shown in the figure. (When the comet is far awa, its path is close to one of the two asmptotes of the hperbola.) Assume that a comet is on a path whose polar equation is r H 8( D cos θ) D where distances are in thousands of miles.. Plot the path on our grapher. Use a window with [D55, 55] for and equal scales on the two aes. Does our graph agree with the figure? 2. How far is the comet from Earth s center at the point shown in the figure where θ H.8 radian? 3. Find the area swept out b the line segment from Earth s center to the comet from the time θ H.8 to the time θ H 5.5. Show our work. 5. Perform a quick check to show that the distance ou calculated in Problem is reasonable. 6. What did ou learn as a result of doing this Eploration that ou did not know before? 28 / Eploration Masters Calculus: Concepts and Applications Instructor s Resource Book
Eploration 7-6c. Year t Millions P/ t ( P/ t)/p 9 3.7 95 5. 2.38.57... 96 2 79.3 2.59... 97 3 23.2 2.36.6... 98 226.5 2.275... 99 5 28.7 2. dp /dt P M P / t P M (D7.927... R D5 )P C.282... b linear regression, with correlation r HD.9853..., graphed here. 6. The population is stable if dp/dt H. This happens (triviall) at P H and again if (D7.927... R D5 )P C.282... H. D.282... P H M 353.5 million people D7.927... R D5 The slope-field graph confirms this number with the horizontal slope lines at P M 35. 7. Using the prediction P M 286.2 at t = 7 (ear 2), the new initial condition is P H 286.2 C 2 H 86.2. 7 86.2 8.6 9 7.8 399.8 387.2.3.2. ( P/ t)/p 95 96 97 98 The values of P decrease because of overcrowding. 8. See the graph in Problem, showing the decreasing population. The curve does follow the slope lines. 9. 9 76,9, 9 92,7, 2 3 d P dt H P[(D7.927 D5 )P C.282...] This is an eample of a logistic differential equation. 3. Using the differential equation, the slopes for P H 2 are 2.33... million people per ear. Because the vertical scale is /5 of the horizontal scale, the geometric slopes should be (/5)(2.332...) M.5, which agrees with the slope field. The slope field plotted b grapher agrees with the one given here.. t P t P D5.7 3 2.7 D 57. 228.3 D3 7.8 5 25. D2 89.3 6 269. D 9. 7 286.2 3.7* 8 3.3 55.5 9 3.8 2 8. 32.2 5 3 P P # #8 92 6,6, 93 23,76,7 www.census.gov Answers will var.. Answers will var. Chapter 8 Eploration 8-a. fq() H 2 3 D 56 2 C 6 D 672 H 2( 3 D 3 2 C 5 D 56) Factor b repeated snthetic substitution. 2. 2 D3 5 D56 2 D22 56 D 28 D28 D7 fq() H 2( D 2)( D )( D 7) fq() H H 2,, or 7 f() f () min. ma. min. + + 2 2 3 5 6 7 8 9 5 5 5. The calculated values match the given values within about %. t 3. See the graph in Problem 2, showing intervals where fq() is positive or negative, and thus where f() is increasing or decreasing, respectivel. Calculus: Concepts and Applications Instructor s Resource Book Solutions for the Eplorations / 95
. See the graph in Problem 2, showing ma where f() stops increasing and starts decreasing, and min where f() stops decreasing and starts increasing. These etrema (maima and minima) agree with the given figure. 5. f P() H 36 2 D 32 C 6 B quadratic formula, f P() H H 2.883... or 5.7862... The graph shows these zeros, and the intervals where f P() is positive or negative, and thus the graph of f() is concave up or down, respectivel. f() f () 6. Yes, the arcs agree with the concavit of the given graph. 7. See the graph in Problem 5, showing points of inflection where the sign of f P() changes. These points of inflection agree with the given figure. 8. Possible answer: f() p.i. p.i. + + 2 3 5 6 7 8 9 3. Q is infinite at H because D/3 is equivalent to / /3, which involves division b zero. The slope becomes infinite as approaches zero from either direction. As shown on the given graph, there is a cusp at H, where the value of is a local minimum.. Q H 3 D/3 D 5 3 2/3 PHD 9 D/3 D D/3 9 HD 9 D/3 ( C ) 5. At H, PH D 2 9 G. 6. P is zero onl when HD. 7. P(D2) HD 9 3 6 D H 2 3 3 2 I Thus the graph is concave up at HD2. P(D.5) HD 9 3.625.5 HD 58 3.5 G Thus the graph is concave down at HD.5. 8. Answers will var. Eploration 8-2b. The graph accounts for hq() undefined at H with the open circles on the ends of the branches. h() 2 3 5 6 7 8 9 f() f () ma. min. + + und. + 2 3 5 6 7 8 9 f () f () 9. Answers will var. Eploration 8-2a. H 5 2/3 D 5/3 p.i. p.i. + und. 2 3 5 6 7 8 9 2. hq() is not infinite at H. The graph stas bounded on a neighborhood of. 3. h is continuous, but not differentiable, at H.. 3 g () g () Q H 3 D/3 D 5 3 2/3 H 5 3 D/3 (2 D ) 2. Q is zero onl when H 2. At this, the tangent line is horizontal, and the graph stops increasing and starts decreasing. As shown on the given figure, the graph has a maimum at H 2. 96 / Solutions for the Eplorations Calculus: Concepts and Applications Instructor s Resource Book
5. 2. 6 3 L 2 2 3. L H 6 C H H N L H 6 C 6. 3 L 2 2 3 6. (Note that the graph is concave up for 2 G G 3 and concave down for 3 G G.) 6 Minimum of L H 95.959... at H 6.3299... 5. LQ H6 D 6 D2 LQ H 6 H 6 D2 6 2 H 6 HJ6.3299... (But D6.3299... is out of the domain.) For G 6.3299..., LQ is negative. For I 6.3299..., LQ is positive. N L is a minimum at H 6.3299..., which agrees with Problem. 6. If H 6.3299, then H 6.3 299... H 2.98... (Note that for minimum wall length, 6 H, and there are 6 segments of length and segments of length. This observation was pointed out b student James Guess [his real name!] in 2.) Total breadth H 3(6.3299...) H 8.9897... M 9. ft Total depth H 2.98... M 2.5 ft 7. Answers will var. Eploration 8-3b 7. Answers will var. Eploration 8-3a. 2 R 2: walls, 2 ft each 2 H 2 ft total R : 6 walls ft each, walls ft each 6 C H 22 ft total R : 6 walls ft each, walls ft each 6 C H 28 ft total. V H πr 2 h H π 2 At H, H 2, V H in. 3 At H, H 9, V H 9π in. 3 At H 2, H 6, V H 2π in. 3 At H 3, H 3, V H 27π in. 3 At H, H, V H in. 3 2. For the point (, ) shown in the diagrams, H 2 D 3. So the radius of the inscribed clinder is, the altitude is =2 D 3, and the volume is V H πr 2 h H π 2 (2 D 3) H 2π 2 D 3π 3. 3. d V d H 2π D 9π 2. H 2π D 9π 2 H 9π 2 9 D d V d H at H and H 2 9 H 8 3. Calculus: Concepts and Applications Instructor s Resource Book Solutions for the Eplorations / 97
5. V H 2π 2 D 3π 3 : V 8 6 2. e.b H D b 2 b M.895... (solving numericall close to b H.5) 3. (The graph shows the back half of the solid and the clindrical shell formed b rotating a representative strip.) 2 Alt. Maimum volume occurs at H 8. 3 The other solution, H, is a local minimum. 6. At H 8 3, r H 8 3, h H, V H 25 6 9 π H 28... π in. 3 7. Find an equation for the quantit to be maimized. Get the equation in terms of one independent variable. Find the derivative. Find critical points where the derivative is zero (or undefined). Find out whether the critical points correspond to maimum or to minimum values. Answer the question that was asked. 8. Answers will var. Eploration 8-a dv H 2π(3 D )(( D 2 ) D e. ) d V H 2π b (3 D )( D 2 D e. ) d H 3.89...π H 3.827.... (The graph shows the back half of the solid and the washer slice formed b rotating a representative strip.) Radius. The circumference is 2π, so dv D C h d = 2π d. 2. H D 2, so dv H 2π ( D 2 ) d H 8π 2 D 2π 3. 3. V H H3 (8π 2 D 2π 3 ) d H H 8 3 π3 D 2 π H3 H 3.5π H r r 2 As goes from to 3, the shells generate the whole figure. The curve shown for negative values of is just the image of the graph as it rotates, not the graph itself.. Slice the region parallel to the -ais. Pick a sample point (, ) on the graph, within the slice. As the region rotates, the slice generates a clindrical shell with radius, altitude, and thickness d. Thus, dv H 2π d Because H ln, H e. So, dv H 2πe d V H H2 2πe d H H 6.268... (numericall) (Eact answer is 2πe 2, integrating b parts.) dv H π 2 2 D 2 d V H π [( D 2 ) 2 D e.8 ] d H 3.62...π H 2.799... 5. The main difference is that with clindrical shells ou slice the rotated region parallel to the ais of rotation, rather than perpendicular as with washers. 6. Answers will var. 5. Answers will var. Eploration 8-b. Slicing horizontall makes the lengths of the strips a piecewise function of, whereas slicing verticall makes the lengths a simple function of. 98 / Solutions for the Eplorations Calculus: Concepts and Applications Instructor s Resource Book
Eploration 8-5a Eploration 8-6a.. The graph shows a narrow slice whose surface resembles a frustum of a cone. Doublcurved surface 2 Distance (D, ) to (, ) H 2 C 2 H 2 Distance (, ) to (, ) H 2 C 2 H 2 Distance (, ) to (2, ) H 2 C 3 2 H Length of curve is about 22 C H 5.997... This method underestimates the length of the graph, because it fails to account for the curvature of the graph between the sample points. 2. Distance (D, ) to (D.5,.25) H.5 2 C.75 2 H.825 Distance (D.5,.25) to (, ) H.5 2 C.25 2 H.325 Distance (, ) to (.5,.25) H.5 2 C.25 2 H.325 Distance (.5,.25) to (, ) H.5 2 C.75 2 H.825 Distance (, ) to (.5, 2.25) H.5 2 C.25 2 H.825 Distance (.5, 2.25) to (2, ) H.5 2 C.75 2 H 3.325 Length of curve is about.325 C.825 C.825 C 3.325 H 6.87... This estimate is better because the man shorter line segments better approimate the curve than the few long line segments. 3. The estimate gets better as gets smaller, so to get the eact length, take the limit as.. L H 2 C 2 H / ] C [ 2, but the mean value theorem states that for some c in the sample interval, fq(c) H /. 5. Because H d and M d H fq() d, dl can be written dl H ()] C [fq 2 d H d [fq()] 2 C 2 d 2 = d d 2 C, 2 where d is evaluated at the sample point H c. 6. H 2 d H 2d dl H d d 2 C 2 H d (2d 2 C) 2 H C 2 d L H H2 dl H H2 C 2 d HD HD 7. L H 6.257... (numericall) 7 C ln7 C C 2 5 D ln5 D 2 is eact. 8. Answers will var. Rotate a curved graph, get a doubl-curved surface. 2. The arc length of the base of the cone is 2πR, so if ou slit the cone down one side, and unroll it, the resulting section of a circle will have radius L and arc length 2πR. A full circle of radius L would have arc length 2πL and area πl 2. But this section is onl 2 πr 2πL of a full circle, so it has area 2 πr 2πL πl 2 H πrl, Q.E.D. 3. First note that r l H R L because of similar triangles, so Rl D rl H. Then, using Problem 2, the frustum has area A H πrl D πrl H πrl D πrl C πrl D πrl C πrl D πrl H πrl D πrl C π() C πrl D πrl H πr(l D l) C πr(l D l) = π(r C r)(l D l) or: A H πrl D πrl H πr r L D R l H πr l L D L l H π R L (L2 D l 2 ) H π R (L L C l)(l D l) H π R C R L l (L D l) H π R C R R r (L D l) H π(r C r)(l D l) But the average radius is (R 2 C r), the circumference at this radius is π(r C r), and the slant height of the frustum is (L D l), Q.E.D.. The slice is approimatel a frustum of slant height dl and average radius. So the surface area is about 2π dl. 5. H 3 d H 3 2 d dl H d d 2 C 2 H C 9 ds H 2π dl H 2π 3 C 9 S H 2π 3 C 9 d d d H π 8 ( C 9 ) /2 (36 3 d) H 8 π 2 3 ( C 9 ) 3/2 π H 2 ( 7.5 D ) = 3.563... 6. Answers will var. Calculus: Concepts and Applications Instructor s Resource Book Solutions for the Eplorations / 99
Eploration 8-7a. The graph agrees with the given figure. 2. r(.3) H 3 D 2 cos.3 H 9.799... (.3) H r(.3) cos.3 H 8.7699... (.3) H r(.3) sin.3 H 2.7286... The sample point shown is at about (8.8, 2.7). The angle measures about 7, which is.296... radian. So all three values agree with the graph. 3. At θ H.3, the radius (of the sector of circle that the wedge approimates) is r(.3), and the central angle is dθ H., so the approimate area is. 2π π 9.799... 2 H.589... unit. q. The sector is approimatel 2 dπ of a circle of radius r and area πr 2 q, so da H 2 dπ πr 2 H 2 r2 dθ. 5. A H θh2π 2 θh 2 3 D 2 cos θ dθ H 8.2977... (numericall) 6. The -radius is approimatel.5, so the area formula predicts that the area will be π 6.5 H 8.823... Close! (-radius is eactl 6 sin (cos D 2 ) 3 H.72..., making the area π 6.72... H 8.2977..., which is precisel the numerical answer.) 7. Answers will var. Eploration 8-7b. The graph agrees with the given figure. 7. The graph shows that at an apparent intersection point in the second quadrant, r, is positive, but r 2 is negative, meaning that the rose is being traced in the fourth quadrant when the circle is being traced in the second quadrant. B the time the rose gets to the second quadrant points, the circle is being traced in the fourth quadrant. Apparent intersection r is positive. 8. Answers will var. Eploration 8-7c. The graph agrees with the given figure. 2. r(.8) H 35.955... M 36. thousand miles 3. da H 2 r 2 dθ H 3528( D cos θ) D2 dθ A H 5.5 3528( D cos θ) D2 dθ H 38.9953....8 M 39 million square miles. dl H (dr/dθ ) dθ 2 C r 2 H [D8( D ) cos θsin 2 θ] ( 2 C [8 D ) cos θ ] 2 dθ L H 5.5 dl H 82.852... M 82.5 thousand miles θh.8 r 2is negative. 5. Check: r(.8) M 36 as the comet approaches Earth, and r(5.5) H 38.5... M 38 as the comet recedes from Earth. So the distance traveled on the curved path should be a bit greater than 36 C 38 H 72 thousand miles. Thus the 82.5 thousand miles is reasonable. 6. Answers will var. 2. 6 sin 2θ H θ H.5 sin D (/6) H.5(.7297... C 2πn) or.5(π D.7297... C 2πn) H.368... or.259... in Quadrant I (Store as a and b.) 3. da H.5r 2 2 D r 2 dθ A H.5 b (36 sin 2 2 D 6) dθ a H 5.332... square units. Counting squares in the region gives about 5.3, thus confirming the answer. 5. A H.5 π/2 36 sin 2 θ dθ H.37... H.5π, an interesting multiple of π. Chapter 9 Eploration 9-3a. 2 sin 3d u dv 2 + sin 3 2 3 cos 3 2 + 9 sin 3 27 cos 3 HD 3 2 cos 3 C 2 9 sin 3 C 2 27 cos 3 C C 6. dl H (dr) 2 C(rdθ) 2 H s co) 2 (2θC 36θ) sin 2 (2 dθ L H π/2 dl H.5326 units θh.5326.../π H.6258..., so the answer is not an interesting multiple of π. 2 / Solutions for the Eplorations Calculus: Concepts and Applications Instructor s Resource Book