Solutions to the Exercises of Chapter 5

Transcription

1 Solutions to the Eercises of Chapter 5 5A. Lines and Their Equations. The slope is 5 5. Since (, is a point on the line, y ( ( is an ( equation of the line in point-slope form. This simplifies to y Using the slope-intercept form of the equation of a line, we get y +.. The point-slope form of the equation is y ( (. In simplified form it is y.. Rewriting the equation as y and then as y form. So the slope is and the y-intercept is. puts it into slope-intercept 5B. Computing Slopes of Tangents 5. Take Q (+, + on the graph. So + (+ + +( and hence +( ( +. Dividing both sides by, gives + and hence lim. Som P. 6. Take ( +, 8+ on the graph. So lim 8+ (+ + + ( +(, and + ( +( ( + +(. Dividing both sides by, gives + +( +6 +( and hence. So m P.. Take ( +, y + on the graph. So y + + common denominators, we get y y y + +. Pushing to zero, we see that lim. Because y Note: The hint supplied for the solution of Eercise 8 is not relevant. and hence + y, it follows that y.som P is equal to y. y. Taking. So + 8. Take ( +, y + the graph. So (y + +, and after multiplying out, y +y +y +( +. Since (, y is on the graph, y and hence y +y +(. Factoring out a and dividing by, now gives us that (y +y +( and (y +y +(. So Now push to zero. In the process goes to zero, and therefore, lim y, y, and y.som P is equal to y or.. Take ( +, y + on the ellipse. So (+ 5 + (y+. Therefore, + +( 5 + y +y +(, y +y+(. y. Since

2 and hence + +( + y+( + y. Since + y, +( + y+( So y+( +( and hence (y + ( +. Therefore, y+ 5 y 5 y. So m P 5 y.. Since goes to when is pushed to, we now find that lim 5C. Derivatives. Note that f(+ f( (+ + +( +( +( +( + +(. Pushing to zero, tells us that f (.. Observe that h(+ h( It follows that (+. Working with the numerator, we get ( + ( + ( ( + ( + ( ( + (+ ( (+ f f( + f( ( lim ( ( +. (+. Pushing to zero, gives us.. i. f (. The slope in question is f ( (. ii. g(.sog (. The slope is g ( ( iii. f(. So f (. The slope is f (. 5. iv. f(.sof (. The slope is f ( 8.. i. f (. ii. dy. d iii. f ( 5. iv. y + π.so dy d +π. v. g( + 6. So g ( +. vi. f( + +.Sof ( i. dy dy +8. So, when. Therefore the point in question is (, 6. d d ii. Since y +8 ( 8, the parabola crosses the ais at and again at 8. In reference to Archimedes s theorem note that the area of the inscribed triangle is The area of the parabolic section is, therefore, 56 6.

3 5D. Definite Integrals 5. Inserting the given points on the ais we have Proceeding from left to right and adding as in Eamples 5. or 5., we get the following (. +. (. +.5 (. +. (.5 +. (. +.6 ( This is a rough approimation of the area under the graph of y over the interval from toonthe ais. The actual value of this area can be shown to be.6 (up to three decimal accuracy. This computation involves the logarithm function that will be studied in Chapter.. 6. Inserting the given points we get 6 Proceeding from left to right and adding, we get: ( This is a rough approimation of the area under the graph of y from to. The actual value is d d (.8.. In each case, the symbol represents the area under the curve in question as obtained by the addition of small rectangles process described in the tet. For d the meaning of the symbol is illustrated by drawing the graph of y from to and inscribing under it narrow rectangles as in Firgure 5.(c. For d, the graph is different but the task is identical. 8. The graph of the hyperbola 5 y is obtained by the process eplained in Section 5.. It is sketched below. Solving for y, we get y ± 5 5. So the graph of y 5 5

4 5 is the upper portion (both parts of this graph. The definite integral 5 5 d is the area under the upper part of the hyperbola that falls between and (as obtained by the process of adding small rectangles. 5E. The Tractri Correction: In Figure 5., delete B and then replace (, y by B(, y.. The slope of the string is z. Note that it is negative because the string slopes downward from left to right. By Pythagoras s theorem, z a. So the slope is m a. Since the string is always tangent to the curve, the slope is also given as f (. It follows that f ( a. Recall that Perrault asked Leibniz about the eact nature of this curve. The answer is: the graph of y f(. So the answer involves finding an antiderivative of a. This is not easy. See Section... The formula for the length L of a curve from a point (a, c to a point (b, d, where a b, is Since f ( a L b a +f ( d. and the arc runs from (c, d to (a,, where <c a, weget L a + a d a +a d c c a c a d a c a d. 5F. Maimum and Minimum Values. In view of the diagram

5 the function that needs to be analyzed is f( ( ( + +. Using the rules already established, we get f ( 6 +. Set f ( and solve for. Because is not the answer to the question, we can take. After canceling, weget (6 +(6. By the quadratic formula, (6 ± (6 (6 (6±6 or 6. Because, the maimum occurs when 6. So to achieve the maimum, both pieces have the same length of 6 units. This answer is the same as that of Eample 5.. Is this surprising?. Subdivide the segment into two pieces and y as shown: y Since the product y, the sum of the lengths is f( + +. So f (. So the minimum must occur at. This means that the smallest length that the segment can have is +.. Let the two sides of the rectangle be and y. Then +y, so + y 5, and y 5. The area is f( y (5 +5. So f ( 5. It follows that 5 and y 5 give the required dimensions.. Let (, y be a random point on the parabola. The distance between (, y and (, is d ( +(y ( +( + ( 6 + +( The point (, y for which the distance d is a minimum is the same point for which the distance squared d is a minimum. It remains, therefore, to analyze the function f( Note that f ( + 6. This polynomial has as a root. So divides it. By polynomial division, + 6( ( Use of the quadratic formula shows that + +6 is never zero. So is the only of f (. The point on the parabola corresponding to is (,. 5. The circular base of the cylinder has radius y and hence area πy. Because the volume of a cylinder is base height, it follows that the volume is V ( πy π( π( 6 + π( 6 +. Observe that V ( π( + π( + π( (. So the maimum is achieved for or. Since gives V (, it follows that V ( π( π is the largest volume the cylinder can have. 5

6 5G. Areas and Definite Integrals 6. By the Fundamental Theorem of Calculus: d. 8 d 8 d d d 8 ( ( ( (( ( ( ( (. The area represented by the first integral is that under the curve y and above the segment from to on the ais; the area represented by the second integral is that under y y _ y 8 the curve y third integral. and above the segment from 8 to on the ais; and similarly for the. Solving + y for y gives y ±. So the graph of y is the upper half of the circle and the graph of y is the lower half. Convince yourself that the definite integral d represents the area of one quarter of the circle of radius. So d π( π. In the same way, a a d πa. 8. Since π5 5 5 d d, we get by using Eercise, that 5 5 π. 8 5 d. As above, a b a a d b a a a d b a πa πab. Recall that + y is a b an ellipse. By solving this equation for y and considering positive y only, check that the graph a is the upper half of the ellipse. So the definite integral a of y b a the area of the upper right quadrant of the ellipse + y. a b b a a d is. i. The slope of the segment is 8 8. The slope of the tangent at any point (, y on the parabola is. If this tangent is to be parallel to the segment, then.so and Q (,. ii. The slope of the line through the point Q perpendicular to OP is. The 6

7 point-slope form of the line in question is y (. iii. Take the segment OP as base of the triangle OQP. The length of the segment is + ( Let T be the point of intersection between the line of (ii and the segment OP. Then QT is the height of the triangle OQP. Note that T is the intersection of the lines y and y (. Solving ( for, we get +, or.so and T (, 6. The length of QT is ( ( + ( ( The area of the triangle OQP is.. iv. By Archimedes s theorem, the area of the parabolic section OPQ is 6. Substracting this from the triangular area under the segment OP gives for the area under the parabola from to. v. By the Fundamental Theorem of Calculus, this area is also given by 6 d H. Definite Integrals as Areas, Volumes, and Lengths of Curves. Notice that the rotation of this triangular region produces the required cone. The equation of the line through and (h, r isy r h.so V. The volume is given by h V πf( d h πf( d π r h d π r h h πr h. ( π d π π.

8 . The length of the arc on the parabola y from the point (, to the point (5, 5 is L 5 5 +f ( d + d. Refer to the last part of Section 5. to see that the graph of y is a hyperbola. Solving this equation for y shows that the upper half of this hyperbola is the graph of the function f( +. It follows that the area under the upper half of the hyperbola from to 5 is also equal to 5 + d.. Solve + y for y to see that the upper half of the ellipse is the graph of f( i. A 5 f( d d. ii. V 5 5 πf( d 5 6π 5 5 (5 d. iii. Since f ( is the slope of the tangent at the point (, y, we see from Eercise that f ( 5 y It follows that the length L of the arc is given by L +f 6 ( d + 5(5 d. Correction: In the statement of Eercise 5(v the y-coordinates of the two points are incorrectly listed. They should be and respectively. 5. Consider the hyperbola y. So + y. Therefore, and. It follows from the considerations towards the end of Section 5. that the general shape of this hyperbola is - i. To apply Leibniz s tangent method let ( +, y + be a point on the graph that lies on the same part of the curve as P. Notice that 8

9 Divide both sides by to get (+ (y+ + +( y +y+( y + +( y+( +( y+( (+ (y+. y + +. So +. Pushing to zero, gives lim. This is the slope at any point y+ y P (, y on the hyperbola. Note that we need to have y. ii. To find a function whose graph is precisely the upper half of the hyperbola, solve y for y. Since y, we get y ( and y ±. Note that y f( is the required function. By part (i, the derivative is equal to f ( f(. iii. The area under the upper half of this hyperbola and over the interval from to is f( d d. iv. The volume of the solid obtained by rotating the region of one complete revolution about the -ais is πf( π d ( d. v. Using (ii, we get that the length of the hyperbolic arc from the point (, to the point (, is +f ( d + ( d. 5I. Theorems of Pappus of Aleandria 6. We need to use Pappus s theorem A because the concern in this eercise is surface area. Consider the circle on the right in Figure 5.5 and drop a perpendicular segment from C to the ais shown. If the segment with the circle attached is rotated one revolution around the ais, the circle traces out the surface of a donut. Incidentally, the surface of such a perfect donut is called a torus in mathematics. Now turn to the specifics of the eercise and in

10 particular to Figure 5.6. What was just described is now viewed from the top. The ais is perpendicular to the page and goes through the point of intersection of the segments labeled r and R. The circle being rotated is now being viewed from the top as the thicker black segment; a frontal view of this circle is also shown in black (in this eercise only the boundary counts and not the inside. Pappus s theorem A tells us that the surface area of the donut is the product of the circumference of the rotated circle and the distance traveled by the centroid C of the circle. Consider Figures 5.5 (the diagram on the right and 5.6 together. We see that the diameter of the rotated circle is R r, so its radius is (R r, and hence its circumference is π ( (R r π(r r. The centroid of the circle is its center C. Observe that the distance from C to the ais is (R r+r (R + r. So in one complete rotation, C traces out a circle of radius (R + r. Because this circle has a circumference of π ( (R + r π(r + r, we see that this is the distance traveled by the centroid C. We can therefore conclude that length of arc distance traced out by centroid π(r r π(r + r π (R r. So the surface area of the torus in question is π (R r.. This is another application of Pappus s theorem A. Refer to diagram in Figure 5.5 (on the left. Let r be the radius of the semicircular arc and let C be its centroid. By the vertical symmetry of the semicircular arc, its centroid C lies on the radius that is perpendicular to the ais. Its general position is shown in the figure. But where precisely is it located? When the semicircle is rotated one revolution about the ais a sphere of radius r is produced. Pappus s theorem A says that Surface area length of arc distance traveled by centroid. So πr πr distance traveled by centroid. Therefore, the distance traveled by centroid is r. Let d be the distance from the centroid to the ais and note that the centroid traces out a circle of radius d. So the distance traveled by the centroid is πd. Therefore πd r, and hence d r.6r. The location of the centroid of the semicircular arc has been π determined. 8. Consider the semicircular region on the left in Figure 5.5. This is no longer the arc itself, but the entire semicircular area. Let C be the centroid of the region and let r be the radius of the semicircle. The solid that is traced out by one full rotation of the region is a sphere of radius r. Now apply Pappus s theorem B. Since the volume of the sphere of radius r is πr, and the area of the semicircle that is being rotated is πr,wenowget πr πr distance traveled by C. By the symmetry of the region, C lies somewhere on the radius that is perpendicular to the ais. Let its distance from the ais be d and note that the distance traveled by C during one complete rotation is πd. Therefore, πr πr πd π r d. Solving for d gives us d r.r. π