# Section 4.1. Math 150 HW 4.1 Solutions C. Panza

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1 Math 50 HW 4. Solutions C. Panza Section 4. Eercise 0. Use Eq. ( to estimate f. Use a calculator to compute both the error and the percentage error. 0. f( =, a = 5, = 0.4 Estimate f: f ( = 4 f (5 = 9 f f (a f f (5( 0.4 9( actual f = f(4. 5(5 = 7.8 Eercise 3. Estimate y using the differentials [Eq. (4]. 3. f( = cos, a = π, d = 0.04 f ( = sin ( π y sin (0.04 ( Error: E = actual f f (a E = 7.8 ( 7. = 0.3 Percentage error: = error actual f 00% = % y dy = 4.40% f (ad (for small d 30. The stopping distance for an automobile is F (s =.s s ft, where s is the speed in mph. Use the Linear Approimation to estimate the change in stopping distance per additional mph when s = 35 and when s = 55. when s = 35: a = 35, s = F (s = s F F (35 s 4.88( 4.88 The change in stopping distance from 35 to 3 mph is about 4.88 feet. Eercises 4 and 49. Find the linearization at = a. 4. f( =, a = when s = 55: a = 55, s = F (s = s F F (55 s 7.04( 7.04 The change in stopping distance from 55 to 5 mph is about 7.04 feet. 49. y = ( + /, a = 0 f ( = L( = f(a + f (a( a L( = f( + f (( L( = ( 4 L( = 4 + y = ( + 3/ L( = y(a + y (a( a L( = y(0 + y (0( 0 L( = L( = +

2 Math 50 HW 4. Solutions C. Panza Section 4. Eercises, 0, and. Find all critical points of the function.. f(t = 8t 3 t f (t = 4t t = t(t 0 = t(t t = 0, Critical points: t = 0,. R(θ = cos θ + sin θ R (θ = sin θ + sin θ cos θ = sin θ( cos θ 0 = sin θ( cos θ 0. f( = f ( = ( ( 4 ( = ( = = 4( 4 = 0, 4 Critical points: = 0, 4 sin θ = 0 or cos θ = θ = 0 θ = π 3 θ = 5π 3 Critical points: πk, π 3 + πk, 5π 3 + πk for k Z Eercises 3, 44, and 58. Find the minimum and maimum value of the function on the given interval by comparing values at critical points and endpoints. 3. y = 3 +, [0, ] y = = 3( = 3( 7( 0 = 3( 7( =, 7 Note: 7 is not in [0, ] -value y( 0 (endpoint 0 min (endpoint (critical point 0 ma Minimum: y(0 = 0 Maimum: y( = 0

3 Math 50 HW 4. Solutions C. Panza 44. y = +, [0, ] y = + = 4( + = = 0 No solution + = = y = 3 4 ln, [, 3] -value y( 0 (endpoint min (endpoint 0.58 ma Minimum: y( = 0.58 Maimum: y (0 = y = 3 4 = = = 3 = -value y( (endpoint.7 ma 3 (endpoint 0.3 (critical point 8.4 min Minimum: y( 8.4 Maimum: y (.7 Eercises and 7. Verify Rolle s Theorem for the given interval by checking f(a = f(b and then finding a value c in (a, b such that f (c = 0.. f( = sin, [ π 4, 3π ] 4 ( π ( 3π f = and f = f ( = cos 4 4 ( π ( 0 = cos 3π f = f π 4 4 = f is continuous on [ π 4, 3π ] ( 4 and differentiable on π 4, 3π ( 4. Since f π ( 4 = f 3π 4, c = π ( π 4, 3π 4, and f ( [ π = 0, Rolle s Theorem is verified for f( on π 4, 3π ] f( =, [3, 5] 8 5 f (3 = and f (5 = f (3 = f (5 f ( = (8 5 (8 (8 5 = 8 30 (8 5 0 = = (4 5 = 0, 5 4 f is continuous on [3, 5] and differentiable on (3, 5. Since f (3 = f (5, c = 5 4 (3, 5, and f ( 5 4 = 0, Rolle s Theorem is verified for f( on [3, 5].

4 Math 50 HW 4.3 Solutions C. Panza Section 4.3 Eercises and 4. interval.. y =, [9, 5] y = y(5 y(9 = c 5 9 c = 5 3 c = 8 = c 4 c = At c =, y (c = Find a point c satisfying the conclusion of the MVT for the given function and y(5 y( y =, [, 4] + y = ( + f(4 f( = (c + 4 (c + = (c + = 9 (c + = 8 c + = 3 c = 3 c.4 At c = 3.4, y (c = y(4 y(. 4 Eercise 3, 4, 48, 54, and 55. Find the critical points and the intervals on which the function is increasing or decreasing. Use the First Derivative Test to determine whether the critical point is a local min or ma (or neither. 3. y = y = = = (3 ± 3 4(5( (5 = 3 ± 0 There are no critical points of y. Increasing: (,

5 Math 50 HW 4.3 Solutions C. Panza 4. y = + + y = + ( + 0 = + 0 = + = ( ± 4(( ( = ± 5 Critical points: ± 5 f ( + Decreasing: Increasing: Minimum: y Maimum: y ( 5, 5 ( 5, + 5 ( 5 ( ( + 5, 48. y = θ cos θ, [0, π] y = + sin θ 0 = + sin θ sin θ = θ = 7π π + πk, + πk Critical points: 7π, π f ( + + 7π, π Decreasing: ( 7π Increasing: ( 0, 7π ( π, π Minimum: y ( π 4.03 Maimum: y ( 7π 5.40 π 54. y = ( e y = ( e 0 = ( e = ± Critical points: ± f ( + + Decreasing: (, Increasing: (, (, Minimum: y ( 3.4 Maimum: y ( y = ln, ( > 0 y = = 0 = = Critical points: f ( + 0 Decreasing: (0, Increasing: (, Minimum: y ( =

6 Math 50 HW 4.4 Solutions C. Panza Section 4.4 Eercises 4, 4, and 7. Determine the intervals on which the function is concave up or down and find the points of inflection. 4. y = t 3 t + 4 y = 3t t y = t 0 = t t = y + Inflection point: t = Concave up: (, Concave down: (, 4. y = + 9 y = 9 ( + 9 y = ( 7 ( = ( 7 = 0, ±3 3 y Inflection points: = 0, ±3 3 Concave up: ( 3, 0 ( 3, Concave down: (, 3 ( 0, y = + ln ( > 0 y = 4 + y = 4 0 = 4 0 = 4 = ± y + 0 Inflection point: = Concave up: (, Concave down: (, Eercises 8, 35, and 37. Find the critical points and apply the Second Derivative Test (or state that it fails. 8. f( = f ( = = 4( 4 = 0, ± Critical points: = 0, ± f ( = f (0 < 0 f ( > 0 f ( > 0 f( and f( are local minima f(0 is a local maimum

7 Math 50 HW 4.4 Solutions C. Panza 35. f( = sin + cos, [0, π] f ( = sin cos sin 0 = sin ( cos = 0, π 3, π Critical points: = 0, π 3, π f ( = cos sin cos f ( (0 > 0 f π < 0 3 f (π > 0 f(0 and f(π are local minima f ( π 3 is a local maimum 37. f( = e f ( = e e 0 = e ( 0 = = ± Critical points: ± f ( = (4 3 e 4e f ( > 0 ( f < 0 ( f is a local minimum ( f is a local maimum Eercises 45 and 5. Find the intervals on which f is concave up or down, the points of inflection, the critical points, and the local minima and maima. 45. f( = 8 / ( 0 f ( = 4 / 0 = 4 / 0 = ( 3/ 0, /3 = 0 is an endpoint, so not a critical point f ( f ( 0 + /3 5. f(θ = θ + sin θ, [0, π] f ( = + 3/ 0 = + 3/ 0 < + 3/ for all 0, so no inflection points + + Critical points: = 0, /3 Concave up: (0, Concave down: none Inflection points: none f( /3 is a local minimum No local maimum f (θ = + cos θ f (θ = sin θ 0 = + cos θ 0 = sin θ π = θ 0, π, π = θ f (θ + + θ f (θ 0 π + π Critical points: θ = π (saddle Concave up: (π, π Concave down: (0, π Inflection points: θ = π No local etrema

8 Math 50 HW 4.5 Solutions C. Panza Section 4.5 Eercises 4 and 0. Use l Hôpital s Rule to evaluate the limit, or state that L Hôpital s rule does not apply lim 5 H = 4( 3 + 5( 4 = 3 cos sin 0. lim = 0 sin (L Hôpital s Rule does not apply. Eercises and. Show that L Hôpital s Rule is applicable to the limit as ± and evaluate.. lim sin sin(/ /. lim e L Hôpital s Rule applies lim sin sin(/ / H ( / cos(/ / cos(/ = cos 0 = Eercises 8,, 3, and 34. Evaluate the limit. [ 8. lim 4 ] [ ] [ ] H / ( 4 = 4 L Hôpital s Rule applies H lim e e. lim 0 tan 4 tan 5 H = 0 e H 0 4 sec 4 5 sec 5 = 4( 5( = 4 5 = 4 3. lim 0 cos H 0 sin H 0 cos = = 44. lim ( e e H 3 H H = 0 e e e

9 Math 50 HW 4.7 Solutions C. Panza Section A wire of length m is divided into two pieces and the pieces are bent into a square and a circle. How should this be done in order to minimize the sum of their areas? Let = the length of the piece formed into a circle (circumference. So the radius of the circle is π m and one side of the square is 4 m. A( = area of circle + area of square ( ( = π + π 4 = 4π A ( = π = π = 4 + π π = π 4 + π 5.8 The sum of their areas is minimized when 5.8 m is used for the circle and.7 m is used for the square. 7. Find the dimensions of the rectangle of maimum area that can be inscribed in a circle of radius r = 4 (Figure. = half the width of the rectangle y = half the height of the rectangle r = + y y = A( = 4 A ( = 4 0 = 4 0 = Critical points: = ±4, ± = (Note: if = 4, there wouldn t be a rectangle y = ( = The area of the rectangle is maimized when the dimensions are 4 by 4.. Find the point P on the parabola y = closest to the point (3, 0 (Figure 4. D = the square of the distance between the point P = (, and (3, 0 D( = ( 3 + ( 0 = D ( = = ( = ( ( Critical points: = The point P = (, is the closest point on the parabola y = to (3, 0.

10 Math 50 HW 4.7 Solutions C. Panza 30. According to postal regulations, a carton is classified as oversized if the sum of its height and girth (perimeter of its base eceeds 08 in. Find the dimensions of a carton with a square base that is not oversized and has maimum volume. h = the height of the carton = one side of base 4 = girth of carton V ( = gives the volume of the carton 08 = 4 + h h = 08 4 V ( = (08 4 = V ( = 8 0 = = 0, 8 Critical points: = 0, 8 (Note: 0 doesn t make sense The volume of the carton is maimized (without being oversized when the dimensions of the base are 8 in. by 8 in. and the height is 3 in. 49. All units in a 00-unit apartment building are rented out when the monthly rent is set at r = \$900/month. Suppose that one unit becomes vacant with each \$0 increase in rent and that each occupied unit costs \$80/month in maintenance. Which rent r maimizes monthly profit? = number of \$0 rent increases (number of unoccupied units 00 = number of units rented ( = 0 corresponds to \$900/month rent = monthly profit from one unit P ( = monthly profit from all rented units P ( = (00 ( = P ( = = = 9 Critical points: = 9 The monthly profit is maimized when the rent is (0 = \$990.

11 Math 50 HW 4.8 Solutions C. Panza Section 4.8 Eercises, 4, and. Apply Newton s Method to f and initial guess 0 to calculate,, 3.. f( = 3 +, 0 = 3 4. f( = 3 + +, 0 = f ( = 3 n+ = n n n f ( = 3 + n+ = n n n f( = sin, 0 = 7 f ( = sin cos n+ = n n n n n /3.70, sin sin cos Eercises and 4. Approimate to three decimal places using Newton s Method and compare with the value from a calculator.. 5 / /4 = ( /4 Let f( = 3 3 5, 0 = /4 = 3 f ( = 3 Let f( = 4 3, 0 = n n 0 f ( = Calculator: 5 / /4 0.70, Calculator: 3 /4 =