Math 130: PracTest 3 Answers Online Frida 1 Find the absolute etreme values of the following functions on the given intervals Which theorems justif our work? Make sure ou eplain what ou are doing a) 1 4 4 3 [1, 5] b) d) arctan ln(1 + ) (, ) 4 + 1 (, 0) c) 3 1/3 [ 8, 8] a) (From a previous eam) A 10 meter tall flag pole topples to the ground in such a wa that the angle θ between the ground and the pole decreases at 0 rad/s (see below) How is the distance between the tip of the flag pole and the ground changing when = 8 meters? b) How is the bottom edge of the triangle changing at the same moment? Pole = 10m θ 3 (From an eam last ear) The photo above shows road salt being unloaded from a tanker at Port Oswego, NY (NY Times, Oct 07) The salt comes off the conveor and forms a conical pile The height of the pile is changing at 01 m/min and the radius at 0 m/min At what rate is the conveor adding salt to the pile (how is the volume changing) when the pile is 3 m tall and has a radius of 6 m? (Hint: Look up the volume formula for a cone in the front cover of our tet, if ou need to) 4 In an animated cartoon, a bo with a square base is growing so that its height is increasing at 4 cm/s and its bottom edges are decreasing at cm/s Determine how the volume of the bo is changing when the height is 8 cm and the edge length is 5 cm Is the volume (V = h) increasing or decreasing? #5 #4 3 θ 1 H 5 a) (See figure above) From an eam last ear: A baseball diamond is a square with 90 ft sides David Ortiz hits the ball and runs towards first base at a speed of 4 ft/s At what rate is his distance from third base changing when he is halfwa to first base? b) How is the angle θ changing at this same moment? 6 a) Carefull state the Mean Value Theorem and draw a diagram which illustrates it b) Carefull state the CIT c) State the definitions of a critical point and of an absolute ma/min 7 a) Let f() = 3 3 on [1, 3] Does the MVT appl to this function? (Does it satisf the two assumptions of the MVT? Eplain wh) If so, find the point c where f (c) = f(b) f(a) b a b) Does the MVT appl to the function f() = +1 not, eplain wh on [, 3]? If so, find the point c where f (c) = f(b) f(a) b a If
8 Draw the graph of a differentiable function on [0, 5] which has no absolute ma or eplain wh this is impossible 9 Draw a CONTINUOUS function that has a critical number for each set of conditions given, or eplain wh it is impossible a) A critical number at = and f () 0 b) A critical number at = which is not a relative etreme point even though f () = 0 c) A critical number at = where f () DNE which is not a relative etreme point 10 Do a complete graph of f() = 1 5 5 4 3 3 (To receive full credit on the test, ou must indicate critical points, relative etrema, increasing, decreasing, ections, and concavit Use number lines for the derivatives) WARNING: Mess values: Use our calculator to compute function values 11 Graph f() = e e Same directions as above 1 The graph below is the graph of f () Draw a possible graph of the original function = f() Determine where there are critical points, relative etremes, points of ection, where the function is increasing, decreasing, concave up, and concave down Use number lines to help organize our information a) Where is f () = 0? Positive? Negative? Mark this on the number line b) Where is f () = 0? (You can answer this question b remembering that f is the derivative (slope) of f So where is the slope of the function below 0? Where is it positive? Negative? Mark this on the number line c) Now draw the graph of f start at the dot at ( 5, 4) This f, NOT the original function f 5 3 1 1 3 f 5 4 3 1 0 1 3 4 f 5 4 3 1 0 1 3 4 13 Use the information about the first and second derivatives of f to sketch a CONTINUOUS function that has derivatives like those given Indicate on our graph which points are local etrema and which are ections Make up -values for the critical and ection points consistent with the information supplied ++ 0 + + + DNE f 1 0 ++ DNE + + + f 1 14 Find the derivatives of the following functions Use logarithmic differentiation where helpful a) = (cos 9) b) = tan c) ln d) (arcsin ) e) ln(4 7+1 )
15 Find the derivatives of these functions using the derivative formula for a general eponential function a) 5 6 b) 4 4 c) 3 +tan d) 4 [ +1] 3 e) For which values of does 4 4 have a horizontal tangent? 16 Graph f() = 3 4 4 3 17 Another graph: Do a complete graph of f() = ( + 1)e 18 A child s sandbo is to be made b cutting equal squares from the corners of a square sheet of galvanized iron and turning up the sides If each side of the sheet of galvanized iron is meters long, what size squares should be cut from the corners to maimize the volume of the sandbo? 19 A garden is to be created in si sections as shown in the design below using 400 meters of fencing (the solid lines) An eisting rock wall is to be used as one side What is the maimum area that can be enclosed in this design? (Justif our answer) 0 (From Da 33) President Gearan has $600 to spend to enclose his ard to keep out students One side of the ard will use an eisting rock wall while the other 3 sides will use electric fencing The fence parallel to the wall costs $3 per meter while the other fence costs $1 per per meter What dimensions of the ard maimize the enclosed area Free (eisting wall) at $1/m at $3/m at $1/m 1 (From Da 33) Most shipping cartons are constructed with double laers of cardboard at the top and bottom (where the carton folds over on itself) If such a carton with a square base is to hold 16,000 cubic cm, what dimensions minimize the material used (surface area, include double top and bottom)? laers on top and bottom A rectangular invitation card is to have an area of 64 square inches The top and bottom edges are trimmed with gold ink (0 cents per inch) and silver on the sides (5 cents per inch) What dimensions of the rectangle minimize the cost of the ink? Justif our answer
3 Use the information about the first and second derivatives of f to sketch a function that would have derivatives like those given Indicate on our graph which points are local etrema and which are ections Make up values for the critical and ection points consistent with the information supplied ++ 0 + + + 0 f 1 0 + 0 f 0 4 A manufacturer makes aluminum cups in the form of right circular clinders (no top) The volume is to be 8π cu in Find the dimensions that use the least material Hint: Use the volume formula for a clinder as the constraint equation The surface area of a cup includes the side and bottom Justif our answer 5 A horizontal gutter is to be made from a piece of sheet aluminum 100 inches long and 8 inches wide b folding up equal wihs along the dashed lines into a vertical position How man inches should be turned up on each side to ield the maimum carring capacit (ma volume)? (Ans: inches) 100 8 6 Graph f() = 6 Indicate everthing + 3
Practest 3A Answers 1 a) Use CIT p () = 3 3 4 = ( 3 4) = ( + 1)( 4) Critical # s at = 1, 0, 4 Onl = 4 is in the interval Check critical # s: f(4) = 3 Endpts: f(1) = 75 and f(5) = 1875 Abs ma at = 1 with f(1) = 75; Abs min at = 4 with f(4) = 3 b) Tr to use SCPT since the interval is NOT closed First show that there is onl one CP f () = 4( +1) 8 ( +1) = 4 4 ( +1) = 0 at = ±1 Onl crit # in the interval is = 1 So SCPT applies Use First Derivative Test (FDT) to classif the point RMin 0 + + + f 1 0 There is a relative min at = 1 B the SCPT, there is an absolute min at = 1 with f( 1) = c) Use CIT f () = 1 1 = 0 at = ±1, DNE at = 0 At crit # s: f( 1) =, f(0) = 0, f(1) = Endpts: /3 f( 8) = f(8) = Abs ma at = and = 8 with f() = f( 8) = ; Abs min at = and = 8 with f( ) = f(8) = d) Tr to use SCPT since the interval is NOT closed First show that there is onl one CP f () = 1 1+ 1+ = 0 at = 1 There is onl one CP So SCPT applies Use First Derivative Test (FDT) to classif the point RMa f + + ++ 0 There is a relative ma at = 1/ B the SCPT, there is an absolute ma at = 1/ with f(1/) 041 Pole = 10m θ 1/ dθ Given: = 0 rad/s d Find: =8 Relation: 10 = sin θ or = 10 sin θ d Rate: = d (10 sin θ) d dθ = 10 cos θ When = 8, = 10 8 = 6 and so cos θ = 6 10 d = 10(06)( 0) = 1 m/s a) Find d Relation: + = 10 = 100 So d d + = 0 or d = d = 8 ( 1) = 16 m/s 6 = 06 Thus, 3 dh Given: dv Find: dr = 01 m/min and = 0 m/min r=6, h=3 Relation: V = 1 3 πr h Rate: dv dv = π 3 = d ( 1 3 πr h) ( rh dr dh + r ) When r = 6 and h = 3: dv = π r=6, h=3 3 [(6)(3)(0) + 6 (01)] = 36π m 3 /min
4 a) Find dv given dh d = 4 cm/s and = cm/s Relation: V = dv h Rate: h=8, =5 dv = (5)(8)( ) + 5 (4) = 160 + 100 = 60 = cm 3 /s h=8, =5 d dh = h + So 5 a) Find dc given da a=45 c = 45 + 90 = 45 5 So = 4 ft/s Relation: c = a + 90 Rate: c dc da = a dc or = a c da When a = 45, θ c a dc = 45 45 4 (4) = 10733 = ft/s 5 5 b) Now find dθ Relation (use the constant side!): tan θ = a 90 Rate: sec θ dθ a=45 cos θ 90 4 = 1 90 da dθ or = 1 da 90 sec θ = da Use the triangle: When a = 45, from part (a) we have c = 45 5, so cos θ = ( 90 c ) = ( 90 45 5 ) = ( 5 ) = 5 = 08 So dθ = cos θ da a=90 90 = 08 (4) = 013 rad/s 90 6 a) MVT: The Mean Value Theorem Let f be continuous on a closed interval [a, b] and differentiable on (a, b) Then there is some point c between a and b so that f (c) = f(b) f(a) b a b) CIT: Let f be a continuous function on a closed interval [a, b] Then the absolute etrema of f occur either at critical points of f on the open interval (a, b) or at the endpoints a and/or b c) Assume that f is defined at c Then c is a critical number of f if either (1) f (c) = 0 or () f (c) does not eist d) Let f be a function whose domain D contains the point c f has an absolute minimum at c if f(c) f() for all in D The number f(c) is the minimum value of f 7 a) Yes Since f is a polnomial it is both continuous and differentiable so the MVT applies to this function f(3) f(1) 3 1 = 18+ = 10 f () = 3 3 = 10 or = 13/3 So = ± 13/3 But onl = 13/3 is in the interval b) No f is not defined at 0, so f is not continuous 8 Impossible Differentiable continuous The interval is closed, so EVT f has an abs ma and min 9 All are possible a) b) c)
10 Graph f() = 1 5 5 4 3 3 r ma f () 1/ 0 () 1/ r min Locate critical numbers: f () = 4 4 = ( 4) at = 0, ± inc rma dec Neither dec rmin inc ++ 0 0 0 ++ f 0 Check concavit with the second derivative f () = 4 3 8 = 4( ) at = ± concave dn inf up inf dn inf concave up 0 ++ 0 0 + + + f () 1/ 0 () 1/ Plot the critical numbers and the ections: f(0) = 0 and f() = 64/15, f( ) = 64/15 f( ) 64 and f( ) 64 11 Do a complete graph of = f() = e e f 1 r min f Locate critical numbers: f () = e + e e = e = 0 at = 0 dec l min inc f 0 + + ++ 0 Check concavit with the second derivative f () = e + e = ( + 1)e = 0 at = 1 concave dn ect concave up 0 + + ++ 1 Plot the critical numbers and the ections: f(0) = 1 and f( 1) = e 1 0736 1 The graph below is the graph of f () Draw a possible graph of the original function = f() f is dashed RMa Infl Infl Infl 5 3 1 1 3 RMin Inc RM Dec CP Dec RMn Inc f + 0 0 0 ++ 5 4 3 1 0 1 3 4 f C Dn Infl C Up 0 ++ Infl 0 C Dn Infl 0 C Up ++ 5 4 3 1 0 1 3 4
13 Here s one eample: 1 14 a) ln = ln(cos 9) = ln(cos 9) = 1 d d 9 tan 9) = ln(cos 9) sin(9) 9 cos 9 b) ln = ln tan = tan ln = 1 d d = sec ln + ( tan ) 1 c) ln = ln ln = ln ln = (ln ) = 1 d 1 = (ln ) d = d d = (cos 9) (ln(sin 9) ( d tan = = sec ln + tan ) d = d d = ln ln d) ln = ln(arcsin ) = ln(arcsin ) = 1 d d = ln(arcsin ) + 1 1 = d ( ) = (arcsin ) ln(arcsin ) + d 1 e) = ln(4 7+1 ) = (7 + 1) ln 4 = = 7 ln 4 15 When b is a positive constant, use the derivative rule d d (bu ) = b u ln b du d d d [5 6 ] = 5 6 ln 6 = 5 ln 6(6 ) a) d d [4 4 ] = 4 3 4 + 4 4 ln 4 = 3 4 [4 + ln 4] b) = 3 +tan ln 3 ( + sec ) c) = 6[ + 1] 4 [ +1] 3 ln 4 d) From part (b), the slope is 0 when 3 4 [4 + ln 4] = 0 Therefore = 0 or = 4 ln 4 16 Graph f() = 3 4 4 3 f 1 r min Locate critical numbers: f () = 1 3 1 = 1 ( 1) at = 0, 1 dec Neither inc rmin inc f 0 ++ 0 ++ 0 1 Check concavit with the second derivative f () = 36 4 = 1(3 ) = 0 at = 1 concave up inf dn inf concave up f + + ++ 0 0 + + + 0 /3 Plot the critical numbers and the ections: f(0) = 0 and f(1) = 1 f(/3) = 16/7 = 059851837 17 Do a complete graph of f() = ( + 1)e
18 Solution 3 1 Maimize the area A = 1 Subject to 400 = + 4 where 0 00 3 Eliminate = 100 Locate critical numbers: f () = e + ( + 1)e f () = ( + + 1)e = ( + 1) e = 0 at = 1 inc Neither inc f ++ 0 ++ 1 Check concavit with the second derivative f () = ( + )e + ( + + 1)e f () = ( + 4 + 3)e = ( + 1)( + 3)e at = 1, 3 conc up inf conc dn inf conc up f ++ 0 0 + + + 3 1 Plot the critical numbers and ections: f( 1) = /e 0736 and f( 3) = 10/e 3 0498 4 Rewrite A A() = (100 ) = 100 on [0, 00] 5 Use CIT (or possibl SCPT) Find the critical numbers A () = 100 = 0 = = 100 19 Solution Use CIT: Check critical point: A(100) = 100(50) = 5000m Endpts: A(0) = 0 and A(00) = 0 So b CIT, Absolute ma at = 100 and area 5000 square meters 1 Maimize the volume V = Subject to = + where 0 1 3 Eliminate = 4 Rewrite V V () = ( ) = 4 8 + 4 3 on [0, 1] 5 Use CIT (or possibl SCPT) Find the critical numbers Use the CIT V (1/3) = 1 3 ( 4 3 ) = 16 7 V (0) = 0 = V (1) V () = 4 16 + 1 = 4(1 4 + 3 ) = 4(1 )(1 3) = 0 = = 1, 1/3 So b the CIT, the absolute ma occurs when = 1/3 and the volume is 16 7 0 Maimize: Area A = Constraint: Cost C = 3 + = 600 at $3/m at $1/m cubic meters Eliminate: = 300 3 Smallest = 0, largest = 00 since 3(00) = 600 Substitute: A = (300 3 ) = 300 3 Domain: [0, 00] Differentiate: A = 300 3 = 0 = 100 Justif: Since the interval is closed, we can use the CIT Evaluate A at the critical and end points At the Critical point: A(100) = 100(300 3 100) = 15, 000 (Abs Ma) At the End points: A(0) = 0, A(00) = 0 So b the CIT, the Abs Ma of 15, 000 square meters occurs at = 100 m and = 300 3 100 = 150 m
1 laers on top and bottom Minimize: Materials S = Constraint: Volume V = = 16000 tops, bottoms + 4 sides {}}{ 4 + 4 Eliminate: = 16000 can be an value larger than 0 Substitute: S = 4 + 64000 Domain: (0, ) Differentiate: S = 8 64000 = 0 8 3 = 64000 So 3 = 8000 so = 0 Justif: There is onl one CP, so use the SCPT Classif the CP with the First Derivative Test S + + + R Min 0 0 There is a Relative Min at = 0 which must be an Abs Min b SCPT = 40 Minimize: Cost C = 0() + 5() 0 (top and bottom) plus 5 sides Constraint: Area A = = 64 Eliminate: = 64 Domain: 0 < < Substitute: C = 40 + 640 (0, ) Differentiate: C = 40 640 = 0 4 = 640 = 16 so = 4, 4 Justif: 1st Der TestC + + + R Min at = 4 is an Abs Min b SCPT 0 4 since there is onl one critical point and it is a rel min = 16 3 4 Minimize: SA = πr + πrhcr Constraint: Vol V = πr h = 8π Eliminate: h = 8π πr = 8 r Domain: 0 < r < Substitute: SA = πr + 16π r (0, ) Differentiate: SA = πr 16π r = 0 r 3 = 8 r = Justif: 1st Der TestSA + + + R Min at r = in is an Abs Min b SCPT 0 3 since there is onl one critical point and it is a rel min h = in
5 Maimize: V = 100 Constraint: + = 8 Eliminate: = 8 Smallest = 0, largest = 4 since (4) = 8 Substitute: V = 100(8 ) = 800 00 Domain: [0, 4] Differentiate: V = 800 400 = 0 = Justif: Since the interval is closed, we can use the CIT Evaluate V at the critical and end points At the Critical point: V () = 1600 800 = 800 At the End points: V (0) = 0, V (4) = 0 So b the CIT, the Abs Ma of 800 square inches occurs at = in and = 4 in 6 Graph f() = 6 + 3 f r ma (0, ) ( 1, 15) 15 (1, 15) 1 1 Locate critical numbers: f () = 1 ( +3) at = 0 inc r ma dec + + + 0 f 0 Check concavit with the second derivative f () = 1( +3) +1()( +3)() = 36 36 = 0 at = ±1 ( +3) 4 ( +3) 3 conc up inf down inf conc up f + + + 0 0 + + + 1 1 Plot the critical numbers and the ections: f(0) = and f( 1) = f(1) = 15