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# Calculus 1 Chapter 3 notes MTH 161

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This 18 page Class Notes was uploaded by Jack Magann on Sunday February 15, 2015. The Class Notes belongs to MTH 161 at University of Miami taught by Dr. Beverly in Fall2014. Since its upload, it has received 97 views. For similar materials see 161 Calculus 1 in Mathematics (M) at University of Miami.

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Date Created: 02/15/15

Chapter 31 The most used application of derivatives is optimization Optimization best design or rate of change to minimize cost or effort Maximum and Minimum values let c be a number in the domain D of f The fx is absolute max Value offon D is fc 2 fx on all x on D absolute min Value offon D is fc S fx on all x on D Absolute values can occur anywhere on a graph and endpoints local mins and maxes the lowest or highest part on a single section off local mins and maxes can not occur on endpoints Extreme Value Theorem If f is continuous on a closed interval ab then fattains an absolute max value fc and an absolute min value fd at some numbers c and d in ab Critical number of a function fis a number c in the domain of f such that either f c O orf c DNE undefined critical numbers are points on graph Fermat s Theorem If f has a local maxmin at c and if f c exists then f c 0 it could also be f c DNE Fermat s Theorem however doesn t say that every time f c O or DNE there is a max or min value Need to look on an interval for the absolute values find critical points and plug them back into fx highest point is the absolute max and lowest is the absolute min Ex Find the absolute min and max values of the function fx x3 3x2 1 on the interval 34 step 1 Find values of f at critical numbers f x 3x2 6x 3xx 2 The critical numbers are at x0 and x2 and including the endpoints X 12 x4 Step 2 Find the values of fx at critical points f0 1 f2 3 f f4 17 Step 3 Find the absolute max and min Maxfx 17 min fx 3 Chapter 32 Mean Value Theorem MVT Hypotheses memorize Let function f satisfy the following hypotheses 1 f is continuous on closed interval ab 2 fis differentiable on open interval ab Then there is a number c in ab such that I fbfa f C ba This is the average of all the tangents in the interval so one tangent with the same slope in problems 1verify function satisfies hypotheses of MTV on ab 2Find all numbers c in the interval that satisfy conclusion of MVT EX fx 3x2 2x 5 on 11 Verify 1 Fx is continuous on 11 since it is a polynomial and is continuous on the interval 00 oo 2 FX is differentiable on 11 since it is a polynomial and is differentiable on the interval 00 oo Calculate 2 2 6C2 31 215 2 1 2 15 6c 2 2 c0 EXZ Verify that f x xx satisfies the MVT on 14 Verify fx is continuous on 00 2U 2 00 on the interval 14 and differentiable on 00 2U 2 00 on the interval 14 Calculate 2 L1 1 1 2 3 3 i Hm 3 3 9 18 c 2 ci3V2 2 The negative answer is not in the interval and so can not be included in the answer Chapter 33 The derivatives of function fx f x and f x can be used to find features of fx if one doesn t know what fx looks like f x shows When fx is increasing or decreasing When fx has a min or max value f xshows When fx is concave up or concave down the points of in ection on fx Ex y x4 4x3 y 4x3 12x2 2 4x2 3 y 12x2 24x 2 12xx 2 The critical values on y are x0 and x3 put these values on a number line and plug a value from each of the three regions into f x to see where f xis positive and negative Where f x is negative fx is decreasing 00 3 Where f x is positive fx is increasing 3 00 Where f x goes from negative to positive fx has a local or absolute min Where f x goes from positive to negative fx has a local or absolute max The critical values for yquot are x0 and x2 put these values on a number line and plug a value from each of the three regions into f x to see where f xis positive and negative Where f x is negative fx is concave down 02 Where f x is positive fx is concave up 00 OU2 00 Critical points on f x are points of in ection on fx plug the critical numbers into fx 00 and 216 Second Derivative Test Take the critical numbers from f x and plug them into f x if f x is positive there is a minimum value on fx if f x is negative there is a maximum value on fx if f x is 0 then it is neither a max nor a min and that s all this test tell us Chapter 34 Sketching a curve Steps 1 Determine the Domain of the original function 2 Find the x and y intercepts x intercept is when y0 yintercept is when x 0 3 Find if there is symmetry of the function Odd fx f x even fx f x or neither 4 Find Asymptotes Horizontal limx4 00 f x L where L is the asymptote Vertical limxa f x 2 oo and limxa fx 00 or limxa f x 00 and limxa f x 00 5 Find intervals of increasing and decreasing Using first derivative to find this 6 Find Minimum and Maximum values Using critical values of first derivative to find this 7 Find concavity of graph and in ection points Use second derivative and its in ection point to find this Using these steps along with finding values using the original function one can accurately sketch the graph of a function 2x2 Ex Sketch y x21 Domain 00 1U 11U1 00 Intercepts both are 00 Symmetry f x f x so the graph is function Asymptote limx gt 1 fx 0 and limx gt 1 00 Vertical limx gt1 00 and limx gt1 0 2x2 11mx4 00 m 2 Horlzontal Increasing and decreasing intervals I x214x 2x22x 4x f x x212 LUZ increaSingi 000 decreasing 000 Max at x0 intervals of concave up and concave down fx 4x2 12 4x2x2x2 1 4x2416x2 12x24 x2 14 x2 13 x2 13 concave up 00 1U 1 00 concave down 11 no in ection points since f x has critical points that don t exist on fx This function can now be accurately sketched using these qualities ex2 Sketch y 2 x x Domain 0 00 Intercepts y intercept 00 x intercept 00 and 40 Symmetry neither Asymptotes lim2 x x oo and lim2 x xoo x gtoo x gt oo Therefore there are no asymptotes Increasing and decreasing intervals increasing 01 decreasing 1 00 Max 11 Concave up and concave down intervals concave up DNE concave down 0 00 then there is no in ection points With these qualities the function could be sketched Chapter 35 optimization in context problems 1 A box with an open top is made by taking a 16 by 30quot piece of cardboard cutting out the same square piece on each corner and then folding the remaining figure What does the length of the cut out squares need to be to maximize the volume of the box With this question approaching it with a diagram is the most helpful So draw a rectangle that s 16 by 30 designate the square areas on each corner that s taken out since both sides of the square are equal we can designate that x Now we have to look at the formula the question is asking us to optimize Volume The volume of the box is VLWH using the diagram L 30 2x W 16 2xH x So V 30 2x16 2xx 4x3 92x2 480x Now find the derivative of V and its critical point to find when the volume reaches a max V 12x2 184x 480 43x 10x 12 Critical numbers are 103 and 12 12 however is not in the domain of the original function which is 08 this is because x is the length of the square cut out can t be negative since that would mean adding to the sides can t be 0 since something does have to be cut out to form the box can t be eight or higher since that would make the shorter side nonexistent and there could be no box Checking the intervals using V we see that V has a max at 13 0 1o So the answer 1s x 1nches For problems like this it may also require you to relate one part of the shape being looked at to the one you want 0 optimize to make the later one in terms of a single variable This could mean putting surface area if it s a known value in terms of y and plug that into the volume to make it in terms of x The derivative then can be taken and volume optimized This can work the other way around too where surface area is the aspect eing optimized Graphing practice x graph y rm Use the steps from prev10us chapter Domain 00 00 Intercepts yintercept00 x intercept00 Symmetry odd Asymptotes Horizontal y1 and y1 use limits Vertical None 2 y 3 1ncreas1ng1nterval oooo decreas1ng1nterval None x2 No maxes or mins 3 quot 6 965 Concave up 00 0 concave down 0 oo x22 in ection point 00 Using these qualities one can accurately graph this equation Extra credit problem Review chapter 3 1 A company manufactures a box with V 24 f t3 with an open top and a length twice as long as its width Find width of the box so minimum amount of material is used Here we are minimizing surface area and so we have SA 2 LxW 2LgtltH ZWXH and 24 LWH Now we define the dimension as variable L2x Wx Hy so 24 2 2x231 and y g Therefore SA 2 2x2 24x 1 48x 1 2 2x2 72x 1 Take the derivative of the surface area SA 2 4x 72x 2 2 4x 2 x3 18 The critical points are then x O and x W O is not in the domain of the original function though since the width of the box x can not be zero checking both intervals on either side of x Vl B in SA it goes from negative to positive meaning there is a minimum there The Answer then is 3V 18ft 26ft 2 A 23inch piece of string is cut in two parts One is used to make a circle and the other a square How should the string be cut so the sum of the two areas is a minimum set the length of the string equal to x and y then put it in terms of x Lxy23 y23 x set each part of the string to the corresponding perimeter of the shape they make x 2mquot r being the radius of the circle made 23x4a a being the length of each side of the square made set these equal to the strings length and solve for one of the variables 23 2m 4a r 23 4 21139 We can now substitute this into the Area formula which is A nrz a2 23 4a 2 A7T a2 2n Take the derivative 23 4a 87 AI 2n gt lt 27 4 92 16a 4cm 2a 2 23 4a 2a 2 Set the top equal to zero to find the critical numbers 27 472 18 O 9216a4cm a 4TL39 plug this back in to solve for the lengths 23 x 4 x 23 101m 4n 4n 72 y m 3 A company is making an open top square base rectangular tank with a volume of 53 f t3 What dimensions yield the minimum surface area Define the variable 53 LWH L x W x H y 53 xzy Put y in terms of x and plug it into the surface area formula y SAx24xyx24x x2212x 1 Take the derivative SA 2 2x 212x 2 2 2x 2x3 106 the critical number that fits in the domain is x W Using a number line analysis we see that there is a minimum at x 3V 106 Plug this back in to get other dimensions of the box 53 24 y 3 1062 The tank is 47 by 47 by 24 4 From a 10 inch by 10 in piece of cardboard squares are cut out of each corner so that the sides fold up to make a box What dimensions yield the maximum volume What is the Maximum volume Draw the 10 by 10 square and label the squares being cut out Make each square x by x for the length and width Makes it so when folded up the height ofthe box is x Define the variables V LWH 10 2x10 2xx 100x 40x2 4x3 Take the derivative V 100 80x 12x2 2 43x 5x 5 The critical numbers are x 35 and x5 x5 though is not in the domain of the original function since 5 would make one of the sides nonexistent which can not happen for this problem Test intervals to get that there is a maximum at x 3 5 Plug into original equation for dimensions and maximum volume the box needs to be 67 by 67 by 17 V 741 in3 5 The velocity of a particle in feet per second is given by v t2 2t 7 where t is the time the particle has traveled Find the time at which the velocity is at a minimum ust as the first derivative can be used to find the maximum and minimum values of the original function the second derivative can be used to find the maximum and minimum values of the first derivative Find the second derivative acceleration a 2 2t 2 with critical numbers at t1 Testing intervals gives us that there is a minimum at t1 Answer t 1 second 6 Using a rectangular sheet with a perimeter of 27 cm and dimensions x and y a cylinder is created by rolling the sheet up so that the side with length x makes the perimeter of the circular base What values of x and y give the largest volume State the formulas being used x 2 272x2y x27tr sor V7Ir2y7t y Solve perimeter for y and plug it into volume 27 2x x 2 27 2x 27x2 x3 y so V n 2 2n 2 81 41 Take the derivative V 81t54x 121m2 5496 6962 6412 161t2 81 Find critical points 6x9 x O x9 and x0 x0 is not in the domain of the original function Testing the intervals x9 is a maximum Plug into perimeter to solve for length of original paper 27 18 3 2 2 Answer X 9 and y 3 Chapter 37 Antiderivatives Antiderivatives use the derivative to find the original function ex f x 3x2 Using the inverse process of finding the derivative f x can be fx x3 or fx x3 20 or fx x3 100 etc So we just write f x x3 C where C is a constant Without more information we cannot determine the value of C To do this we would need points on the function given Notation styles The example above uses prime notation where the derivative is marked as f x and the original is f x The notation you may also see is where the derivative is marked as f x and the original function is F x Read the question carefully for any specific notation Ex f x 8x9 3x6 12x find the antiderivative Fx 2 110x10 3x7 3x4 C double check by taking the derivative of Fx 5 EXZ f x 3 sinx 2x 5 find the antiderivative The algebra here can make taking the antiderivative confusing so you have to simplify the equation 1 fx 3 sinx 2x4 96 5 so 1 Fx 2 3 cosx 3x5 2x3 The original function can also be found with second derivatives This requires we add another different constant when we take the second antiderivative Since we do not know if it is the same as C we will call it D Ex f x 12962 6x 4 f x 4x3 3x2 4x C fx x4x3 2x2CxD If in the equation above f 1 1 and f O 4 what is fx We start by plugging in f O fO 0403 202C0D 4 so D 4 Now plug in f1 knowing D 4 f11413 2121C41 so C 3 Second antiderivatives are commonly seen with position velocity acceleration problems Ex a 2 6t 4 initial velocity is 6 and initial displacement is 9 v 2 3t2 4t C butwe know 120 2 6 6 302 40 C so C 6 s t3 2t2 6x D butwe knows0 9 903202 60D so D9 making st32t2 6t9 All these notes are from a Calculus 1 class run by Dr on Beverly at the University of Miami A majority of the problems used by the teacher and copied in these notes are taken from the textbook Essential Calculus Second edition By James Stewart Citation Stewart James Essential Calculus 2nd ed Np Cangage Learning 2013 Print Extra credit problem Review chapter 3 1 A company manufactures a box with V 24 f t3 with an open top and a length twice as long as its width Find width of the box so minimum amount of material is used Here we are minimizing surface area and so we have SA 2 LxW 2LgtltH ZWXH and 24 LWH Now we define the dimension as variable L2x Wx Hy so 24 2 2x231 and y g Therefore SA 2 2x2 24x 1 48x 1 2 2x2 72x 1 Take the derivative of the surface area SA 2 4x 72x 2 2 4x 2 x3 18 The critical points are then x O and x W O is not in the domain of the original function though since the width of the box x can not be zero checking both intervals on either side of x Vl B in SA it goes from negative to positive meaning there is a minimum there The Answer then is 3V 18ft 26ft 2 A 23inch piece of string is cut in two parts One is used to make a circle and the other a square How should the string be cut so the sum of the two areas is a minimum set the length of the string equal to x and y then put it in terms of x Lxy23 y23 x set each part of the string to the corresponding perimeter of the shape they make x 2mquot r being the radius of the circle made 23x4a a being the length of each side of the square made set these equal to the strings length and solve for one of the variables 23 2m 4a r 23 4 21139 We can now substitute this into the Area formula which is A nrz a2 23 4a 2 A7T a2 2n Take the derivative 23 4a 87 AI 2n gt lt 27 4 92 16a 4cm 2a 2 23 4a 2a 2 Set the top equal to zero to find the critical numbers 27 472 18 O 9216a4cm a 4TL39 plug this back in to solve for the lengths 23 x 4 x 23 101m 4n 4n 72 y m 3 A company is making an open top square base rectangular tank with a volume of 53 f t3 What dimensions yield the minimum surface area Define the variable 53 LWH L x W x H y 53 xzy Put y in terms of x and plug it into the surface area formula y SAx24xyx24x x2212x 1 Take the derivative SA 2 2x 212x 2 2 2x 2x3 106 the critical number that fits in the domain is x W Using a number line analysis we see that there is a minimum at x 3V 106 Plug this back in to get other dimensions of the box 53 24 y 3 1062 The tank is 47 by 47 by 24 4 From a 10 inch by 10 in piece of cardboard squares are cut out of each corner so that the sides fold up to make a box What dimensions yield the maximum volume What is the Maximum volume Draw the 10 by 10 square and label the squares being cut out Make each square x by x for the length and width Makes it so when folded up the height ofthe box is x Define the variables V LWH 10 2x10 2xx 100x 40x2 4x3 Take the derivative V 100 80x 12x2 2 43x 5x 5 The critical numbers are x 35 and x5 x5 though is not in the domain of the original function since 5 would make one of the sides nonexistent which can not happen for this problem Test intervals to get that there is a maximum at x 3 5 Plug into original equation for dimensions and maximum volume the box needs to be 67 by 67 by 17 V 741 in3 5 The velocity of a particle in feet per second is given by v t2 2t 7 where t is the time the particle has traveled Find the time at which the velocity is at a minimum ust as the first derivative can be used to find the maximum and minimum values of the original function the second derivative can be used to find the maximum and minimum values of the first derivative Find the second derivative acceleration a 2 2t 2 with critical numbers at t1 Testing intervals gives us that there is a minimum at t1 Answer t 1 second 6 Using a rectangular sheet with a perimeter of 27 cm and dimensions x and y a cylinder is created by rolling the sheet up so that the side with length x makes the perimeter of the circular base What values of x and y give the largest volume State the formulas being used x 2 272x2y x27tr sor V7Ir2y7t y Solve perimeter for y and plug it into volume 27 2x x 2 27 2x 27x2 x3 y so V n 2 2n 2 81 41 Take the derivative V 81t54x 121m2 5496 6962 6412 161t2 81 Find critical points 6x9 x O x9 and x0 x0 is not in the domain of the original function Testing the intervals x9 is a maximum Plug into perimeter to solve for length of original paper 27 18 3 2 2 Answer X 9 and y 3 Chapter 37 Antiderivatives Antiderivatives use the derivative to find the original function ex f x 3x2 Using the inverse process of finding the derivative f x can be fx x3 or fx x3 20 or fx x3 100 etc So we just write f x x3 C where C is a constant Without more information we cannot determine the value of C To do this we would need points on the function given Notation styles The example above uses prime notation where the derivative is marked as f x and the original is f x The notation you may also see is where the derivative is marked as f x and the original function is F x Read the question carefully for any specific notation Ex f x 8x9 3x6 12x find the antiderivative Fx 2 110x10 3x7 3x4 C double check by taking the derivative of Fx 5 EXZ f x 3 sinx 2x 5 find the antiderivative The algebra here can make taking the antiderivative confusing so you have to simplify the equation 1 fx 3 sinx 2x4 96 5 so 1 Fx 2 3 cosx 3x5 2x3 The original function can also be found with second derivatives This requires we add another different constant when we take the second antiderivative Since we do not know if it is the same as C we will call it D Ex f x 12962 6x 4 f x 4x3 3x2 4x C fx x4x3 2x2CxD If in the equation above f 1 1 and f O 4 what is fx We start by plugging in f O fO 0403 202C0D 4 so D 4 Now plug in f1 knowing D 4 f11413 2121C41 so C 3 Second antiderivatives are commonly seen with position velocity acceleration problems Ex a 2 6t 4 initial velocity is 6 and initial displacement is 9 v 2 3t2 4t C butwe know 120 2 6 6 302 40 C so C 6 s t3 2t2 6x D butwe knows0 9 903202 60D so D9 making st32t2 6t9

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