Review sheet Final Eam Math Calculus I Fall 5 UMass Boston The eam is closed tetbook NO CALCULATORS OR ELECTRONIC DEVICES ARE ALLOWED DURING THE EXAM The final eam will contain problems of types similar to the problem types given here EACH PROBLEM TYPE given in this review sheet will be REPRESENTED BY AT LEAST ONE EXAM QUESTION The final will contain or more problems in turn consisting in a number of sub-problems Problem Evaluate the it if it eists 5 + 6 5 + 5 6 DNE 5 + 5 + DNE + + 6 5 8 + h 9 7 h h + h + 8 8 h h 6 + 9 + 7 7 h + h h 5 8 + + 9 + + 6 5 + 6 + 7 9 9 5 + h 8 h h 9 + + 6 5 + 5 + h h h h +h h Solution h +h h + h h h + h + h + h h h + h h + h h h + h + +
Problem Find the it or show that it does not eist If the it does not eist, indicate whether it is ±, or neither The answer key has not been proofread, use with caution + 9 66 + + 7 + + 5 66 + + 8 + 9 + 5 + + 5 + + + 6 + + 5 + 6 + + 7 + 5 + a + b sin 8 + 6 cos a b sin DNE DNE DNE Solution + + 5 + Divide top + 5 + and bottom by highest term + 5 in denominator + + Solution + + + + + + + + + + + + The sign highlighted in red arises from the fact that, for negative, we have that Problem Evaluate the it if it eists + + + + 5 + 6
5 6 + + DNE DNE 5 + 5 + + + 6 5 8 + h 9 7 h h + h + 8 8 h h 6 + 9 + 7 7 h + h h 5 + 5 8 9 + + 6 5 + 6 + 7 9 9 5 + h 8 h h 9 + + 6 5 + 5 + h h h h +h h Solution h +h h + h h h + h + h + h h h + h h + h h h + h + + Problem Find the horizontal and vertical asymptotes of the graph of the function Check your work by plotting the function using the Internet y + + 5 y + y + + 5 Vertical:,, horizontal: y, y 6 y + vertical:,, horizontal: y y + Vertical:, 5, horizontal: none 7 y 7 + 6 vertical:,,, horizontal: y y vertical:, horizontal: y 8 y vertical:,, horizontal: y 9 + + no vertical asymptote, horizontal: y ± vertical: 6, no horizontal asymptote
+ 9 y vertical:, horizontal: y, y, Solution Vertical asymptotes A function f has a vertical asymptote at a if a f ± The function is algebraic, and therefore, if it is defined, has a finite it ie, no asymptote Therefore the function can have vertical asymptotes only for those for which f is not defined The function is not defined for + +, which has two solutions, and These are precisely the vertical asymptotes: indeed, + + + + + and + + + Horizontal asymptotes A function f has a horizontal asymptote if say, N, then y N is the equation of the corresponding asymptote Consider the it We have that + + ++ ++ + + + + + + + ± + f eists If that it eists, and is some number, Therefore y is a horizontal asymptote The case, is handled similarly and yields that y is a horizontal asymptote A computer generated graph confirms our computations when < y ++ y y Problem 5 Differentiate tan csc sec sec cot sec tan sin cos
csc 5 sec tan 6 sec + tan csc cot cos sin sec tan + sec 9 sin sin e cos sin cos sin e 7 sec sec tan + sec sin e 8 csc cot csc e sin + cos + sin tan sec e tan e cot csc Problem 6 Compute the derivative f + ft t + t y y + + 6 + ++ y c + c y tt + t + t +t +t c + c y 5 y 6 y t t t + t t + 7 gt t t t 8 y a + b + c t+ t t 8t t 5 t +t 6 t 5 6 + t y u6 u + 5 u f a + b c + d 5 f + + 9 y A + B + C b + a B C 6 f + + +6+ + t 5 + t + 7 t u + u ad bc d+c ++ + 7 f + c c c+ Solution 65 This can be differentiated more effectively using the chain rule, however let us show how the problem can be solved
directly using the quotient rule t t t t t t t t t t t + t t t t t t t t t t t t + t Problem 7 Compute the derivative using the chain rule f + y cos + f + ± + y cos f sin cos sin sin sec sin cos csc cos cos ln ln cos sin 5 y + cos 6 f sin cos sin sin sin sin 5 y sec ln ln 7 f + tan cos sin + tan sec 6 y tan5 sec tan sec sin 8 f cos + sin 9 y sin cos + sin cos sin 7 y + sin + cos tan5 5 sec 5 Solution 7 + cos + + + + + + + + + + + + + + + +cos +sin +cos Please note that this problem can be solved also by applying the transformation + + ± + ± +
before differentiating, however one must not forget the ± sign arising from ± Our original approach resulted in more algebra, but did not have the disadvantage of dealing with the ± sign Solution 7 Let u cos Then y u Chain Rule: d du du d u sin sin cos Solution 75 Let u + cos Then y u Chain Rule: d du du d u sin sin + cos sin sin cos sin sin This last step is optional Solution 79 Let u Then y sin u Chain Rule: d du du d cos u cos u Solution 75 Chain Rule: Chain Rule: d sec d sec sec sec tan sec d d sec sec tan d d sec tan There are many ways to simplify this answer, including both of the following sec tan sec sin
Solution 76 Product Rule: d d d tan5 + tan5 d d Use the Chain Rule to differentiate tan5 in the first term d 5 sec 5 + tan5 tan5 5 sec 5 Solution 77 Quotient Rule: + cos d d + sin + sin d d + cos + cos By the Chain Rule, d d + cos sin and d d + sin cos By the Pythagorean Identity, cos + sin + cos cos + sin sin d + cos cos + cos + sin + sin + cos d cos + sin + cos + sin + cos + cos + sin + cos + cos + sin + cos Problem 8 Verify that the coordinates of the given point satisfy the given equation Use implicit differentiation to find an equation of the tangent line to the curve at the given point The answer key has not been proofread, use with caution y sin cosy, π, π y 7 8 y y 5,, sin + y y, π, π y + π + y + y,, circle y + y y +,, hyperbola 5 + y + y,, 6 + y,, y + y + 7 + y 5 y,, y 9 + 9 + y at, y y + y + y at, y 5 + 6
Solution 8 y First we verify that the point, y π, π indeed satisfies the given equation: y sin π,y π π cosy π,y π π cos π sin π left hand side right hand side so the two sides of the equation are equal both to when π and y π Since we are looking an equation of the tangent line, we need to find d π,y π - that is, the derivative of y at the point π, y π To do so we use implicit differentiation d y sin cosy d d sin + y d d sin cosy + d d cosy d π,y π sin + y cos cosy siny d d d sin + siny cosy y cos Set π, y π π π sin π + π sin cos π cos π π d π,y π d π,y π π cos π Therefore the equation of the line through π, y π is y π π y Problem 9 What is the -coordinate of the point on the hyperbola y 6 that is closest to the point,? What is the -coordinate of the point on the ellipse + y 6 closest to the point,?
A rectangular bo with a square base is being built out of sheet metal pieces of sheet will be used for the bottom of the bo, and a single piece of sheet metal for the sides and the top of the bo What is the largest possible volume of the resulting bo that can be obtained with 6m of metal sheet? cubic meters Recall that the volume of a cylinder is computed as the product of the area of its base by its height Recall also that the surface area of the wall of a cylinder is given by multiplying the perimeter of the base by the height of the cylinder A cylindrical container with an open top is being built from metal sheet The total surface area of metal used must equal m Let r denote the radius of the base of the cylinder, and h its height How should one choose h and r so as to get the maimal possible container volume? What will the resulting container volume be? Solution 9 The distance function between an arbitrary point, y and the point, is d + y On the other hand, when the point, y lies on the hyperbola we have y 6 In this way, the problem becomes that of minimizing the distance function dist + y + 6 This is a standard optimization problem: we need to find the critical endpoints, ie, the points where dist As the square root function is an increasing function, the function + 6 achieves its minimum when the function l dist + 6 does l is a quadratic function of and we can directly determine its minimimum via elementary methods Alternatively, we find the critical points of l: l + 5 5 On the other hand, 6 + y and therefore 6 Therefore, ] [, As 5 is outside of the allowed range, it follows that our function either attains its minimum at one of the endpoints ± or the function has no minimum at all It is clear however that as tends to, so does dist Therefore dist attains its minimum for or and y ± ± 6 Direct check shows that dist + 6 and dist + 6 5 so our function dist has a minimal value of achieved when, which is our final answer Notice that this answer can be immediately given without computation by looking at the figure drawn for 9 Indeed, it is clear that there are no points from the hyperbola lying inside the dotted circle centered at, Therefore the point where this circle touches the hyperbola must have the shortest distance to the center of the circle Solution 9 Let B denote the area of the base of the bo, equal to the area of the top Let W denote the area of the four walls of the bo the four walls are all equal because the base of the bo is a square Then the surface area S of material used will be S B }{{} two pieces for the bottom + W }{{} walls + }{{} B B + W the bo lid Let denote the length of the side of the square base and let y denote the height of the bo Then and As the surface area S is fied to be 6 square meters, we have that B W y S B + W 6 + y As y is positive, the above formula shows that 6 and so Let us now epress y in terms of : + y 6 y 6 y 6
The problem asks us to maimize the volume V of the bo The volume of the bo equals the area of the base times the height of the bo: V B y y 6 6 As is non-negative, it follows that the domain for is: [, ] To maimize the volume we find the critical points, ie, the values of for which V vanishes: 6 V 6 9 9 6 ± As measures length, is not possible outside of the domain for Therefore the only critical point is Direct check shows that at the endpoints and, we have that V Therefore the maimal volume is achieved when : V ma V 6 Problem Find the the implied domain of f, and y intercepts of f, horizontal and vertical asymptotes, intervals of increase and decrease, local and global minima, maima, intervals of concavity, points of inflection Label all relevant points on the graph Show all of your computations y f + + + y-intercept: -intercept: Horizontal asymptote: y, vertical: none local and global min, and global ma at + Intervals of decrease:, +, of decrease, at local +, intervals, + Concave down on,, concave up on,, Inflection points at:,, y f 5 + 9 + y-intercept: horizontal asymptote: y, vertical: none increasing on, +, decreasing on, local and global min at, local and global ma at + +, concave up on,,, concave down,, inflection points at,,
y f + + + +, f + + The two functions are plotted simultaneously in the, y-plane Indicate which part of the graph is the graph of which function For f ++ y-intercept:, no intercept no asymptotes decreasing on [, ], increasing on [, ] global and local min at, global and local ma at ± concave up on [, ] no inflection points : For f ++ : ++ y-intercept:, no intercept no asymptotes increasing on [, ], decreasing on [, ] global and local ma at, global and local min at ± concave down on [, ] no inflection points y f e + e e e y 5 f e + e e + e y 6 f ln + +
y 7 f + + + y-intercept: none, -intercepts: 5 horizontal asymptote: y, vertical: and always decreasing no local/global minima/maima concave down on, cup,, concave up on,, inflection point at 8 f + + + y-intercept:, -intercept: horizontal asymptote: y, vertical: none increasing on, +, decreasing on, local and global min at, local and global ma at + concave up on, cup,, concave down,, inflection points at,, +, Solution Domain We have that f is not defined only when we have division by zero, ie, if + equals zero However, the roots of + are not real numbers: they are ± ±, and therefore + can never equal zero Alternatively, completing the square shows that the denominator is always positive: Therefore the domain of f is all real numbers + + 9 9 + + >, y-intercepts The y-intercept of f equals by definition f 5 + 9 + The intercept of f is those values of for which f The graph of f shows no such, and that is confirmed by solving the equation f : 5 + 9 + 5 + 9 f Mult by +, 5 ± 5 9 5 ± 9 so there are no real solutions the number 9 is not real Asymptotes Since f is defined for all real numbers, its graph has no vertical asymptotes To find the horizontal asymptotes, we, 9
need to compute the its f and 5 + 9 ± + f The two its are equal, as the direct computation below shows: 5 + 9 Divide by leading ± + monomial in denominator 5 + 9 ± + + + Therefore the graph of f has a single horizontal asymptote at y Intervals of increase and decrease The intervals of increase and decrease of f are governed by the sign of f We compute: f 5 + 9 + 5 + 9 + 5 + 9 + + + + As the denominator is a square, the sign of f is governed by the sign of + To find where + changes sign, we compute the zeroes of this epression: + Mult by 6 +, 6 ± 6 6 ±, ± Therefore the quadratic + factors as + and The points, split the real line into three intervals:,,, + +,, and each of the factors of has constant sign inside each of the intervals If we choose to be a very negative number, it follows that is a negative, and therefore f is negative for, For, +, eactly one factor of f changes sign and therefore f is positive in that interval; finally only one factor of f changes sign in the last interval so f is negative on +, Our computations can be summarized in the following table Interval f f,, + + +, Local and global minima and maima The table above shows that f changes from decreasing to increasing at and therefore f has a local minimum at that point The table also shows that f changes from increasing to decreasing at + and therefore f has a local maimum at that point The so found local maimum and local minimum turn out to be global: there are two things to consider here First, no other finite point is critical and thus cannot be maimum or minimum - however this leaves out the possibility of a maimum/minimum at infinity This possibility can be quickly ruled out by looking at the graph of f To do so via algebra, compute first f and f : f f f f + + + 5 5 + + + 9 + + 9 + +
On the other hand, while computing the horizontal asymptotes, we established that ± f This implies that all sufficiently far away from, we have that f is close to Therefore f is larger than f and smaller than f for all sufficiently far away from This rules out the possibility for a maimum or a minimum at infinity, as claimed above Intervals of concavity The intervals of concavity of f are governed by the sign of f The second derivative of f is: f f + + + + + + second differentiation: + chain rule + + + + + + + + + 6 + + factor out + + + 6 + + + + 6 + + + + + When computing the domain of f, we established that the denominator of the above epression is always positive Therefore f changes sign when the terms in the numerator change sign, namely, at, and Our computations can be summarized in the following table In the table, we use the symbol to denote that the function is concave up in the indicated interval, and to denote that the function is concave down Interval f f,, +,, + Points of inflection The preceding table shows that f changes sign at,, and therefore the points of inflection are located at, and, ie, the points of inflection are, f,,, f,,, f, 5 We can ask our graphing device to use the so computed information to label the graph of the function Finally, we can confirm visually that our function does indeed behave in accordance with our computations, y +, + infl:, infl:, infl:, 5 Solution 8 This problem is very similar to Problem We recommend to the student to solve the problem first with closed tetbook and only then to compare with the present solution Domain As f is a quotient of two polynomials rational function, its implied domain is all ecept those for which we get division by zero for f Consequently the domain of f is all for which + + However, the polynomial + + has no real roots - its roots are ± 6 ±, and therefore the domain of f is all real numbers Alternatively, we can complete the square: + + + + and so + + is positive for all values of, y-intercepts The y-intercept of f equals by definition f which f We compute + + + The intercept of f is those values of for
f + + + +, and the -intercept of f is Asymptotes The line a is a vertical asymptote when f ± ; as f is defined for all real numbers, this implies that a ± there are no vertical asymptotes The line y L is a horizontal asymptote if f eists and equals L We compute: f ± + + + + + + + + + Therefore y is a horizontal asymptote for f An analogous computation shows that f and so y is the only ± horizontal asymptote of f Intervals of increase and decrease The intervals of increase and decrease of f are governed by the sign of f We compute: f + qutotient rule + + + + + + + + + + + + + + + + + + + + + + + + + As + + is positive, the sign of f is governed by the sign of + + To find out where + + changes sign, we compute the zeroes of this epression: + + use the quadratic formula, ± Therefore the quadratic + + factors as + The points, split the real line into three intervals:,,, + and +,, and each of the factors of has constant sign inside each of the intervals If we choose to be a very negative number, it follows that is a negative, and therefore f is negative for, For, +, eactly one factor of f changes sign and therefore f is positive in that interval; finally only one factor of f changes sign in the last interval so f is negative on +, Our computations can be summarized in the following table Interval f f,, + + +, Local and global minima and maima The table above shows that f changes from decreasing to increasing at and therefore f has a local minimum at that point The table also shows that f changes from increasing to decreasing at + and therefore f has a local maimum at that point The so found local maimum and local minimum turn out to be global: indeed, no other finite point is critical and thus cannot be maimum or minimum; on the other hand ± f and this implies that all sufficiently far away from have that f is close to, and therefore f is larger than f and smaller than f for all
Intervals of concavity The intervals of concavity of f are governed by the sign of f The second derivative of f is: f f + + + + + + + use chain rule + + + for second differentiation + + + + + + + + + + + + + + + + + factor out ++ + + + + + + + + + + + + + + + + + + 8 + + factor + 8 + + + + As we previously established, the denominator of the above epression is always positive Therefore the epression above changes sign when the terms in the numerator change sign, namely, at, and Our computations can be summarized in the following table Interval f f,, +,, + Points of inflection The preceding table shows that f changes sign at,, and therefore the points of inflection are located at, and, ie, the points of inflection are,,,,, Problem Use the Intermediate Value theorem and the Mean Value Theorem/Rolle s Theorem to prove that the function has eactly one real root f + + 7 f + + + f cos + sin Solution f 9 and f Since f is continuous and has both negative and positive outputs, it must have a zero In other words, for some c between and, fc If there were solutions a and b, then we would have fa fb, and Rolle s Theorem would guarantee that for some -value, f However, f +, which always positive and therefore is never Therefore there cannot be or more solutions The above can be stated informally as follows Note that f +, which is always positive Therefore, the graph of f is increasing from left to right So once the graph crosses the -ais, it can never turn around and cross again, so there can only be a single zero that is, a single solution to f Solution f5 cos 5 + sin 5 5 5 < because cos a, sin b [, ] for arbitrary a, b Similarly f 5 cos 5 + sin 5 + 5 5 > Therefore by the Intermediate Value Theorem f has at least one solution in the interval [ 5, 5] Suppose on the contrary to what we are trying to prove, f has two or more solutions; call the first solutions a, b That means that fa fb, so by the Mean value theorem, there eists a c a, b such that f c fa fb/a b /a b On the other hand we may compute: f + cos cos sin <, where the first inequality follows from the fact that sin, cos [, ] So we got that f c for some c but at the same time f < for all, which is a contradiction Therefore f has eactly one solution Problem Estimate the integral using a Riemann sum using the indicated sample points and interval length
8 + d Use four intervals of equal width, choose the sample point to be the left endpoint of each interval 6 fd 9 + 7 d Use three intervals of equal width, choose the sample point to be the left endpoint + and f 8 + Thus 5 5 85 fd d Use three intervals of equal width, choose the sample point to be the midpoint of each interval + 6 and f + Thus 5 5 fd f + f + f 5 and f + Thus 8 d + + Use and right endpoint sampling points 5 d + + Use and left endpoint sampling points 97 6 69 8 + + 8 7 + 6 6 89 59 d Use four intervals of equal width, choose the sample point to be the left endpoint of each interval + 5 and f + Thus fd f + f + f + 9 + 9 7 + f 69 6 87 7 d Use four intervals of equal width, choose the sample point to be the right endpoint + 6 8595 6596 + f + f + f f fd 5 and f + Thus Solution The interval [, ] is subdivided into n intervals, therefore the length of each is The intervals are therefore [, ], [, ], [, ], [, ] The problem asks us to use the left endpoints of each interval as sampling points Therefore our sampling points are,,, Therefore the Riemann sum we are looking for is f + f + f + f 8 + + 8 + + 8 + + 8 + 9 + 7 5
Solution The interval [ 5, 5] is subdivided into n intervals, therefore the length of each is The intervals are therefore [ 5, 5], [ 5, 5], [ 5, 5] The problem asks us to use the midpoint of each interval as a sampling point Therefore our sampling points are,, Therefore the Riemann sum we are looking for is f + f + f + 5 + 8 y 7 5 Problem Evaluate the definite integral d 9 + + d 8 d ] [ + + + d + d [] [] 9 d [] 5 6 7 8 + d ] + d + + + ] 6 [ + [ 5 + 8 + ] 5 + d ] 96 5 d + d 6 d [ 7 7 + 5 5 + + + ln ] ln ] 5 + 56 5 5 [ [ [ 5 6 7 + d 5 6 + 6 5 d + d + d + d + d d [] [] [] [] [] [] 5 6 π π π π π π π π csc θdθ cos θ cos dθ θ sin θ cos θ dθ tan θdθ sin θ + sin θ tan θ sec dθ θ sin θ cos θdθ sin d [] [] [] [] [] [] 6 ] 6 + [ []
Solution 8 d d + d when when [ d + d + ] [ + ] 8 + + 8 Problem Evaluate the definite integral The answer key has not been proofread, use with caution e e d ln e d e e + e + d lne + + d ln 5 d Solution + d u9 u d + d + + du u [ln u]9 ln ln 9 ln Set u + 6 Problem 5 Differentiate f using the Fundamental Theorem of Calculus part f sin t dt f + t 5 dt + 5 sin
f t dt e f t dt f 5 5 g cos t dt ln f e ln cos d Solution 5 We recall that the Fundamental Theorem of Calculus part states that We can rewrite the integral so it has as the upper it: d a htdt h where a is a constant Therefore d d f + t 5 dt + 5 dt + 5 dt d + t 5 FTC part dt + 5 d Solution 5 f d d d d d d e ln ln ln t dt t dt + e t dt + t dt The Fundamental Theorem of Calculus part I states that for an arbitrary constant a, use this two compute the two derivatives: d d d d ln e t dt d d u t dt u du d ln t dt d w t dt d Finally, we combine the above computations to a single answer w dw d e e e f e e ln t dt d u gtdt gu for a continuous g We du a Set u ln Set w e Problem 6 Find the area of the region bounded by the curves y and y + Find the area of the region bounded by the curves y and y 9
Solution 6 y is a parabola here we consider as a function of y y + implies that y and so the two curves intersect when y y y + y + y + y y or As y, this implies that when y and when y, or in other words the points of intersection are, and, Therefore we the region is the one plotted below Integrating with respect to y, we get that the area is A 9 + [ y + y + y ] y + y + 8 + + + y + y Problem 7 Consider the region bounded by the curves y + and y + What is the volume of the solid obtained by rotating this region about the line? 5 π Consider the region bounded by the curves y and y What is the volume of the solid obtained by rotating this region about the line y? 6π 5 Consider the region bounded by the curves y and y What is the volume of the solid obtained by rotating this region about the line? π
Solution 7 First, plot y + and y + y y r outer r inner r outer r inner z z The two curves intersect when + + or Therefore the two points of intersection have -coordinates between and Therefore we need we need to integrate the volumes of washers with inner radii r inner +, outer radii r outer + and infinitesimal heights d The volume of an individual infinitesimal washer is then πr outer r inner d V π + + d π + + d [ π ] 5 5 + + 5 π