Math 1271 Solutions for Fall 2005 Final Exam

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1 Math 7 Solutions for Fall 5 Final Eam ) Since the equation + y = e y cannot be rearrange algebraically in orer to write y as an eplicit function of, we must instea ifferentiate this relation implicitly with respect to, in orer to fin an epression for y : ( + y ) = (ey ) + y y = u (eu ) u setting u = y + y y = y eu ( y ) + y = eu ( ' y + y') + y y = ey ( y + y ) y y ey y = y ey y = y ey y e y (A) ) The two parts of the Funamental Theorem of (Integral) Calculus can be put together to state that, for a function h() continuous on [ a, b ], the efinite integral " h(t) t, with a < < b, equals f() f(a), where f () = h() If we moify the a upper it of the integration to make it a function u() in the same interval, we then u() have " h(t) t = f (u()) # f (a ) The erivative of this integral is then a u() " h(t) t = a [ f (u()) # f (a ) ] = f (u()) # f (a ) = u u f (u ) " # = f u() " u f(a) is a constant or h(u()) " u' ( ) The integran given in the problem is continuous everywhere (since it is efine everywhere), so the Funamental Theorem an its consequences can be applie here For the integral function efine by t, we must first make an alteration sin(e t ) t + to accommoate the conitions for the Theorem: we move the constant it of integration to become the lower it by reversing the irection of integration:

2 sin(e t ) t + t = sin(e t ) t t + With h() = sin(e ) + sin(e t ) t + an u() =, the erivative of our integral function is t = h(u( )) u' ( ) = sin(e ) ( ) + ( )' = sin(e 4 + ) (C) ) There are many possible ways to choose bouns for the value of a efinite integral; in view of the choices given, which involve integer results, we will want to look at something fairly simple to calculate With the interval being [, ], the smallest value the integran itself, sin, coul possibly have woul be if the secon term within the raical were zero, reucing the function to 4 =, which is equal to at = (in fact, the sine-square term is not zero there) With the with of the interval of integration being Δ = =, we can efinitely say that sin Δ = At the other etreme, the largest value the integran coul possibly have woul be when sin is as large as possible, making the function equal to = ; this takes on its largest value on the interval at =, where = = = 5 This result provies an upper boun for our integral of sin 5 Δ = 5 (B) 4) In some situations, fining a it involving a multiply-stage composition function coul be ealt with by using substitutions Here, however, a substitution base on, say, u = sin t, woul not make the epression less complicate, nor woul it einate the troublesome zero in the enominator A more useful approach is to apply L Hôpital s Rule, along with the Chain Rule: sin(sin(sin t ))) t t = " " [ sin(sin(sin t ))) ]' t t' t u sin(u ) u t setting u = sin(sin t) cos(u ) (sin(sin t )) t t (continue)

3 cos(u ) t v (sin v ) (sin t ) t setting v = sin t t cos(u ) cos(v ) cos t cos(sin(sin t)) cos(sin t ) cos t = cos(sin(sin )) cos(sin ) cos t = cos(sin ) cos = cos() = (C) 5) A efinite integral has a efinition in terms of the so-calle Riemann sum, epresse as a b f () n n i= f ( i * ) Δ n n [ ] b a f (a + i b a n ) n, i= in which we have employe the right-enpoint rule for the n subintervals If we put this epression into corresponence with the sum specifie in the problem, n n n i= (i [ ] n ) 7, we can see that b a = an that the function being use here is f() = 7 Since there is no constant term ae to the i/n in the argument of the function, we can surmise that a =, an therefore b = The efinite integral is thus 7 We can immeiately evaluate this to fin 7 = 8 8 = 8 ( ) = 8 (D) 6) This problem is nearly the same as Problem from the Spring 5 final eam, so a iscussion for most of the solution can be foun in that solution set As for the one revise choice, (), since the slope of the secant line connecting the enpoints of the curve for the function f() on the interval [ -, ], is equal to 4, there must certainly be (a great many) places on the curve where the slope is positive ( f () > ) The other choices, (a), (b), an (c) are also true; consequently, the only false choice here is (E) 7) From the given equation for the volume of a sphere, V = 4 π r, we can establish a relation between the rate of raius change an the rate of volume change for the (spherical) balloon by ifferentiating the volume implicitly with respect to time: t V = t ( 4 π r ) = 4 π t (r ) = 4 / π ( / r r t ) = 4π r r t Since we are tol the rate of volume change (how fast air is being blown into the balloon), we can solve this equation for the rate of raius change to fin: (continue)

4 r t = V t 4π r = in /min 4π r In this problem, however, we are not given the raius at the current moment, but rather the volume of the balloon If we apply the equation for the volume, we can etermine that V = 6 / π in = 4 / π r r = 6 in = 7 in r = in 4 This leas us to conclue that the rate of raius change is r t = V t 4π r = in /min 4π ( in) = 4π 9 in/min = π in/min (E) 8) To etermine the intervals of concavity for a function, we will nee to eamine its secon erivative: f () = f ' () = 5 f '' () = We are interesting in learning where the curve of this function is concave ownwar, which is to say where f () < : f '' () = = 6 ( ) < Since 6 is positive, this conition is only possible if < < < (E) 9) To estimate the value of a function by the use of linear approimation, it is necessary to know the eact value of the function at a point as close as possible to the point at which we wish to make the estimate Since we wish to estimate 77, for the function in this problem, f () = = /, the nearest value that will serve the purpose for making the estimation is = a = 8 We will also nee the first erivative of the function, which is f ' () = / = ( ) The estimate from the linear approimation is then f () f (a) + f ' (a) ( a) f (77) f (8) + f ' (8) (77 8) = 8 + = + ( ) = () ( 8) (77 8) = 5 = 975 (C) A more precise value from a calculator is , so the linear approimation is quite goo; this is because the slope of the cube-root function is not changing particularly quickly near = 8

5 ) We are aske to work out the maimization of the area of a rectangle that can be place within a right isosceles ( ) triangle having legs of length In orer for calculus to be useful here, we nee to formulate this problem in terms of functions of a variable We can arrange the triangle so that the two legs are places along the - an y-aes; this places two of the vertices at (, ) an (, ) The hypotenuse then connects these points an thus lies along the line y = (The slope of the line containing these points is m = =, an the y-intercept is at (, ) ) The rectangle woul also have two of its sies along the coorinate aes; if we place one verte (, y ) of the rectangle on the line we ve etermine, the other vertices are at (, ), (, ), an (, y ) The area of the rectangle we wish to maimize is thus A = y = ( ) = If we set the erivative of this area function equal to zero, we fin that A = = = =, y = = The rectangle of greatest area that fits insie the isosceles triangle is thus a square of sie, with area A = y = We can also quickly verify that this is the maimum area, since (B) A = < ) If we attempt to apply the Limit Laws irectly, we fin that this it gives the ineterminate prouct + ln = " ( )" In orer to become able to apply L Hôpital s Rule, we must moify a prouct of two functions, f() g(), into either of the ineterminate ratios, f () g() or g() f () (continue)

6 When ealing with an ineterminate prouct involving a logarithmic factor, it is almost always avisable to put that factor into the numerator, as the require ifferentiation of /(ln ) in the enominator will generally make the situation worse for evaluating the it So saying, we can now apply L Hôpital s Rule to obtain ln + + ln = " " + / / + + = (A) ) This integral is not one for which we have a technique of solving it; it is simply necessary to recognize the integran as the erivative of the arctangent function We thus have ( ) + = tan = tan tan = π 4 = π 4 (E) ) While it is true that sin oes not eist, it is possible that the it of a + prouct f () sin coul equal zero if f () =, as woul be foun by + + application of the Squeeze Theorem Here, we will nee to separately investigate the it, which gives us an ineterminate ratio: + ln + ln = " " + ln = + ' (ln )' + / = + + Hence, this factor grows without it, so the prouct as increases ln sin grows without it (D) 4) The integran in this efinite integral is not one for which we know an antierivative function, so we will nee to use another approach The only other metho we ve learne at this stage is the u-substitution metho, but there oesn t seem to be anything to base such a substitution upon Something to consier when working with the trigonometric functions is to see whether the integran can be re-written in terms of factors of sin an cos ; the new form of the integran may then suggest a substitution that will be useful In our integral, we will get π / 4 tan = π / 4 sin cos u = cos u = sin (continue)

7 ( )/ u ( )/ = ( ln u ) u : π/4 u = cos : ( )/ = (ln u ) ( )/ So the value of the integral is not ln = (ln ) (ln ) = (ln ) = ln = ln or ln /, but rather ½ ln 5) This is an eample of the sort of trigonometric it problems presente in Section We can work out many of those very easily now that we are familiar with L Hôpital s Rule, but that metho woul be very messy to apply here, since the numerator of the rational function is a three-term prouct which woul have to be ifferentiate (more than once!) Instea, we will break up the quotient into three factors first: (sin ) (sin 4) (sin 6) sin sin 4 sin 6 We coul now use the it law that was evelope in that Section, which can be etene to any function u() which is continuous at =, thus u sinu u sin (A) =, = The ratios nee to have the same argument for the sine function as the term appearing in the enominator in orer to apply this it law, which can be easily arrange: sin sin sin = sin 4 sin 6 sin sin 4 sin sin sin sin 6 6 sin 6 = 4 6 = 48 (D) 6) We are aske to fin the erivative of the function f() =, using the it efinition of ifferentiation It will be easier to work this out using the h-efinition of erivative: f ' () h f ( + h) f () h h ( + h) h h ( + h + h + h ) h multiply out the three factors or use the Binomial Theorem h h + h + h h h + h + h = + + =

8 7) The function f() = ( ) ( ) = + is a polynomial with the inicate -intercepts at =,, an an large- tails etening to negative or positive infinity beyon < or >, respectively So the regions boune by this curve an the -ais must lie in the intervals (, ) an (, ) Without actually rawing a graph, we can also fin that f() is negative if either one or all three factors are negative, which can only happen for the intervals < < or <, respectively So the total area for the boune regions must be foun from the sum of two integrals: A = f () = ( ) = ( ) = ( ) + ( ) ( ) + ( ) ( ) ( ) = 4 / + / / 4 / + 4 / + / = 8) The average value of a function f() on an interval [ a, b ] is calculate from the b a f () formula f ave = we fin f ave = 4 b a ( + ) = 4 4 = + / 4 / + For the function f () = ( + ) over [, 4 ], 4 4 ( + + ) = / + 4 ( + + ) = = = 4 68 = 7

9 9) To etermine the equation for the tangent line to a point (, f( ) ) on the curve for the function y = f(), we nee to fin the first erivative at that value of, f ( ) For the given function, f () =, this will require logarithmic ifferentiation: ln y = ln (ln y ) = ( ln ) y y = ( ) ln + (ln ) = ln + y = y ( ln + ) = ( ln + ) The slope of the tangent line at = is then m = y = = ( ln + ) = ( + ) = Since the value of the function at = is y() = =, the point-slope form of the equation for the tangent line is ( y ) = ( ), which can be reuce to just y = ) This problem is essentially the same as Problem 8 in the Spring 5 final eam A iscussion of the calculation can be foun in that solution set (in which you are also irecte to Problem 6 of the Spring final eam solutions) ) We are aske to investigate the function y = f () = e i) To fin where the etrema of this function an its intervals of increase an ecrease, we must calculate its first erivative, which is f ' () = ( e ) = () e + = e + e = e ( 4 ) (e ) ( ) = e + e ( 4) Since the eponential factor is always positive, the erivative function can only equal zero where 4 = = ¼ = ±½ ; these are the locations of the etrema of f() The function is increasing where 4 > > 4 < ½ or ½ < < ½ ; thus, the function is ecreasing for 4 < > ½ This tells us then that the interval of increase for f() is ( ½, ½ ), the intervals of ecrease are (, ½ ) an ( ½, ), a local minimum occurs at = ½, an a local maimum occurs at = ½

10 ii) To stuy the concavity of the function, we will nee its secon erivative: f '' () = [ e ( 4 )] = (e ) ( 4 ) + e = e ( 4 ) ( 4 ) + e ( 8 ) ( 4 ) = e ( ) = e (6 ) Again, the eponential factor can only be positive, so the secon erivative is zero for 6 = 4 (4 ) = ; this requires that either = or = ¾ = ± The first erivative is not zero at any of these values, so these all represent points of inflection in the curve for the function These points ivie the real numbers into four intervals: (, ), (, ), (, ), an (, ) We coul simply test values of in each interval to fin the sign of the secon erivative there or solve the inequalities 4 (4 ) > an 4 (4 ) < irectly (which involves some little work) We conclue that the intervals of upwar concavity are (, ) an (, ), while those of ownwar concavity are (, ) an (, ) iii) The it at infinity cannot be etermine through use of the Limit Laws for this function, as this gives us + e = " e = ", which is an ineterminate prouct We can bring L Hôpital s Rule to bear on the problem by first rewriting the prouct as an ineterminate ratio: + e = " + e " + ' ( e ) ' + e 4 = " " = (In the other irection, we fin basically the same result: + e = " " = ) Thus, y = is the horizontal asymptote for our function

11 iv) Because the eponential factor e is always positive, the sign of f() is entirely etermine by the sign of Thus, f() > for > an f() > for > Note that this also means that f() is an o function Knowing this woul allow us to fin the properties of the function just for > an then just reverse them appropriately for < For instance, by fining that = ½ is a local maimum, that (, ½ ) is an interval of increase, an that ( ½, ) is an interval of ecrease, we learn immeiately that = ½ is a local minimum, that ( ½, ) is an interval of increase, an that (, ½ ) is an interval of ecrease This o symmetry can also be applie to the intervals of concavity, locations of inflection points, the it at negative infinity, an so forth v) A graph of this function is presente in the Answer Key G Ruffa 6/9