MATH 205 Practice Final Exam Name:

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1 MATH 205 Practice Final Eam Name:. (2 points) Consier the function g() = e. (a) (5 points) Ientify the zeroes, vertical asymptotes, an long-term behavior on both sies of this function. Clearly label which is which, an if any features are not present, say so. We note that in orer for e to be zero, its numerator e must be zero, but e is always positive, so g() has no zeroes. Vertical asymptotes occur when the enominator is zero; this occurs when =. To calculate long-term behavior, we consier the behavior on the right, where a form calls for an invocation of L Hôpital s rule: lim + e = lim e + = + while on the left, e 0, so we have the form 0, which is not ineterminate: lim g() = 0. (b) (5 points) Ientify the critical points of this function, an inicate whether each is a local maimum, local minimum, or neither. We calculate g (): e = ( )e e = 2 ( ) 2 ( ) 2 e Now we fin places where g () is zero or noneistent; the numerator is zero when = 2, an the enominator is zero, inucing noneistence, when =. The latter is neither a local minimum nor a maimum, since it s not even in the omain of g. The former, on the 2 other han, will be a minimum: e transitions from negative when < 2 to positive ( ) 2 when > 2. (c) (2 points) Which if any of the critical points ientifie above are global maima or global minima? Show work or eplain. None of them are global maima, since g() gets arbitrarily large off to the right, an = 2 will not in fact be a global minimum, either, since lim g() < g(2). 2. (20 points) We have a rectangular sheet of carboar which is feet by 8 feet in imensions. By cutting ientical squares out of the corners an foling up the resulting sheet, we can buil a bo without a li. What size on our corner cut maimizes the volume of the bo so generate? 8 Let the epth of the cut be (along the ashe lines epicte above). Then, foling this up gives a bo whose base is of imensions (8 2) ( 2) an whose height is. The total volume is thus V () = (8 2)( 2) = ; we seek to maimize this volume over the interval (0,.5). must clearly be positive; it coul not be.5 or more because then the base imension 2 woul be zero or negative. Now, we fin the critical points of V () Page of 6 December 4, 20

2 MATH 205 Practice Final Eam Name: by calculating V () = = 4( 2 +6) = 4( 2)( ). This is zero when = 2 or = ; the latter is outsie our omain, so our three caniates for area-maimizing behavior are the limiting behavior as approaches 0 from above, the limiting behavior as approaches.5 from below, an = 2. As might be epecte, the first two are terrible: lim V () = V (0) = lim V () = V (.5) = 0.5 so the thir choice is the best (specifically, V ( 2 ) = = ).. (6 points) Calculate the following erivatives: e t (a) (4 points) Fin. t In general, f() = f(), so in this particular eample, e t t = et t. (b) (6 points) Fin arctan The epression being ifferentiate here is the arctangent of a quotient; we thus might reasonably epect to use the chain rule an the quotient rule. We might let u = 2, an +2 then invoke the chain rule: arctan = arctan u = u u arctan u = u arctan u = ( + )(2) (2 )() ( + 2) 2 + u 2 = ( + )(2) (2 )() ( + 2) 2 ( + ( ) 2 2) +2 (c) (6 points) Given g(s) = e s cot(s 2 ), fin g (s). The epression being ifferentiate here is a prouct, one of whose factors is a composition, so we might epect to use the prouct rule an then the chain rule. Defining u = s 2 preemptively, we start with the prouct rule, an introuce u as necessary: g (s) = s ( e s cot(s 2 ) ) = e s cot(s 2 ) + e s s cot(s2 ) = e s cot(s 2 ) + e s s cot(u) = e s cot(s 2 ) + e s u s u cot(u) = e s cot(s 2 ) + e s 2s( csc 2 u) = e s cot(s 2 ) 2se s csc 2 (s 2 ) Page 2 of 6 December 4, 20

3 MATH 205 Practice Final Eam Name: 4. (5 points) The keratoi cusp is a curve satisfying the equation y 2 = 2 y + 5. (a) (0 points) Fin a formula for y on this curve in terms of an y. We ifferentiate both sies with respect to, an use the prouct an chain rule as appropriate: ( ) y 2 = ( 2 y + 5) y y y2 = 2y + 2 y y y y = 2y (2y 2 ) y = 2y + 54 y = 2y y 2 (b) (5 points) Fin the equation of the tangent line to the keratoi cusp at the point (2, 4). Using the result above at this point, we fin that at (2, 4), the slope of the tangent line is y = 2 2( 4)+5 24 = 64 = 6. 2( 4) The equation of the tangent line is thus, in point-slope form, (y + 4) = 6 ( 2). 5. (8 points) Evaluate the following integrals: (a) (6 points) 2 (2 4). Using the Funamental Theorem of Calculus an the antierivative power rule: 2 ( 2 4) = 4] 2 = 8 ( ) = 9 (b) (6 points) ( 2 +) 2 This is not a simple antierivative, so we consier substitution possibilities. Since eponentiation is being applie to the complicate epression 2 +, it is promising to consier u = 2 +, which yiels the pseuo-equation u = 2. Then, translating the above integral: ( 2 + ) = ] 5 u = = u2 u 5 ( ) = 4 5 (c) (6 points) (t 5 t) t 6 t 2. This is not a simple antierivative, so we consier substitution possibilities. Since the square rooteponentiation is being applie to the complicate epression t 6 t 2, it is promising to consier u = t 6 t 2, which yiels the pseuo-equation u = (6t 5 6t) ; note that then (t 5 t) can be replace with u, so we translate the above integral: 6 (t 5 t) t 6 t 2 = u /2 uu = 6 6 /2 + C = (t6 t 2 ) /2 + C 9 Page of 6 December 4, 20

4 MATH 205 Practice Final Eam Name: 6. (6 points) A sentry at Blackgate Prison has turne a spotlight on an escapee who is currently 0. miles to the north an 0.4 miles to the east of the prison. She notices that the escapee is traveling eastwars at three miles per hour. (a) (8 points) How quickly will she nee to rotate the spotlight to keep it traine on the escapee? Let the angle of the spotlight from true north be calle θ, an let the istance the fugitive is to the east of the prison be calle. Since the fugitive is a constant istance 0. to the north of the prison, an a istance easy of the prison, rawing a right triangle makes it clear that tan θ =. 0. We know that is currently 0.4, an that =, so we may use relate-rates techniques to etermine θ, ifferentiating each sie of the above relationship with respect to t: tan θ = θ 0. θ tan θ = θ 0. sec2 θ θ = = cos2 θ an since θ is in a right triangle ajacent to a sie of length 0. an with hypotenuse 0.5, we may specifically epan that to θ = cos2 θ = ( ) = =.6 in raians per hour. (b) (8 points) How quickly is the escapee s istance from the prison changing? ( ) = (s2 ) (2 ) = s s (s2 ) s (2) = (2s) s = s An since an s have current values of 0.4 an 0.5 respectively, s = = (5 points) Determine the following limits. (a) (5 points) Evaluate lim +6 2 or emonstrate that it cannot be evaluate. 2 2 Direct evaluation gives 0, so we might use L Hôpital s rule: 0 lim ( + 6) 2/ = (8 2/ ) = 2 (b) (5 points) Evaluate lim + ln or emonstrate that it cannot be evaluate. 2 5 You might simply look at the ominant term in the numerator an enominator to fin that lim + ln 2 5 times on ineterminate forms evaluating to : 2 + ln lim 2 5 = + ; alternatively, L Hôpital s rule can be use several 2 + ln = Page 4 of 6 December 4, 20

5 MATH 205 Practice Final Eam Name: (c) (5 points) Using the ifference quotient, fin the erivative with respect to of f() = You may not use L Hôpital s rule for this problem. f () 0 f( + h) f() h (( + h) 2 4( + h) + 2) ( )) 0 h 6h + h 2 4h 0 h 6 + h 4 = (5 points) Let f() = (a) (5 points) Where is f() increasing? Where is it ecreasing? Label which is which. We consier the sign-changes of f () = = 6( 2)( + ). This is clearly zero at = 2 an =. Probing between an among the zeroes inicates that when <, f () is positive, an when < < 2, f () is negative, an when > 2, f () is positive again. Thus f() is increasing on (, ) an (2, ), an is ecreasing on (, 2). (b) ( points) What are the critical points of f()? Is each a local maimum, a local minimum, or neither? Since = is a transition from increase to ecrease, it is a local maimum, an since = 2 is a transition from ecrease to increase, it is a local minimum. (c) (7 points) Determine where f() is concave up an where it is concave own, an ientify points of inflection. We consier the sign-changes of f () = This is zero at =, an probing to 2 the left an right of this zero inicates that f () < 0 when <, an f () > 0 when 2 >. Thus, f() is concave own on (, ) an concave up on (, ) (8 points) Answer the following questions about the function h() = (a) (4 points) What is the omain of h()? h() is efine as long as the argument of the square root is non-negative, so wherever , whihc is when 2 25, which is in the interval [ 5, 5]. (b) (4 points) Where oes the erivative of h() eist? Using the chain rule, g () = 2 2. The argument of the square root must eist, limiting 25 2 our options to the omain foun in part (a), but in aition it must be nonzero, so we must aitionally eclue values such that 25 2 = 0, which are = ±5. Thus, our interval of ifferentiability is ( 5, 5). 0. (5 points) Mirana has just taken a 40mg intravenous ose of G-2 pailon hyrochlorate. In two hours the level of the rug in her system, which is subject to eponential ecay, will have reuce to 25mg. Page 5 of 6 December 4, 20

6 MATH 205 Practice Final Eam Name: (a) (5 points) Construct a function moeling the quantity of the rug in her boy after t hours. We know that this is an eponential ecay system, so the function moeling the quantity of the rug is f(t) = Ce kt for some well-chosen C an k. We also know, base on the situation escribe, that f(0) = 40 an f(2) = 25. Thus, using the first fact: so C = 40. Now, using the secon fact: so f(t) = 40e ln(5/8) 2 t. 40 = f(0) = Ce 0k 25 = 40e 2k 5 8 = e2k ln 5 8 = 2k ln = k (b) (6 points) How quickly is the rug being eliminate after 2 hours? Using the chain rule, f (t) = 40 ln e ln(5/8) 2 t, so specifically f (2) = 20 ln 5 8 eln(5/8) This is approimately 5.87, so we may interpret this result to inicate that the rug is being eliminate at a rate of approimately 5.87 milligrams per hour. (c) (4 points) The eperimental protocol requires that subjects be kept uner careful observation until the level of the rug in their boy is below 0mg. After how many hours can Mirana be release from observation? The question here is asking when f(t) = 0. This can be solve algebraically: 40e ln(5/8) 2 k = 0 e ln(5/8) 2 k = 4 ln k = ln 4 k = 2 ln 4 ln 5 8 This is approimately 5.9, so she can be release after nearly 6 hours. Page 6 of 6 December 4, 20