UBC-SFU-UVic-UNBC Calculus Exam Solutions 7 June 2007

Transcription

1 This eamination has 15 pages including this cover. UBC-SFU-UVic-UNBC Calculus Eam Solutions 7 June 007 Name: School: Signature: Candidate Number: Rules and Instructions 1. Show all your work! Full marks are given only when the answer is correct, and is supported with a written derivation that is orderly, logical, and complete. Part marks are available in every question.. Calculators are optional, not required. Correct answers that are calculator ready, like 3 + ln 7 or e, are fully acceptable. 3. Any calculator acceptable for the Provincial Eamination in Principles of Mathematics 1 may be used. 4. A basic formula sheet has been provided. No other notes, books, or aids are allowed. In particular, all calculator memories must be empty when the eam begins. 5. If you need more space to solve a problem on page n, work on the back of page n CAUTION - Candidates guilty of any of the following or similar practices shall be dismissed from the eamination immediately and assigned a grade of 0: a) Using any books, papers or memoranda. b) Speaking or communicating with other candidates. c) Eposing written papers to the view of other candidates. 7. Do not write in the grade bo shown to the right Total 100

2 UBC-SFU-UVic-UNBC Calculus Eam 007 Name: Page 3 of 15 [4] 1. Find an equation for the line tangent to the following curve at the point where = 1: y = 6 / +. At the point where = 1, one has y =. By the quotient rule y = [ 6 + / ] + ) 6 /) [ + 1 1/] + ). Substituting = 1 gives the slope of the desired line: m = y ) = [6 + ] ) 6 ) [ ] + 1 = 3 4. Using the point 1, ) and the slope 3/, we have the line s equation: y = + 3 1) = [6]. Find an equation for the line tangent to the following curve at the point, 0): y = + siny). Taking the derivative of both sides with respect to gives dy = cosy) d [ y + dy d Plugging in, y) =, 0) and writing m for the corresponding value of dy/d, we get m = 1 4 [0 + m], i.e., 3m = 3, i.e., m = 1. The desired tangent line has equation ]. y = 0 ), i.e., y = +. Continued on page 4

3 UBC-SFU-UVic-UNBC Calculus Eam 007 Name: Page 4 of 15 [6] 3. Find the derivatives of the three functions below. Do not simplify. a) = tan + e ) a ) = sec + e )[1 e ] = 1 e cos + e ). b) = e tan 1 3 ) b ) = [e ] tan 1 3 ) [ + e ) ]. c) = sin 1 ) sin 1 ) c ) = 1 [ sin 1 ) ] [ sin 1 ) ] sin 1 ) ). Continued on page 5

4 UBC-SFU-UVic-UNBC Calculus Eam 007 Name: Page 5 of 15 [6] 4. For each limit below, find the eact value with justification) or eplain why the limit does not eist. a) + ) 4. lim When < 0, = =. Thus A = lim + ) ) ) = lim 4 ) 1 4/ = lim = lim / [ ) 4] 4 ) = 4 =. b) 1 lim ) ) B = lim ) 6 ) ) + 3) 6 1 = lim 9 3 3) + 3) = lim = [6] 5. Use the definition of the derivative as a limit to find f ), given f) = 1 +. Finding f ) using differentiation rules will earn no marks, but it could help you check your work with limits.) { f f + h) f) 1 + h ) = lim = lim h 0 h h 0 h h) } 1 + { ) [ ] } 1 + h h = lim h 0 h h h 1 + { 1 + h + + h } h = lim h 0 h [1 + + h]1 + ) { } 1 h 1 = lim = h 0 h [1 + + h]1 + ) 1 + ). Continued on page 6

5 UBC-SFU-UVic-UNBC Calculus Eam 007 Name: Page 6 of 15 [6] 6. Let ft) = ert e rt e rt where r > 0 is a constant. + e rt a) Derive the small-t approimation ft) rt for t 0. b) a) Eplain why the approimate value rt is greater than the eact value ft) whenever t > 0. What happens when t < 0? The basic idea of differential calculus is the tangent-line approimation: ft) f0) + f 0)t for t 0. ) For the given function f, we have f0) = 0 and by the quotient rule) f t) = [rert + re rt ]e rt + e rt ) e rt e rt )[re rt re rt ] e rt + e rt ). b) Plugging in t = 0 reveals f 0) = r, so the desired statement is an instance of ). Epanding the numerator above allows massive cancellation: f t) = r ert + e rt ) e rt e rt ) e rt + e rt ) e rt + + e rt) e rt + e rt) = r This makes it rather easy to find = 4re rt + e rt ). e rt + e rt ) f t) = 8re rt + e rt ) 3 [ re rt re rt] = 8r [ e rt e rt e rt + e rt ) 3 When r > 0 and t > 0, we have rt > 0 so e rt > 1 > e rt. Thus the bracketed ratio above is positive valued, and we have f t) < 0 whenever t > 0. Thus f is concave down on the interval [0, + ), and therefore the eact value ft) is less than its tangentline approimation throughout this interval. Similarly, f is concave up on the interval, 0], so its eact values are larger than the values predicted by its tangent lines in that interval. ]. Continued on page 7

6 UBC-SFU-UVic-UNBC Calculus Eam 007 Name: Page 7 of 15 [8] 7. A conical reservoir with an open top and verte down holds water for a desert community. The reservoir is 6 metres deep at its centre; its diameter at the top is 10 metres. The rate at which water evaporates from the reservoir is proportional to the area of the water s top surface a circular disk). When the water is 5 metres deep, its depth is decreasing at the instantaneous rate of cm per day. Find, with suitable units,... a) b) the rate of change of depth when there is 3 metres of water in the reservoir, and the rate of change of volume when there is 3 metres of water in the reservoir. Draw a picture please!) and name some unknowns: h r A V H R = depth of water in the reservoir, metres, = radius of the water s top surface, metres, = area of the water s top surface, square metres, = volume of water in the reservoir, cubic metres, = total depth of reservoir as built, metres, = radius of reservoir s top surface as built, metres. Famous geometric formulas, and an argument with similar triangles, give three useful facts: 1) V = 1 3 πr h, ) A = πr, 3) r h = R H = 5 6. Since the radius is not mentioned anywhere in the question, we use 3) to eliminate r: r = 5 6 h = 1 ) V = 1 ) 5π h 3, ) A = 5π h. a) b) Evaporation causes a loss of volume. The proportionality statement means that, for some constant c, ) dv 5π dt = ca, i.e., h dh 36 dt = c5π 36 h. ) This identity is valid for all times: cancellation gives dh dt = c. That is, the water level in the reservoir decreases as a constant rate. This rate is cm/day no matter what the depth is, so this value applies in particular when the depth is 3 metres. Convert c to 0.0 m/day and use it in line ) above: dv dt = ca = 100 m day ) 5π 36 3m) ) = π 8 m 3 day. At the given instant, the volume is decreasing at π/8 m 3 per day. Continued on page 8

7 UBC-SFU-UVic-UNBC Calculus Eam 007 Name: Page 8 of 15 [7] 8. The position of a moving particle at time t is = st), where a) st) = ln1 + t) t 1 + t. Epress the particle s velocity as a function of t. Simplify your answer. b) Eplain why st) > 0 whenever t > 0. c) Find the particle s maimum velocity, and the time when it occurs. a) The particle s velocity at time t is the rate of change of its position: vt) = ds dt = t) t 1 + t 1 + t) = t t) = 1 + t 1 + t) t) = t 1 + t). b) c) Clearly s t) = vt) > 0 whenever t > 0, so function s is increasing on the interval [0, + ). Its minimum value on that interval must occur at the left endpoint, where s0) = ln1) = 0. Thus st) > s0) = 0 for all t > 0. The instant when velocity is maimized must obey v t) = 0. So we calculate v t) = 1 + t) t 1 + t) 1 + t) t 1 + t) 4 = 1 + t) 3 = 1 t 1 + t) 3. The critical point t = 1 provides an absolute maimum for the function v, because v t) > 0 when 1 < t < 1 and v t) < 0 when t > 1. The particle s maimum velocity is v ma = v1) = 1 4. Points where t 1 can be ignored because they lie outside the domain of s.) Continued on page 9

8 UBC-SFU-UVic-UNBC Calculus Eam 007 Name: Page 9 of 15 [5] 9. Let f) = ). By using your findings from the previous question, or otherwise,... a) Show that f is increasing on the interval > 0. b) Find the eact value of lim f). a) One effective way to show that f is increasing on an interval is to prove that f ) > 0 there. Here lnf)) = ln ), so differentiation gives f ) f) = ln ) ) 3/ / = ln ) ) 3/ / b) In terms of the function s from question 8, f ) = f)s3/). Clearly f) > 0 whenever > 0, and we showed earlier that st) > 0 whenever t > 0. These facts together imply f ) > 0 whenever > 0, so function f is increasing on the interval 0, + ). The idea behind logarithmic differentiation works here, too. Recognizing the definition of derivative completes the calculation: ln1 + 3/) lim lnf)) = lim 1/ = 3 lim h 0 ln1 + h) ln1) h/3 let h = 3/) = 3 d d =1 ln) [ ] 1 = 3 = 3. =1 This shows that lnf)) 3 as. Hence f) e 3 as. Continued on page 10

9 UBC-SFU-UVic-UNBC Calculus Eam 007 Name: Page 10 of 15 [8] 10. A cup of coffee at 96 C is set on a table in an air-conditioned classroom. It cools to 60 C in 10 minutes, and then to 40 C in another 10 minutes. What is the temperature of the room? Let A denote the ambient temperature we seek. To find A, let ut) be the coffee temperature at time t, and recall Newton s law of cooling: the temperature difference yt) def = ut) A obeys dy dt = kyt) for some constant k. It follows that, for some constant C, ut) A = yt) = Ce kt. The problem statement gives relevant information for three different t-values: 1) C = u0) A = 96 A; ) Ce 10k = u10) A = 60 A; 3) Ce 0k = u0) A = 40 A. Using 1) to eliminate C reduces ) 3) to the pair of equations Recognizing e 0k = e 10k ) then gives 40 A 96 A = 96 A)e 10k = 60 A, 96 A)e 0k = 40 A. ) 60 A, i.e., 40 A)96 A) = 60 A). 96 A Epanding this equation and simplifying the result leads to A + A = A + A, i.e., 40 = 16A i.e., 15 = A. The ambient temerature is 15 C. Continued on page 11

10 UBC-SFU-UVic-UNBC Calculus Eam 007 Name: Page 11 of 15 [10] 11. Let f) = 4. i) Find the domain of f. ii) Find the intervals in which f is increasing and decreasing. iii) Find the absolute maimum and minimum values for f on its domain, and all the points where these are attained. iv) Find the -coordinates of all inflection points for f. v) Sketch the graph of f. A pair of aes is supplied on the net page.) i) ii) The value f) is defined if and only if 4 0, i.e., 4. So the domain of f is the closed interval [ 4, 4]. The product rule gives f ) = 4 + = ) 34 )4 + ) =. 4 = 4 ) Thus f ) > 0 iff 4 < < 4 or 0 < < 4. Since f is continuous at the endpoints of these intervals, f is increasing on the intervals [ 4, 4] and [0, 4]. Likewise, f ) < 0 iff 4 < < 0 or 4 < < 4. endpoints of these intervals, f is decreasing on the intervals [ 4, 0] and [4, 4]. Since f is continuous at the iii) Absolute etrema may happen only at critical points CP), singular points SP), or endpoints of the domain EP). Here there are no SP s, three CP s = 0, ±4), and two EP s = ± 4). In this case, each one gives an absolute etremum of some kind: One has ) iv) Using the quotient rule, we find abs min: f min = 0, attained at = 0, ± 4, abs ma: f ma = 3, attained at = ±4. f ) = 3 d 16 3 [16 3 d = 3 ] 4 [16 3 ] 4 4 [16 3 ]4 ) + [16 3 ) ] = 3 4 ) 3/ = ) 4 ) 3/ This has zeros at points where 4 ) = 36 ± ) = 18 ± 13 = 18 ± 33. ) Continued on page 1

11 UBC-SFU-UVic-UNBC Calculus Eam 007 Name: Page 1 of 15 Notice that > = 8 > 4 = 4), so choosing the + sign in ) generates -values outside the domain of f. It follows that there are just two inflection points, and their -coordinates are = ± ± v) The corresponding y-coordinates are about 7.8.) In the sketch below, a green dot highlights each local etremum, and a red dot is shown at each inflection point. 50 y = sqrt4 ) y Continued on page 13

12 UBC-SFU-UVic-UNBC Calculus Eam 007 Name: Page 13 of 15 [6] 1. The nonlinear equation 1 sin) = cos) has a solution near the point 0 = 3π/. Use the tangent lines at = 0 to the two curves y = 1 sin), y = cos) to find a better approimation call it 1 ) to the solution near 0. Let f) = sin. The quotient rule gives f ) = cos ) sin, so f 0 ) = f 3π/) = 0 1) 3π/) = 4 9π. The line tangent to the curve y = f) at 0, f 0 )) is y = f 0 ) + f 0 )[ 0 ] = 1 3π/ + 4 [ 3π/] = 9π 3π + 4 9π [ 3π ]. 1) Net, let g) = cos. Then g ) = sin, so the line tangent to the curve y = g) at 0, g 0 )) is y = g 0 ) + g 0 )[ 0 ] = 0 + 1[ 3π/)] = 3π. ) The point of intersection for the tangent lines in 1) and ) above can be found by using ) to eliminate y from 1): 3π = 3π + 4 [ 9π 3π ] [ = 3π ] = /3π)) 4 9π 1 = 6π 4 9π. This gives the improved estimate 1 = π4 7π ) 4 9π ). The following sketch not required) shows the two given curves with their tangent lines at = 3π/ and the point 1, y 1 ) , 0.1) where the tangent lines meet. The curves meet when , accurate to si digits.) 1 y=cos), y=sin)/ 0.5 y Continued on page 14

13 UBC-SFU-UVic-UNBC Calculus Eam 007 Name: Page 14 of 15 [8] 13. Every tangent line of negative slope for the curve y = 3 can be used to construct a rectangle, as shown below: put one corner at the origin O), one at the line s -intercept A), one at the line s y-intercept B), and one in the first quadrant to complete the figure P ). Find the coordinates of P for the rectangle of smallest perimeter that can be constructed this way. The sketch shows the construction, but not the minimizing configuration.) B P O A For y = 3 we have y = 3. Every tangent line of negative slope has a point of tangency obeying > 3/. Suppose the point of tangency is = z. Then the equation of the tangent line is y = 3z z ) + 3 z)[ z] = 3 z) + z. This line s y-intercept set = 0) is B = z. Its -intercept set y = 0) is A = z /z 3). So the perimeter of the constructed rectangle when the point of tangency is = z is To minimize this, calculate fz) = A + B = 1 f z) = zz 3) z ) z 3) z = z z z 3 + z, z > 3. [ ] z 3 z 3) zz )4z 3) z 3) z 3) = = z 3). There is only one critical point in the interval of interest, namely, z =. Clearly f z) < 0 for 3/ < z < and f z) > 0 for z >, so the point z = gives an absolute minimum for f over the interval 3/, + ). The corresponding intercepts give the coordinates of the minimizing point P : P = A, B), where A = z z 3 = 4, B = z = 4. z= z= Continued on page 15

14 UBC-SFU-UVic-UNBC Calculus Eam 007 Name: Page 15 of 15 [8] 14. Use the three properties below to identify the function f and sketch its graph: i) f0) = 0, and ii) the graph of f has an inflection point at which the tangent line is horizontal, and iii) f ) = 6 + for all. Fact iii) gives f ) = C for some C. An inflection point for f must obey f ) = 0, i.e., = 1/3. Fact ii) requires f ) = 0 at this point, so C = 1/3. Detail: ) 1 0 = f 1/3) = ) + C = C Thus f ) = /3. This implies y y = /3 1/3, 1/7) 0,0) f) = K 0.5 for some K. Fact i) requires 0 = f0) = K, so f) = [6] 15. Make a rough sketch of the following figures on the same set of aes: C : y = , L : y = 5. Then find the area of the region lying below C and above L. The curve C and the line L meet above the points where = 5, i.e., 1 + =, i.e., = ±1. Between these points, the curve C lies above the line L. The desired area is 1 ) [ ] 1 10 A = d = 10 tan 1 5 = 10 tan 1 1) 5 ) = 5π 10. = 1 = 1 10 y = 101+ ) 1, y=5 y The End