Practice Problem Solutions: Exam 1

Transcription

1 Practice Problem Solutions: Exam 1 1. (a) Algebraic Solution: Te largest term in te numerator is 3x 2, wile te largest term in te denominator is 5x 2 3x Tus lim x 5x 2 2x 3x 2 x 5x 2 = 3 5 Numerical Solution: Plugging in 100, 1000, and 10,000 for x gives , , and , respectively. Tus, te limit is 0.6, or 3/5. (b) Algebraic Solution: Te largest term in te numerator is x 3, wile te largest term in te denominator is x 2 1 x 3. Tus lim x x 2 + 5x x 3 x x 2 x = x Numerical Solution: Plugging in 100, 1000, and 10,000 for x gives 95.2, 995.0, and , respectively. Tus te limit is. (c) Algebraic Solution: Since 3 x is exponential and x is only polynomial, te denominator is muc larger tan te numerator, so te limit will be 0. Numerical Solution: Plugging in 5, 10, and 15 for x gives , , and , respectively, so te limit is 0. (d) Algebraic Solution: lim (2 + ) = 12 ( ) Numerical Solution: Plugging in 0.1, 0.01, and 0.01 for x gives , , and , respectively, so te limit is 12. (e) Algebraic Solution: lim x 3 x 2 9 x 3 x 3 (x 3)(x + 3) x 3 x 3 x + 3 = 6 Numerical Solution: Plugging in 3.1, 3.01, and for x gives 6.1, 6.01, and 6.001, respectively, so te limit is 6. (f) Algebraic Solution: We multiply te numerator and denominator by te conjugate of te radical x 2 ( x 2)( x + 2) expression. lim x 4 x 4 x 4 (x 4)( x 4 x + 2) x 4 (x 4)( x + 2) x 4 1 x + 2 = 1 4 Numerical Solution: Plugging in 4.1, 4.01, and for x gives , , and , respectively, so te limit is 0.25, or 1/4. 2. (a) Te temperature decreased by 7.5 C in 200 m, wic is 3.75 C every 100 m. Terefore, te temperature at 600 m sould be (14.0 C) (3.75 C) = C

2 (b) As stated in part (a), te temperature is decreasing at a rate of (3.75 C)/(100 m), wic gives a slope of C/m. Using te point-slope formula, we get T (D 500) Tis can be simplified to T D Ceck: To ceck tis formula, we can plug in 300, 500, or 600 for D. Tis results in values of 21.5, 14.0, and for T, respectively. Terefore, tis formula agrees wit te given data, as well as our answer to part (a). (c) We set T equal to 16.0 and solve: D Te solution is D m. Ceck: We can ceck tis by plugging into out formula from part (c), wic yields a temperature of 16.0 C. Also, it makes sense tat te answer is between 300 m and 500 m, since 16.0 C is between 21.5 C and 14.0 C. 3. Using Pairs of Points: We find te slope between pairs of points close to x = 7. points slope We conclude tat f (7) 2. (7,5) and (7.1, ) (7,5) and (7.01, ) (7,5) and (7.001, ) (7,5) and (7.0001, ) Using Te Limit Definition of te Derivative: Tis is really te same ting. We know tat f (7) f (7 + ) f (7) f (7 + ) 5. Te given data tells us te value of f (7 + ) for = 0.1, 0.01, 0.001, and We can use tis to estimate te limit: f (7 + )

3 Based on tis table, te limit is Using Pairs of Points: We find te slope between pairs of points close to x = 1. points slope (1,3) and (1.01, ) (1,3) and (1.001, ) (1,3) and (1.0001, ) (1,3) and ( , ) It seems clear tat te limit is between 3.29 and Using Te Limit Definition of te Derivative: Again, tis is really te same ting. We ave f (1) f (1 + ) f (1) We can estimate tis limit using a table of values: Tus te limit is between 3.29 and We ave: lim (3 + ) 2 9 ( ) = 6 6. In pysics, te energy stored in a stretced spring is determined by te equation E = 1 2 k x2, were E is te energy, k is a constant (te spring constant), and x represents te distance tat te spring as been stretced. (a) Since 1 2 k is a constant, te formula is de = 1 ( 2 k 2x dx ), wic simplifies to de = k x dx

4 (b) Plugging in k = 0.20 Joules/cm 2, x = 10 cm, and dx = 1.5 cm/sec gives de = 3.0 Joules/sec Quick Ceck: We can ceck tis answer quickly by tinking about te next second. Rigt now, te energy is 1 2 k x2 = 1 2 (0.20)(10)2 = 10 Joules A second from now, te spring will be 11.5 cm long, so te energy will be 1 2 k x2 = 1 2 (0.20)(11.5)2 = Joules Tus, te energy will increase by Joules over te next second. Tis agrees fairly well wit our instantaneous rate of 3.0 Joules/sec. If we wanted a more accurate ceck, we would consider te energy increase over 0.1 sec or 0.01 sec instead. 7. (a) It looks like te population was about 5.5 million in 1950, and 27million in Terefore, te average annual population growt was (27 million) (5.5 million) = 21.5 million 55 years = 0.39 million/year Any answer between 0.38 million/year and 0.40 million/year would receive full credit. (b) Te best metod is to carefully draw a tangent line to te grap and ten estimate te slope of te tangent line using two faraway points. Te answer is about 0.23 million/year, and any answer between 0.17 million/year and 0.29 million/year would receive full credit. (c) Again, te best metod is to carefully draw a tangent line to te grap and ten estimate te slope of te tangent line using two faraway points. Te answer is about 0.92 million/year, and any answer between 0.80 million/year and 1.05 million/year would receive full credit. (d) Assuming te population is 27 million in 2005, and is increasing at a rate of 0.92 million/year, te population in 2015 sould be (27 million) + (10 years)(0.92 million/year) = 36.2 million Any answer between 35 million and 38 million would receive full credit. 8. (a) f (x) = 15x 2 6x + 3 Ceck: Plugging x = 2 into our formula gives f (2) = 51. We can confirm tis number using a slope estimate: f (2) f (2.01) f (2) = =

5 (b) Since y = 5x 1/2 + 3, we ave dy dx = 5 2 x 1/2 Ceck: Plugging x = 4 into our formula gives dy dx slope estimate: = We can confirm tis number using a ( ) ( ) = = (c) Tere are a few different ways to proceed. One is to cange te fraction into a product: u = (5x 3 x 1/2 )x 1. Using te product rule, we get du dx = (15x 2 12 ) x 1/2 x 1 (5x 3 x 1/2 )x 2 However, it works muc better to start by simplifying te fraction: Ten Eiter of te above answers is correct. u = 5x3 x 1/2 x = 5x 2 x 1/2 du dx = 10x x 3/2 Ceck: Again, we can ceck our derivative wit a slope estimate. According to our answer, du = 10.5 wen x = 1. Te slope between x = 1 and x = 1.01 is: dx (d) We ave g(t) = 5t 2 t + 3, so g (t) = 10t 3 1. = Ceck: According to our formula, g (2) = We can ceck tis wit a slope estimate: g (2) g(2.01) g(2) = = (e) Using te cain rule, dx = 3 ( t ) 2 ( (2t) = 6t t ) 2 Ceck: According to our formula, te derivative is 24 wen t = 1. We can ceck tis wit an estimate: =

6 (f) We ave y = (x 3 + x) 5. Using te cain rule, dy dx = 5(x3 + x) 6 (3x 2 + 1) = 5(3x2 + 1) (x 3 + x) 6 Ceck: According to our formula, te derivative is wen t = 1. We can ceck tis wit an estimate: = (a) We ave ds = 5 + 6t cm/sec (b) Setting te velocity to 35 cm/sec gives us te equation 5+6t = 35. Solving for t yields t = 5 sec. Ceck: We can ceck parts (a) and (b) by estimating te velocity of te ball at t = 5 sec. According to te formula we were given, te total distance rolled by te ball in te first 5 seconds is s = 5(5) + 3(5) 2 = 100 cm Similarly, te total distance rolled by te ball in te first 5.01 seconds is s = 5(5.01) + 3(5.01) 2 = 100 cm = cm Tus te ball rolls cm in te extra 0.01 sec, wic translates to cm/sec. Tis verifies our answer to part (b), wic in turn verifies our answer to part (a) 10. (a) Since y = x 1/3, we ave dy dx = 1 3 x 2/3. Plugging x = 8 into tis formula gives dy dx = 1. Tus te 12 tangent line as slope 1, and goes troug te point (8,2). We can use te point-slope formula to 12 get te equation: y = 2 1 (x 8) 12 Ceck: We can grap y = 3 x and y = 2 1 (x 8) on our calculator to verify tat te line is 12 tangent to te curve at te point (8,2). (b) Tis is a combination of te product rule and te cain rule. By te product rule, f (x) = ( ) ( ) derivative of x 3 (4x + 1) 7 + x 3 derivative of (4x + 1) 7 Te derivative of x 3 is 3x 2, and te derivative of (4x+1) 7 is 7(4x+1) 6 (4) (by te cain rule). Ten f (x) = 3x 2 (4x + 1) x 3 (4x + 1) 6

7 Ceck: According to our answer, f (0.25) = 52. We can ceck tis wit a slope estimate: f (0.25) f (0.251) f (0.25) = = Relatively speaking, our estimate is pretty close to te predicted slope, so our formula is probably correct. (c) Since y = x 2 (1 x 2 ) 1, we can use te product rule and te cain rule to differentiate. By te product rule, dy dx = ( ) ( ) derivative of x 2 (1 x 2 ) 1 + x 2 derivative of (1 x 2 ) 1 Te derivative of x 2 is 2x, and te derivative of (1 x 2 ) 1 is (1 x 2 ) 2 ( 2x), wic simplifies to 2x(1 x 2 ) 2. Ten Ceck: estimate: dy dx = 2x(1 x2 ) 1 + 2x 3 (1 x 2 ) 2 According to our answer, dy = wen x = 2. We can ceck tis wit a slope dx ( ) ( ) = 0.44 (d) Tis problem is a little bit different tan any oter problem we ave seen. We know tat f (x) = x 2 g(x), but we don t know wat te function g(x) is. Noneteless, we can take te derivative of f using te product rule: f (x) = 2x g(x) + x 2 g (x) Plugging in x = 2 gives: f (2) = 4g(2) + 4g (2) Since g(2) = 4 and g (2) = 3, we conclude tat f (2) = (a) Using te product rule, we get V dp + P dv = 0 (b) We are given tat V = 250 cm 3, P = 100 kpa, and dv = 50 cm 3 /min. Plugging tese into our formula from part (a) gives (250) dp + (100)( 50) = 0 Solving for dp gives dp = 20 kpa/min

8 Ceck: We know tat PV is constant. To start, we ave PV = (100)(250) = 25,000 One minute later, te volume will ave decreased to 200. Ten P(200) = 25,000 so P = 125. Ten P increases by 25 kpa in te first minute, wic seems reasonable for an initial rate of increase of 20 kpa/min. If we wanted a more accurate estimate, we could investigate te cange in pressure over 0.1 minutes or 0.01 minutes. 12. (a) We ave tat f (x) (x 2), so f (2.1) (b) Tis is a bit difficult, and is not very similar to any oter questions we ave considered. Te linear approximation assumes tat te function is te same as te tangent line. In reality, te tangent line will lie below te grap of te function, because of te way tat te grap curves. It follows tat te linear approximation is an underestimate.