California State University Northridge MATH 255A: Calculus for the Life Sciences I Midterm Exam 3

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1 California State University Northrige MATH 255A: Calculus for the Life Sciences I Mierm Exam 3 Due May Instructor: Jing Li Stuent Name: Signature: Do not write your stuent ID number on this front page. Please write your stuent ID number in the space provie on the secon page. Take your time to rea the entire paper before you begin to write, an rea each question carefully. Remember that certain questions are worth more points than others. Make a note of the questions that you feel confient you can o, an then o those first: you o not have to procee through the paper in the orer given. This exam is a take-home exam. The correct answer requires justification written legibly an logically: you must convince me that you know why your solution is correct. Answer these questions in the space provie. Use the backs of pages if necessary. Please recor how much time you nee to finish the test. MINS Where it is possible to check your work, o so. Goo Luck! 1

2 Stuent number: Problem Total Points Your Marks Question 1. [18 points] Fin any critical numbers for f an then use the secon erivative test to ecie whether the critical numbers lea to relative maxima or relative minima. If f (c) = 0 for a critical number c, then the secon erivative test gives no information. In this case, use the first erivative test instea. (a)[8 points]f (x) = 3x 3 3x

3 (b)[10 points] f (x) = (x + 3) 4 3

4 Question 2. [34 points] For the function f (x) = x 2 e x (a) [5 points] Fin the omain, an the x an y intercept where possible, an asymptotes where applicable; (b) [12 points] Fin the critical points, the intervals where f (x) increases or ecreases an the relative extrema of f (x); (c) [12 points] Fin the inflection points an the intervals of concavity/convexity; () [5 points] Using parts (a), (b) an (c) sketch the graph of f (x). 4

5 5

6 Question 3. [10 points] Sktech the graph of a single function what has all of the properties liste. a. Continuous for all real numbers b. f (x) > 0 on (, 2) an (0,3) c. f (x) < 0 on ( 2,0) an (3, ). f (x) < 0 on (,0) an (0,5) e. f (x) > 0 on (5, ) f. f ( 2) = f (3) = 0 g. f (0) oesn t exist h. Differentiable everywhere except at x = 0 i. An inflection point at (5,1) 6

7 Question 4. [20 points] Absolute Extrema (a)[10 points] Fin the locations of all absolute extrema for f (x) = (x + 1)(x + 2) 2 on [ 4,0] f (x) = (x + 1)(2)(x + 2)(1) + (x + 2) 2 (1) = 2(x + 1)(x + 2) + (x + 2) 2 = [2(x + 1) + x + 2](x + 2) [2(x + 1) + x + 2](x + 2) = 0 (3x + 4)(x + 2) = 0 x = 4 or x = 2. 3 x f(x) Absolute minimum Absolute maximum (b)[10 points] Fin the locations of all absolute extrema for f (x) if they exist, f (x) = x 4 4x 3 + 4x The critical numbers are 0, 1, an 2. f (x) = 4x 3 12x 2 + 8x = 4x(x 2 3x + 2) = 4x(x 2)(x 1) x f(x) There is no absolute maximum, as can be seen by looking at the graph of f. There is an absolute minimum at x = 0 an x = 2. 7

8 Question 5. [12 points] A fence must be built to enclose a rectangular area of 20,000 ft 2. Fencing material costs $3 per foot for the two sies facing north an south, an $6 per foot for the other two sies. Fin the cost of the least expensive fence. Let x = the length at $3 per foot y = the wih at $6 per foot Perimeter= 2x + 2y = 2x + 40,000 x Cost= 2x(3) + 40,000 x (6) = 6x + 240,000 x Minimize cost. x y = 20,000 y = 20,000 x C (x) = 6 240,000 x ,000 x 2 = 0 6 = 240,000 x 2 6x 2 = 240,000 x 2 = 40,000 x = 200 y = 20, = ft at $3 per foot will cost $1, ft at $6 per foot will cost $1,200. The entire cost will be $2,

9 Question 6. [16 points] Fin y by implicit ifferentiation for each of the following. (a)[4 points] 2 x y = 1 (b)[4 points] x 2 e y + y = x 3 (c)[4 points] y ln x + 2 = x 3/2 y 5/2 (2x1/2 y 1/2 ) = (1) x 1/2 1 y 2 y 1/2 1 y 2 y 1/2 y = 0 = x 1/2 = 2y 1/2 ( x 1/2 ) = 2y 1/2 x 1/2 (x2 e y + y) = (x3 ) (x2 e y ) + (y) = 3x2 2xe y + x 2 e y y + y x 2 e y y + y (x 2 e y + 1) y y = 3x 2 = 3x 2 2xe y = 3x 2 2xe y = 3x2 2xe y x 2 e y + 1 9

10 (y ln x + 2) = (x3/2 y 5/2 ) ln x y + y x + 0 = 3 2 x1/2 y 5/ x3/2 y 3/2 y ln x y 5 2 x3/2 y 3/2 y = 3 2 x1/2 y 5/2 y x y ( 5 ln x ()[4 points] tan y + x = 4 2 x3/2 y 3/2) = 3 2 x1/2 y 5/2 y x 3 y 2 = x1/2 y 5/2 y x ln x 5 2 x3/2 y 3/2 = 3x 3/2 y 5/2 2y x(2ln x 5x 3/2 y 3/2 ) [tan(y) + x] = (4) sec 2 (y) y + 1 = 0 sec 2 (y) y y = 1 1 = sec 2 (y) = cos 2 (y) 10

11 Question 7. [10 points] Many populations grow accoring to the equation = r x(n x) where r is a constant involving the rate of growth an N is the carrying capacity of the environment, beyon which the population ecreases. Show that the graph of x has an inflection point where x = N /2. = r x(n x) = r xn r x = 0 when x = 0, x = N, or x = N 2. 2 x 2 = r N 2r x = r [N 2x] = r [r x(n x)][n 2x] = r x 2 (N x)(n 2x) On (0, N 2 ), 2 x 2 > 0; therefore, the curve is concave upwar. On ( N 2, N ), 2 x 2 < 0; therefore, the curve is concave ownwar. Hence x = N 2 is a point of inflection. 11

12 Question 8. [4 points] The average aily metabolic rate for captive animals from weasels to elk can be expresse as a function of weight by m = 140.2w 0.75 where w is the weight of the animal (in Kg) an m is the metabolic rate (in kcal/ay). w (a)[2 points] Suppose that the weight of a weasel is changing with respect to time at a rate w. Fin. m 0.25 w = 140.2(0.75)w 0.25 w = w (b)[2 points] Determine m for a 250 kg elk that is gaining weight at a rate of 2 kg/ay. m = (250) 0.25 (2) The rate of change of the average aily metabolic rate is about kcal/ay 2. 12

13 Question 9. [10 points] The concentration of a certain rug in the bloostream x hr after being aministere is approximately C (x) = 5x 9 + x 2 Use the ifferential to approximate the changes in concentration for the following changes in x. C = 5x 9 + x 2 (a) [5 points] 1 to 1.5. x = 1, x = 0.5 C = 5(9 + x2 2x(5x)) (9 + x 2 ) 2 = x2 10x 2 (9 + x 2 ) 2 = 45 5x2 (9 + x 2 ) x2 (9 + x 2 ) 2 x 45 5(1)2 C (9 + 1) 2 (0.5) = (0.5) = 0.2 (b) [5 points] 2 to x = 2, x = 0.25 C 45 5(2)2 (9 + 4) 2 (0.25) =