Mat 135 Summer Midterm Solutions

MAT 135: Calculus Midterm Exam, Summer 2009 Semester SOLUTIONS

Section I: Multiple Choice For each item below, respond with the ONE letter that is most correct. Do not attempt to write down a justification for your work; only your answer will be graded. Please your responses on a separate page.

1. [2 points] Which term best describes the function y = (1.2)x ? A. Linear function B. Power function C. Polynomial function D. Exponential function Exponential functions are functions with a constant base and a variable exponent. E. Logarithmic function 2. [2 points] Which of the following statements is/are true for all polynomial functions p(x)? A. p(x) is continuous over its entire domain B. p(x) is increasing over its entire domain C. The domain of p(x) is the interval (−∞, ∞) D. lim p(x) = p(a) for all a in the domain of p x→a

E. All of the above except (b) Not every polynomial is always increasing; for example y = x2 is increasing only on [0, ∞). 3. [2 points] What is the domain of the function y = ln x? A. All real numbers B. All real numbers except x = 0 C. The interval [0, ∞) D. The interval (0, ∞) 0 is not included.

This is actually the domain for any logarithmic function. The number

E. None of the above 4. [2 points] Suppose y = f (x) is continuous at x = a. Which of the following must be true? A. lim f (x) exists x→a

B. f (a) exists C. lim f (x) = f (a) x→a

D. All of the above

This is is the definition of continuity.

E. Both (A) and (B) but not (C) 5. [2 points] Suppose the cost of a bottle of water (C, in dollars) is a function of the temperature outside (T , in degrees Fahrenheit). What are the units of the derivative, C 0 (T )? A. Dollars B. Degrees C. Degrees per dollar 1

D. Dollars per degree The units of C 0 (T ), or dC/dT in Leibniz notation, are the units of C divided by the units of T . E. Dollars per degree per degree

Section II: Problems to Solve 1. In class, we learned that the derivative of a function f at the point x = a, denoted f 0 (a), is given by two different-looking but equivalent formulas: f (a + h) − f (a) h→0 h

f 0 (a) = lim

or

f (x) − f (a) x→a x−a

f 0 (a) = lim

(a) [10 points] Give the three major interpretations of the meaning of f 0 (a). Be as specific and precise as possible. Use correct English and complete sentences. Solution: The three major interpretations are: • The derivative f 0 (a) gives the slope of the tangent line to the graph of f at the point x = a. • The derivative f 0 (a) gives the instantaneous rate at which f is changing right when x = a. • The derivative f 0 (a) gives the instantaneous velocity at time x = a of an object whose position at time x = a is given by f . (b) [8 points] Choose one of the two formulas above and explain why, exactly, the formula is constructed the way that it is. What does the fraction in the formula represent? Why is there a limit present? Again, be as specific and precise as possible, and use correct English sentences. (a) calculates the slope of the Solution: Using the second formula above, the fraction f (x)−f x−a secant line passing through the points (a, f (a)) and (x, f (x)) on the graph of f . To get the slope of the tangent line to the graph of f at x = a, we need to move x closer and closer to a and observe what, if any, limiting or stabilizing value is approached by these secant line slopes. The limit as x → a calculates this value (or tells us if no such value exists). An explanation using the first formula is similar to this one.

(c) [10 points] Use one of the two formulas above to calculate (precisely) the value of f 0 (2) where f (x) = 1/x. DO NOT USE ANY DERIVATIVE RULES from Chapter 3; you will receive no credit if you obtain your answer using some other means besides one of the two formulas above.

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Solution: Here is the calculation using the first formula: f (2 + h) − f (2) h→0 h lim

= =

lim

h→0

lim

1 2+h

−

1 2

h 2−(2+h) 2(2+h)

h −h = lim h→0 2h(2 + h) −1 = lim h→0 2(2 + h) −1 1 = =− . 2(2 + 0) 4 h→0

2. The population of Kenya was 19.5 million people in 1984 and 21.2 million on 1986. Assuming population increases exponentially, a model for the population (P , in millions of people) of Kenya as a function of time (t, which we will measure in years since 1984) is P (t) = 19.5e0.042t (a) [4 points] What would this model predict for the population of Kenya in 2010? Solution: The year 2010 is 2010 − 1984 = 26 years since 1984, so just evaluate the function at t = 24: P (24) = 19.5e0.042·24 = 19.5e1.008 ≈ 19.5 · 2.740015301 ≈ 53.43 million people

(b) [8 points] How long will it be, according to this model, until the population of Kenya reaches 100 million people? Solution: We want to know when P (t) = 100, so we set 19.5e0.042t = 100 and solve using algebra: 19.5e0.042t

=

100

0.042t

=

5.1282

e

0.042t = t =

ln 5.1282 ln 5.1282 ≈ 38.922 0.042

So it will be about 39 years, or until the year 2023, until the population reaches 100 million. (c) [6 points] Find the average rate of change in the population of Kenya from 1986 to 1996 according to this model. State correct units on your answer. Solution: The average rate of change is the total change from 1986 (t = 12) to 1996 (t = 2) divided by the time elapsed: P (12) − P (2) 32.2789 − 21.2 = ≈ 1.107 million people per year 12 − 2 10

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(We used the formula for P to calculate P (12). The value of P (2) is given in the problem statement.) (d) [12 points] Find the instantaneous rate of change in the population of Kenya in 1996 and give a brief explanation of why this rate is different than the average rate you calculated in (c). (Make your explanation understandable to someone who has not had calculus.) Solution: The instantaneous rate of change calls for the derivative, so we calculate P 0 (t) first:  d  d  0.042t  19.5e0.042t = 19.5 e = 19.5(0.042e0.042t ) = 0.819e0.042t dt dt So the rate of change in 1996 (t = 12) is P 0 (12) = 0.819e0.042·12 ≈ 1.356 This is again in millions of people per year. This number is different than the average rate from 1996 to 1986 because the population is not growing at a constant rate, and therefore an average taken over a 10-year interval will produce a different result than an instantaneous rate taken at a single point in time. (e) [8 points] Suppose that the population of Kenya continues to change each year at the rate you calculated in (d). What would the population be in 2010 in that case? Why is this number less than the number you calculated in (a) using the model? Solution: The population in 1996 is 32.2789 million (see work for part (c)), so if we assume that in 1996 the growth continues at a constant rate of 1.356 million people per year, the population in 2010 (which is 14 years later) will be 32.2789 + 14(1.356) = 51.2629 million The model predicts a population of 53.43 million. The number we just calculated is less than this because the rate at which the population grows is increasing. In other words, population is increasing at an increasing rate. So in 1997, the population growth rate will not be 1.356 million people per year but something greater; in 1998, the growth rate will be still greater; and so on. Using a linear projection like this will yield an underestimate because the actual population is growing at a rate faster than linear growth. 3. At time t seconds after it is hit by a golf club, a golf ball is at a height of f (t) = −4.9t2 + 25t + 1 meters in the air. (a) [6 points] Find the average velocity of the ball in the first 2 seconds. Give units on your answer. Solution: The average velocity is the change in height from t = 0 to t = 2 (that is, f (2)−f (0)) divided by the change in time: f (2) − f (0) 31.4 − 1 30.4 = = = 15.2 meters per second 2−0 2−0 2

(b) [10 points] Find (exactly) the instantaneous velocity of the ball at t = 2. Give units.

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Solution: The instantaneous velocity is given by the derivative, f 0 (t). Using the Power Rule and other derivative rules, we get f 0 (t) = −9.8t + 25 So the velocity at t = 2 is f 0 (2) = −9.8(2) + 25 = −19.6 + 25 = 5.4 meters per second. (c) [6 points] What is the acceleration of the ball at t = 2? Solution: Acceleration is given by the second derivative, which is the derivative of f 0 (t): f 00 (t) =

d 0 d [f (t)] = [−9.8t + 25] = −9.8 dt dt

In other words, the acceleration is constant at all times and is equal to −9.8 meters per second per second. (d) [10 points] How high does the golf ball go? For this part, use only algebra and/or calculus. You will receive no credit for answers obtained through graphs or tables. Solution: We are being asked to find the maximum height attained by the golf ball. When this happens, its velocity is zero. So we’ll first set velocity (from (b)) equal to zero and find the time at which this maximium happens: −9.8t + 25 = 0

⇒

t=

25 = 2.551 seconds 9.8

Now find the height at this time using f (t): f (2.551) = −4.9(2.551)2 + 25(2.551) + 1 ≈ 32.89 meters

(e) [10 points] How long is the golf ball in the air? For this part, use only algebra and/or calculus. You will receive no credit for answers obtained through graphs or tables. Solution: An equivalent question is, how long will it be before the golf ball hits the ground again? When it hits the ground, its height is zero. So we set the height formula (f (t)) equal to 0 and solve using the Quadratic Formula: −4.9t2 + 25t + 1

=

0 p −25 ± (25)2 − 4(−4.9)(1) t = 2(−4.9) √ −25 ± 644.6 = −9.8 ≈ −0.04 or 5.14 seconds

The second solution is the one that makes sense here, so the ball is in the air for 5.14 seconds. Note that no calculus is involved in this question. 4. [10 points] The curve traced out by the function y = the tangent line to this curve at the point (3, 0.3). 5

x is called a serpentine. Find an equation for 1 + x2

Solution: To find an equation for the tangent line, we need a point on the line and the slope of the line. We are given the point on the line, (3, 0.3). The slope will be given by the derivative dy/dx at the point x = 3. We get the derivative using the Quotient Rule:   d d (1 + x2 ) · dx [x] − x · dx [1 + x2 ] x d = dx 1 + x2 (1 + x2 )2 2 (1 + x )(1) − x(2x) = (1 + x2 )2 1 − x2 = (1 + x2 )2 At x = 3, this equals: dy 1−9 = = −0.08 dx x=3 (1 + 9)2 We use the point-slope form for a line to get the result: y − y1

=

m(x − x1 )

y − 0.3

=

−0.08(x − 3)

y

=

−0.08x + 0.24 + 0.3

y

=

−0.08x + 0.54

5. The graph below shows the population (P ) of a colony of yeast cells in a laboratory culture, t hours after beginning observations: 800

700

P (number of yeast cells)

600

500

400

300

200

100 0 -1

0

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

t (hours)

(a) [10 points] State the value of lim P (t) and explain briefly what this quantity means in terms of t→∞ the population of the yeast colony. Solution: From the graph we see that lim P (t) = 700. In real terms, this means that over a t→∞ long period of time, the population of the yeast colony will level out at 700 cells and get no

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bigger than that. (b) [12 points] Use the graph to make an accurate graph of P 0 (t) on a separate set of axes. Describe the ways in which the yeast population grows over time using this graph. Solution: We use the graph and the grid marks to estimate the slopes of P at various points. For example, if we draw the tangent line to the graph of P at t = 1, we will get a line that goes up from a height of slightly more than 100 to a height of around 150 when we move from t = 1 to t = 2, meaning that P 0 (1) ≈ 50. We can continue these calculations to create a table of values for P 0 (t), and plotting that table will give us a graph that looks like: dP/dt (rate of change in yeast population, cells per hour)

100 90 80 70 60 50 40 30 20 10 0

0

2

4

6

8

10

12

14

16

18

20

22

t (hours)

Some variation in the estimates for P 0 is expected. However, the key features of any accurate derivative graph are: • The derivative graph must always be positive, because P is always increasing. There should be no points where P 0 crosses the t-axis or goes negative. • The graph of P 0 peaks around t = 5, which is where P is steepest. • The graph of P 0 approaches 0 as t → ∞ because P becomes nearly horizontal as t → ∞.

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