Unit #3 - Differentiability, Computing Derivatives, Trig Review

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1 Unit #3 - Differentiability, Computing Derivatives, Trig Review Some problems and solutions selected or adapted from Hughes-Hallett Calculus. Derivative Interpretation and Existence 1. The cost, C (in dollars, to produce q quarts of ice cream is C = f(q. In each of the following statements, identify the units of the input and output quantities. (a f(00 = 600 (b f (00 = (a The input to f is in quarts, while the output of f is in dollars. (b The input to f is still in quarts, while the output of f is a rate of change of f(x with respect to x, which must be in dollars/quart.. The time for a chemical reaction T (in minutes, is a function of the amount of catalyst present, a (in milliliters, so T = f(a. State the units of the input and the output for the following expressions. (a f(5 = 18 (b f (5 = 3 4. A magnetic field, B, is given as a function of the distance, r, from the center of the wire as follows: B = r r 0 B 0 when r r 0, and B = r 0 r B 0 when r > r 0. Here r 0 and B 0 are positive constants. (a Sketch a graph of B against r. Use your graph to answer the following questions. (As you do so, think about what the meaning of the constant B 0 is. (b Is B a continuous function of r? (c Is B a differentiable function of r? (a A sketch of the magnetic field looks like the following. (athe input to f is in ml, while the output of f is in min. (b The input to f is still in ml, while the output of f is a rate of change of f(a with respect to a, which must be in min/ml. 3. Suppose C(r is the total cost of paying off a car loan borrowed at an annual interest rate of r%. (a What are the units of C(r? (b What are the units of C (r? (b What is the sign of C (r? (a C represents a total cost, so it is in dollars. (b C (r represents the rate of change of that total with respect to the interest rate, so it is in dollars/percent interest. (c If we increase the interest rate on a loan, the total cost to the borrower increases, so the derivative C (r must be positive. (b The graph certainly looks continuous. The only point in question is r = r 0. Using the first formula with r = r 0 gives B = r0 r 0 B 0 = B 0. Then, using the first formula with r approaching r 0 from above, we see that as we get close to r = r 0 B r0 r 0 B 0 = B 0. Since we get the same value for B from both formulas, B is continuous. (c For r < r 0, the graph of B is a line with a positive slope of B0 r 0. For r > r 0, the graph of B looks like 1/x, and so has a negative slope. Therefore the graph has a corner at r = r 0 and so is not differentiable there. 1

2 5. Consider the function { 1 + cos(πx/ for x g(x = 0 for x < or x > Sketch the graph, and use your sketch to answer the following questions. (a Is the function continuous at x = and x =? (b Is the function differentiable at x = and x =? This is a cosine graph, shifted up by one, with period π = 4. We stop the cosine graph after one halfperiod on either side of x = π/ 0: (c For a 65 kg child, the dosage should be f(60 + f (60(65 60 = (5 = 70 mg. 7. Sketch a smooth graph that satisfies each description below (each part is a separate graph. (a The derivative is everywhere positive and gradually increasing. (b The derivative is everywhere negative and gradually increasing in value (smaller negatives, approaching zero. (c The derivative is everywhere positive and gradually decreasing. (a The derivative is everywhere positive function is always increasing. derivative gradually increasing the slopes are getting steeper. Note that the graph continues on at height y = 0 to the left and right of x = and x = respectively. (a This graph is certainly continuous at r =. There is no jump between the cosine graph and the constant graph. (b Looking at the transition from the cosine graph to the constant part, we would expect the graph to be differentiable there. In fact, as r approaches, cos will be approaching its lowest value, or where its derivative = 0. Since the constant part also has derivative zero, we would expect the derivative to exist at x = and equal zero. 6. The dosage (D, in mg of some drugs, like children s Tylenol, depends on the mass of the patient, M in kg. We would express the relationship as a function, D = f(m. (a Express the relationship f(60 = 40 in a sentence, including units. (b Express the relationship f (60 = 6 in a sentence, including units. (c Use the information in parts (a and (b to estimate the dosage for a 65 kg child. (a f(60 = 40 indicates that, for a patient with a mass of 60 kg, the dosage is 40 mg. (b f (60 = 6 indicates that, for a patient with a mass of 60 kg, the dosage should increase by 6 mg for every additional kg above 60 kg. (b The derivative is everywhere negative and gradually increasing in value (smaller negatives, approaching zero. (c The derivative is everywhere positive and gradually decreasing. 8. Consider the function graphed below. (a At what x-values does the function appear to not be continuous? (b At what x-values does the function appear to not be differentiable? (a The given function appears to be continuous at all x-values shown except for x = 7.

3 (b The function appears to be non-differentiable at the x-values x =, 4, 6, 7. (Recall that a function is non-differentiable where it has breaks, sharp corners or vertical tangents. 9. Identify any x-values at which the absolute value function f(x = 8 x + 6, is (a not continuous, and then (b not differentiable. (a The function is continuous everywhere; we can determine this by graphing it. (b The function appears not to be differentiable at x = 6 because the graph has a corner there. This is confirmed by the fact that the derivative, defined by the limit of the difference quotient f(x + h f(x lim h 0 h does not exist for x = 6: h h lim = lim h 0 h h 0 h. (Because of the absolute value, the limit has to be evaluated from each side separately, and you get two different answers, so the overall limit doesn t exist. 10. Let f(t be the number of centimeters of rainfall that has fallen since midnight, where t is the time in hours. Write a one sentence interpretation of each of the following mathematical statements. (a f(8 =.5 (b f 1 (7 = 11 (c f (8 = 0.3 (d (f 1 (7 = 4 (a f(8 =.5: at 8 AM,.5 centimeters of rain have fallen. (b f 1 (7 = 11: 7 centimeters of rain have fallen 11 hours after midnight. (c f (8 = 0.3: at 8 AM, the rain is falling at a rate of 0.3 cm/hr. (d (f 1 (7 = 4: when 7 centimeters of rain have fallen, it is accumulating at a rate of 4 hr/cm (or more colloquially, 1/4 cm/hr. The Derivative as a Function 11. Consider the function f(x shown in the graph below. allows us to both sketch the graph of the derivative and estimate the derivatives that are requested. The derivative (in red is shown with the original function (in blue below. Carefully sketch the derivative of the given function (you will want to estimate values on the derivative function at different x values as you do this. Use your derivative function graph to estimate the following values on the derivative function. at x = the derivative is At each value of x, we estimate the derivative by estimating the slope (=rise/run from the graph. This From the graph and our calculations, it appears that f ( 3 1, f ( 1, f (1 0, and f (3. 3

4 1. Below is the graph of f(x. Sketch the graph of 14. Below is the graph of f(x. Sketch the graph of f (x. The slope on the original function is constant, so the derivative value is constant. f (x. There are two points with slope zero, at x = and x.5, so there must be a derivative value of zero at those two x locations. In between those landmark points, the slopes of f go: negative, positive, negative. 13. Below is the graph of f(x. Sketch the graph of f (x. There is a point with slope zero at x =, so there must be a derivative value of zero at x =. For smaller values of x, the slopes are negative, so the derivative values will be negative. For larger values of x, the slopes are positive, so the derivative values will be positive. Note that the graph of f (x shown here has an accurate vertical scale; however, that would be difficult for you to estimate by eye based simply on the graph of f(x, so we do not expect that level of precision on a test or exam. 15. For the function f(x shown in the graph below, sketch a graph of the derivative. Note that the graph of f (x shown here has an accurate vertical scale; however, that would be difficult for you to estimate by eye based simply on the graph of f(x, so we do not expect that level of precision on a test or exam. Because the derivative gives the slope of the original function at each point x, we know that the derivative is negative where f(x is decreasing and positive where it is increasing. Applying this to f(x, a sketch of the derivative would resemble the graph below: 4

5 16. A child fills a pail by using a water hose. After finishing, the child plays in a sandbox for a while before tipping the pail over to empty it. If V (t gives the volume of the water in the pail at time t, then the figure below shows V (t as a function of t. At what time does the child: (a Begin to fill the pail? (b Finish filling the pail? (c Tip the pail over? The child begins filling the pail when V (t becomes positive, which is when the volume of water in the pail starts increasing. This is at t = 3. The child finishes filling the pail when the volume stops increasing, which is at t = 9. The child tips the pail over when the volume of water in the pail start decreasing, which is where V (t first becomes negative, which is at t = 14. Computing Derivatives Below are a small sample of problems involving the computation of derivatives. They are not enough to properly learn and memorize how to apply all the derivative rules. You should practice with as many problems as you need to become proficient at computing derivatives. Further practice problems can be found in any calculus textbook. From Hughes-Hallett 5th edition, Section (odd Section (odd Section (odd Section (odd Section (odd Section (odd From Hughes-Hallett 6th edition, Section (odd Section (odd Section (odd Section (odd Section (odd Section (odd 17. Let f(x = 4e x 9x + 5. Compute f (x. You can use the product rule here if you like, but it is far easier to rewrite the function before you start, since f (x = 4e x 18x 18. Let f(x = x 6 x + 5 x 3 x. Compute f (x. For this section, simplify products and fractions before you differentiate, rather than using the product rule and quotient rule. For this section, simplify products and fractions before you differentiate, rather than using the product rule and quotient rule. 5

6 all the terms are just powers of x. Differentiating, f(x = x 6 x + 5 x 3 x f(x = x 6 x 1/ + 5 x 3 x 1/ = x x f (x = 13 x 11 ( 7 5 = 13x 5 x + 35 x 4 x 19. Let f(x = 7x + 7x + 5 x. x 9 (a Compute f (x. (b Find f (3. You can use the quotient rule here if you like, but it is far easier to rewrite the function before you start, since all the terms are just powers of x. f(x = 7x + 7x + 5 x = 7x 3/ + 7x 1/ + 5x 1/ (a f (x = 1 x1/ + 7 x 1/ 5 x 3/ (b Evaluating f (x at x = 3, we obtain Let f(t = 7t 7. (a Compute f (t. (b Find f (3. (a f (x = 49 t 8 (b f (3 = 49 ( Let f(x = 4e x + e 1. Compute f (x. Don t be thrown off by the e 1 : that s a constant (equal to e or.7, so the derivative of that term is zero. f (x = 4e x. Let f(x = 4e x + 4x. Compute f (x. f (x = 4e x f(x = (3x (6x + 3. (a Compute f (x. (b Find f (4. You can either expand the product before differentiating (and obtain f(x = 18x 3 + 9x 1x 6, or use the product rule. Both give the same answer. (a f (x = 54x + 18x 1. (b f (4 = Let f(x = x 4 x + 4. Compute f (9. We need to use the quotient rule for this function. 1 f (x = x 1/ (x 1/ + 4 (x 1/ 4 ( 1 x 1/ ( x + 4 = 1 + 4/ x 1 + 4/ x ( x = x( x + 4 f 4 (9 = 3(3 + 4 = Consider f(x = 4x + 3 3x +. (a Compute f (x. (b Find f (5. (a f (x = (b f (5 = (3x + (4x + 3(3 (3x + 6. Consider f(x = 7 x 7 + x (a Compute f (x. (b Find f (1. (a f (x = (b f (1 = x (x Let f(x = x(x 3. (a Compute f (x. (a f (x = 4x + 6 (b f ( 5 = 6 (b Find f ( f(x = 4x3 3 x 4 (a Compute f (x. (b Find f (. (a 4x + 1x 5 (b g(x = ex 5 + 4x. Compute g (x. g (x = (1 + 4xex (5 + 4x 6

7 30. f(x = 4x tan x sec x. (a Find f (x. (b Find f (3. This should definitely be simplified before you differentiate! Recall: tan(x = sin(x cos(x, and sec(x = 1 cos(x, so tan(x sec(x = sin(x cos(x = sin(x. cos(x (a f (x = 8x sin(x + 4x cos(x (b f (3 = 4 sin( cos( f(x = 7 sin x + 1 cos x (acompute f (x. (b Find f (1. (a f (x = 7 cos(x 1 sin(x (b f (1 = 7 cos(1 1 sin( Let f(x = cos x tan x. Compute f (x. f (x = sin(x sec (x 33. f(x = 5 sin x 3 + cos x (a Compute f (x. (b Find f (. Don t forget the identity sin (x + cos (x = 1. (a f (x = (15 cos(x + 5/(3 + cos(x (b f ( = (15 cos( + 5/(3 + cos( f(x = 7x(sin x + cos x (a Compute f (x. (a Find f (3. (a f (x = 7(sin(x + cos(x + 7x(cos(x sin(x (b f ( Let f(x = cos(sin(x. Compute f (x. f (x = sin(sin(x cos(x (x 36. Let f(x = sin 3 x. Compute f (x. f (x = 6 sin (x cos(x 37. Let y = (8 + cos x 6. Compute dy dx. dy dx = 6(8 + cos (x 5 ( cos(x( sin(x 38. Let f(x = 3 ln[sin(x]. Compute f (x. f (x = 3 cos(x = 3 cot(x sin(x 39. Let f(x = ln(4 + x. Compute f (x. f (x = 4 + x 40. [Note: this question should be done in Unit 4] Let f(x = arcsin(x. (a Compute f (x (b Find f (0.4. Recall that the derivative of arcsin(x is: f (x = 1 1 x Evaluating this at x = 0.4, we get: f ( [Note: this question should be done in Unit 4] Let f(x = arccos(14x arcsin(14x. Compute f (x. Applying the quotient rule: f (x = arcsin(14x (arccos(14x arccos(14x (arcsin(14x (arcsin(14x 4. Let 14 arcsin(14x 1 (14x 14 arccos(14x 1 (14x = (arcsin(14x 14 (arcsin(14x + arccos(14x = (arcsin(14x 1 (14x P = V R (R + r. Calculate dp, assuming that r is variable and dr R and V are constant. Note that V is also constant. Let f(r = Using the quotient rule: V R (R + r = V R R + Rr + r. f (r = (R + Rr + r (0 (V R(R + r (R + r 4 = V R(R + r (R + r 4 = V R (R + r 3. 7

8 Trigonometric Functions 43. A population of animals oscillates sinusoidally between a low of 100 on January 1 and a high of 1100 on July 1. Graph the population against time and use your graph to find a formula for the population P as a function of time t, in months since the start of the year. Assume that the period of P is one year. This tells us that ( π cos = 1 ( }{{} 4 x and so x = π 4 or x = π 8 (3 (4 The average population is = 600, the amplitude of the sinusoidal oscillation is = 500, and the period of the oscillation is 1 months. This gives the graph Thus one solution to 1 = cos(x is x = π Find one solution to the equation 3 = sin(3x We first start by isolating x as much as we can. sin(3x + 1 = 3 = 1 (5 sin(3x + 1 = 1 (6 Since the population is at its minumum when t = 0, we use a negative cosine: P (t = 500 cos(bt To find B, we use the fact that the period is 1. Thus B = π 1 = π 6, and P (t = 500 cos( π t Find the period and amplitude of r = 0.4 sin(πt + 1 The period is π/π =, since when t increases from 0 to, the value of πt increases from 0 to π. The amplitude is 0.4, since the function oscillates between 0.6 and Find one solution to the equation 1 = cos(x. We first start by isolating x as much as we can. cos(x = 1 (1 The next challenge is remove the cosine on the left hand side. This can be done in many ways. By sketching a triangle with (adjacent = 1, hypotenuse =. This has an angle of 45 o or π/4 radians. By using your calculator s arccos or cos 1 button and = π/4. Drawing a right-angle triangle with side lengths (opposite = 1, hypotenuse = shows the 30/60 special triangle, and leads to an angle of π 6, because sin = 1 6. This tells us that ( π ( π sin = 1 (7 }{{} 6 3x+1 and so 3x + 1 = π 6 (8 or 3x = π 6 1 (9 x = 1 ( π (10 (11 ( Thus one solution iis x = 1 π Find sin θ and tan θ if cos θ = 8 17, assuming that 0 θ < π/. Consider a triangle where the lengths of the side adjacent to the angle θ and the hypotenuse have been labeled so that cos θ = Calculate the length of the side opposite the angle θ with the Pythagorean theorem: 17 8 = 15. From the triangle we see that sin θ = 15 17, and tan θ = Find sin θ, sec θ, and cot θ if tan θ = If tan θ = , then cot θ = 11. For the remaining trigonometric functions, consider a triangle where the lengths of the sides opposite and adjacent to the angle θ have been labeled so that tan θ = Calculate the length of the hypotenuse with the Pythagorean theorem: = 61. From the triangle we see that sin θ = and sec θ =

9 49. Find a formula for the graph of the function y = f(x given in the figure below. For a cos graph, we need to find a right shift of a that will make cos(x a = 0 when x = 0. This will occur if a = π/, so g(x = 4 cos(x π/. Alternatively, g(x = 4 cos((x π/4. The graph is an inverted cosine curve with amplitude 5 and period π, shifted up by, so it is given by y = f(x = 5 cos x. 50. Find a formula for the graph of the function f(xgiven in the graph below. This graph is a sine curve with period 1/, amplitude 1 and a vertical shift of 3, with no horizontal shift. It is therefore given by f(x = 1 sin(4πx Consider the trig graph below. Find both a cosine and a sine based formula for this graph. This is already like a sine graph, but with an amplitude of 4 and a period of π: f(x = 4 sin(x. 9

10 5. Consider the trig graph below. Find both a cosine and a sine based formula for this graph. The period is 4π, so we need to scale x by 1 to get that horizontal stretch: the base model functions will be cos(x/ and sin(x/. 10

11 To get the cos shift, we note that f(π/ = cos( π/ a = 1, and cos(0 = 1, so comparing those two inputs, we need π/4 a = 0, or a = π/4: f(x = cos ( x π 4 will produce the graph shown. Similarly, to get the sine shift, we note that f(π/ = sin( π/ a = 1, and sin(π/ = 1, so comparing those two inputs, we need π/4 a = π/, or a = π/4. Since we used x/ a as our form, and a turns out to be negative, our sine function will be: g(x = sin ( x + π 4 will produce the graph shown. 53. Sketch the graph of y = x cos(πx on the interval x = [ 4, 4]. Period of the oscillations is π = 1. The boundaries π of the oscillations are given by the straight line y = x (upper bound and y = x (lower bound. 8 y x Sketch the graph of y = x cos(πx on the interval x = [0, 9]. Period of the oscillations is π = 1. The boundaries π of the oscillations are given by y = x (upper bound and y = x (lower bound. 11