Chapter 8: Radical Functions

Transcription

1 Chapter 8: Radical Functions

2 Chapter 8 Overview: Types and Traits of Radical Functions Vocabulary:. Radical (Irrational) Function an epression whose general equation contains a root of a variable and possibly addition, subtraction, multiplication, and/or division Basically, that means an equation with an in a radical. There are two different kinds of irrational functions those with an odd inde, like y = 3 a n n a 0 : and those with an even inde, like y = a n n a 0 : Note that the graphs of the functions with the even inde given by a calculator look like they do not touch the -ais. Radical functions rarely graph perfectly on a calculator because of the window settings and the numerical values of the traits. Know what to epect from the graph and realize what is correct rather than just trusting the calculator. 425

3 The traits that are common to all radical functions are:. Domain 2. Range 3. Zeros 4. y-intercepts 5. Etreme Points 6. End Behavior Asymptotes If the radical has a fraction in it, the traits of a rational function would be included: 7. Vertical Asymptotes 8. Points of Eclusion (POEs) 426

4 8-: Zeros and Domain REMEMBER: The only things that can change the domain from all real numbers are:. a zero denominator, 2. a negative under an even radical, or 3. a non-positive in a log. This section will concentrate on even radicals. The number that indicates the kind of radical 2 for square root, 3 for cube root, etc. is known as the inde. Domain for Radical Functions: Odd inde All Reals Even inde the radicand must be non-negative The radicand being non-negative implies an inequality for the domain. In other words, the domain is found by setting the radicand 0. And because of the = under the inequality, zeros will be found at the same time as the domain. LEARNING OUTCOMES Use sign patterns to determine the domain of radical functions. Find zeros of radical functions. 427

5 EX Find the zeros and domain of y = is the domain and ( 3, 0) is the zero The graph shows a curve only to the left of 3: y y = 3 EX 2 Find the zeros and domain of y = ( 4 + ) So the zeros are ±4, 0 and the domain is 4,

6 The graph shows a curve only between 4 and 4: y y = 6 2 Notice how the curve looks like it does not touch the -ais. But it must because zeros were found. As the technology gets better, this may improve, but for the present, recognize the limitations of the calculator. EX 3 Find the zeros and domain of y = ( 4 + ) So the zeros are ( ±4, 0) and ( 0, 0), and the domain is, 4 ( 0,

7 8 y y = 6 3 EX 4 Find the zeros and domain of y = y + DNE So the zero is (, 0), and the domain is, 2, ). Note that 2 would be a vertical asymptote, not a zero; therefore, it is not included in the domain. 430

8 8- Free Response Homework Find the zeros and domain of each function.. y = y = y = y = y = y = y = 8. y = y = y = y = y = y = y =

9 8- Multiple Choice Homework. Find the domain of the function g( u) = u 9 u. (a) u 0, ) ( (b) u, 0 (c) u 0, 9 (d) u 0, 9 (e) u (, 9 2. What is the instantaneous rate of change at = 3 of the function f given by = f (a) 3 (b) (c) 0 (d) 2.5 (e) DNE ? 3. The derivative of 3 is (a) (b) (c) (d) (e)

10 4. Let f be the function defined by f ( ) = Which of the following is an equation of the tangent line to the graph of f at the point where =? (a) y = 7 3 (b) y = (c) y = 7 + (d) y = 5 (e) y = A particle moves along the -ais so that at time t 0 its position is given by t = 2t 3 2t t 53. At what time is the particle at rest? (a) t = only (b) t = 3 only (c) t = 7 2 only (d) t = 3 and t = 7 2 (e) t = 3 and t = 4 6. Given the graph of y = f ( ) at the right, select a graph that best represents the graph of y = f. 7. The function f ( ) = has as its domain all values of such that (a) > (b) 0 (c) < (d) 0 < (e) > or 0 433

11 8-2: The Chain Rule Just as rational functions needed a separate rule because they could not be written in the form needed for the Power Rule, radical functions often need another rule. Some radicals do not need a new rule namely those that can be written as y = n. Remember that radicals are actually fractional eponents, where the inde is the denominator and the power is the numerator. So d 5 d = d d 5 2 = But most radical functions are not a single variable to a power but are polynomials or rationals to a power. In other words, they are composite functions. Vocabulary:. Composite Function a function made of two other functions, one within 3. the other. For eample, y = 6 3, y = sin 3, y = cos 3, or y = The general symbol is f ( g( ) ). EX If f ( ) = 2 + and g( ) = 2, determine the value of: b) g( f ( 3) ) c) f ( g( ) ) a) f g( 3) a) To find f ( g( 3) ), first find g( 3): = 2, so g( 3) = 3 2 = 9 ( ) = f ( 9) f ( ) = 2 + f ( 9) = 2( 9)+ g f g 3 =9 So, f ( g( 3) ) =

12 b) For g( f ( 3) ), f ( ) = 2 + f ( 3) = 2( 3)+= 7 c) f g( ) = 2 g ( )+= g( f ( 3) ) = g( 7) = 49. EX 2 If f ( ) = and g( ) = 2 +5, find f g( ) f ( g( ) ) = g( ) = 2 +5 A question of how to take the derivative of a composite function arises in Calculus. There are now two (or more) functions that must be differentiated, but, since one is inside the other, the derivatives cannot be taken at the same time. Just as a radical cannot be distributed over addition, a derivative cannot be distributed over a composite function. While this section concerns radical functions, the Chain Rule applies to all composite functions and will come up again and again. It is one of the cornerstones of Calculus. The Chain Rule: d d f g ( ) = f ( g ( ))i g LEARNING OUTCOME Find the derivative of radical functions. 435

13 EX 3 d d d d ( 2 +) 5 ( 2 +) 5 = d d ( 2 +) 5 2 = 5 2( 2 +) = 5( 2 +) 3 2 EX 4 y = 6 3 ; find dy d 2 ( 6 3) 2 ( ) ( ) 2 y = 6 3 = 6 3 dy d = 2 = In this case, the is the f function and the polynomial 6 3 is the g function. Each derivative is found using the Power Rule, but, as 6 3 is inside the, it is inside the derivative of the. 436

14 EX 5 D D = D = ( ) 4 = ( ) 2 3 = Note that the last step of this eample included factoring to determine whether the fraction could be further simplified. Always epress answers in simplest form. The Chain Rule applies to any composite function, not just radicals. EX 6 d d d d = = 20( ) 9 ( 4 ) 437

15 EX 7 F ( ) = ; find F ( ). F ( ) = F = = i 2 = 4 2 i ( 4 ) 2 i4 2 ( 4 ) 2 = = ( 3 + 2) ( 4 ) i ( 4 )

16 8-2 Free Response Homework Find the following derivatives.. d d 5 ; find y y = D d d y = ; find y 3. y = 2; find dy d 0 4. d d y = 3 2; find dy d 5. D y = ; find dy d 6. d d f = 2 5 ; find f ( ) y = 25 ; find y 7. d 2 d 6 ( ) 5 8. d d D D d d d d f ( ) = ; find f ( 5) 439

17 8-2 Multiple Choice Homework. Find a, such that the function f (a) a = 5 (b) a = 25 (c) a = 25 (d) a = 5 (e) a = 5 2. If h( ) = 2 4 (a) 3 (b) 2 (c) (d) 0 (e) Does not eist 34 +, then the value of = 4 + a 2 has the domain 5, 5 is h 2 3. Find the point on the graph of y = between, the tangent to the graph has the same slope as the line through, (a), (b) 2, 2 (c) 3, 3 (d) 4, 2 (e) None of the above 4. What is lim ?. and ( 9, 3) at which and ( 9, 3). (a) 3 2 (b) 3 4 (c) 2 3 (d) (e) The limit does not eist. 440

18 5. Let h be a differentiable function, and let f be function defined by f ( ) = h( 2 3). Which of the following is equal to f ( 2)? (a) h (b) 4 h (c) 4 h 2 (d) h 4 (e) 4 h 4 6. The figure below shows the graph of the functions f and g. The graphs of the lines tangent to the graph of g at = 3 and = are also shown. If B( ) = g( f ( ) ), what is B ( 3)? (a) 2 (b) 6 (c) (d) (e) Given this sign pattern y = f ( ) is , the domain of 4 2 y 2 (a) (, 4) (, 2) (b) (, 4] [, 2] (c) ( 4, ) 2, (e) 4, [ 2, ) (d) 4, [ ] 2, 44

19 2 y Given the graph above, which of the following might be the sign pattern of f? (a) (b) (c) (d) (e) f ( ) f ( ) f ( ) f ( ) f ( )

20 8-3: Etreme Points of Radical Functions Previously, it was found that the critical values of the etreme points were where the derivative equaled zero, where the derivative was undefined, and endpoints of an arbitrarily stated domain. Now this concept will be eplored in the contet of radical functions. REMEMBER: If a value is a critical value, then either: i) dy d = 0; ii) dy d does not eist (i.e., denominator = 0); or iii) that value is at an endpoint of an arbitrarily stated domain. Because a radical function, which usually starts with no denominator, might have a derivative with a denominator, there are numbers that are not ecluded from the domain of y, which can make dy not eist. d LEARNING OUTCOME Find the critical values and etreme points of radical functions. 443

21 EX Find the critical values of y = ( 6 3) 2 ( ) ( ) 2 y = 6 3 = 6 3 dy d = 2 i) = ( ) = 0 = ± 4 3 = 0 Recall from an eample in a previous section the domain is (, 4 0, 4. Therefore, = 4 is not actually a critical value 3 because it is not in the domain. This leaves = 4 3 as a critical value. ii) ( ) = 0 = 0, 4, 4 does not eist iii) There is no arbitrary domain stated. Therefore, the critical values are = 4, 0, 4, and 4. The graph shows 3 that = 4 3 is at the maimum point, while the others are at the minimum points. 444

22 8 y y = 6 3 EX 2 Find the etreme points of y = The domain of the function is (, 2 2, 2 2, ). 2 ( 4 3 2) dy d = = i) dy d = = 0 = = 0 = 0, ± 3, but = ± 3 are not in the domain. This leaves = 0 as a critical value. So 0, 8 point. or ( 0, 2 2) is an etreme 445

23 ii) dy d = DNE ( ) 2 = 0 = ±2, ± 2 iii) As with EX, there is no arbitrary domain. The critical values are = 0, ± 2, and ± 2. The etreme points are ( 0, 2 2), ( ±2, 0), and ( ± 2, 0). 446

24 EX 3 Find the etreme points of f ( ) = on 4, 4. Domain: 4, 3) 2, 2 ( 3, 4 = 2 f i 9 = ( 2 9) 2 0 i 2 ( 2 4) 2 ( 2 9) 2 5 = ( 2 4) 2 ( 2 9) 3 2 ( 2 9) ( 2) ( 2 9) 2 i) ii) dy = 0 means 5 = 0, so = 0 d dy d does not eist means 2 4 = 0 or 2 9 = 0, so = ±2 or = ±3, but = ±3 are not in the domain. This leaves = ±2 as critical values. iii) Endpoints of this arbitrary domain are 4 and 4. The critical values are = 0, ± 2, ± 4. The etreme points are 0, 2 3, ( 2, 0 ), ( 2, 0), 4, 2 3, 4,

25 EX 4 Find the intervals on which y = is increasing. As seen in EX 2 the domain is (, 2 2, 2 2, ). Also seen in EX 2, dy d = and the critical values are = 0, ± 2, ± 2. The sign pattern for the DERIVATIVE would be dy d + DNE DNE 0 + DNE DNE Note that, because of the domain of the radical, there are some intervals on the sign pattern that have no sign. That is because the function does not have real values there. The intervals of increasing are where dy > 0. They would be d (, 2) ( 0, 2) 448

26 8-3 Free Response Homework State the domain and critical values.. y = ( 5 3) y = y = y = y = 9 6. y = y = State the domain, find the critical values, and determine if the critical values occur at a maimum point, a minimum point, or neither. 8. y = y = y = The position of a particle moving along the -ais at time t 0 is described by t = 5+ t 2. Find the velocity at t = Use the equation of the tangent line to f. approimate f 0.9 = +5 at = to 3. Use the equation of the tangent line to f ( ) = 6 at = 2 to approimate

27 4. Use the equation of the tangent line to f approimate f ( 3.). = at = 3 to 5. A person is on an island 8 miles from a straight shore. She needs to go to a town 5 miles down the coast. A boat costs $6 per mile and a car costs $ per mile. How far down the shore should she land the boat in order to minimize the cost? What is that minimum cost? 5 car town 8 boat island problem 3 Find the etreme points y = 2 7. y = y = f ( ) = 20. g 5 = h = F ( ) = A particle s position ( t), y( t) parametric equations t at time 0 t 0 is described by the = t and y( t) = 0t t 2. a) What is the particle s velocity at t = 2? b) What is the particle s acceleration at t = 2? 450

28 24. A particle's position ( t), y( t) parametric equations t at time 0 t 0 is described by the = t and y( t) = 0t t 2. a) When is the particle moving right and down? b) When is the particle at rest? 8-3 Multiple Choice Homework. The function f is differentiable with f 2 of the approimation of f.9 (a) 0.4 (b) 0.6 (c) 0.7 (d).3 (e).4 2. Differentiate the following function f ( ) = 30. (a) f (b) f (c) f (d) f (e) f ( ) = 0 = ( ) = 2 30 ( ) = 2 30 ( ) = 2 30 = and f ( 2) = 4. What is the value using the line tangent to the graph of f at = 2? 3. Find the absolute maimum value of y = 36 2 on the interval 2, 2. (a) 5 (b) 6 (c) 7 (d) 0 (e) 4. The graph of the function f ( ) = is increasing on which of the following intervals I. < II. 0 < < III. < 0 45

29 (a) I only (b) II only (c) III only (d) I and II only (e) I and III only 452

30 5. The table below gives values of the differentiable functions f and g and of their derivatives f and g, at selected values of. If h( ) = f ( g( ) ), what is the slope of the graph of h at = 2? g( ) f ( ) g ( ) f (a) 0 (b) 5 (c) 5 (d) 6 (e) 0 6. The function F is defined by F, where the graph of the = G + G( ) function G is shown in the figure. The approimate value of F is (a) 7 3 (b) 2 3 (c) 2 (d) (e)

31 8 y Which of the following sign patterns apply to the graph above? I. y + 0 DNE + DNE II. + 0 DNE + DNE y 2 III. dy d DNE DNE 0 + DNE DNE (a) I only (b) II only (c) I and II only (d) II and III only (e) I, II, and III dy Given this sign pattern d, at what value of does f have a local minimum? (a) 2 (b) 2 (c) 0 (d) 2 (e) 2 454

32 8-4: General Radical Curve Sketching REMEMBER: Radical Traits. Domain 2. Zeros 3. y-intercept 4. Etreme Points 5. Range 6. End Behavior (EB) And, if there is a rational function within the radical: 7. POE 8. Vertical Asymptotes LEARNING OUTCOME Find all the traits and sketch a fairly accurate radical curve algebraically. EX Given the traits below, sketch a graph of the function. Domain: (, 6 ( 3, 4) 5, ) Zeros: ( 6, 0), ( 5, 0) VA: = 3, = 4 POE: None y-int: ( 0, 3.5) Etreme Points: ( 6, 0), ( 5, 0),., 3.05 End Behavior: y = 2 Range: y 0, 2) 3.05, ) Graph all the asymptotes and points. Note that the only point that is not an etreme point is the y-intercept (in this case). Note also, that because of the domain there will be large sections with no graph. 455

33 y For (, 6 : ( 6, 0) is both a zero and an etreme point. Because of the range this means the zero is a minimum point. There are no other etreme points, so the curve must flatten out towards y = 2. For ( 3, 4): Notice that the etreme point is a little lower than the y-intercept, so it is a minimum point. That means that the curve has to head up to the vertical asymptotes. For 5, ) : 5, 0 is both a zero and an etreme point. Because of the range the zero is a minimum point. There are no other etreme points in this region of the graph, so the curve must head to the end behavior asymptote y = 2. Note that because of the domain, there is no curve graphed outside the domain. 456

34 8 y EX 2 Find the traits and sketch y = Domain: ( + ) y 2, so the domain is, 6 2. Zeros: (, 0) and ( 6, 0) 3. y-int: ( 0, 6) 457

35 4. Etreme Points: dy d = dy = d = 0 = 5 2 ; y = 7 2 An etreme point is 5 2, 7 2. dy d does not eist = 0, 0 and ( 6, 0) 5. Range: y 0, End Behavior: Recall that end behavior describes the behavior of the function at = ±. The domain of this function is, 6. Since neither ± are in the domain, there is no end behavior to consider. 7. Sketch: 6 y y =

36 EX 3 Find the traits and sketch y = Domain: y 2 = ( + 3) ( 3) 0 y , so the domain is 3, 3 4, ) 2. Zeros: ( ±3, 0) and ( 4, 0) 3. y-int: ( 0, 6) 4. Etreme Points: 2 ( ) dy d = dy d = 0 ( ) = 0 ( 9) = 8± = But 3.59 is not in the domain. dy d does not eist = 0 = 3, 3 or 4 The etreme points are ( 0.852, 6.336), ( 3, 0), ( 3, 0), and ( 4, 0). 5. Range: y (, 0 459

37 6. End Behavior: The domain of this function is 3, 3 4, ). Since only + is in the domain, consider only the limit as goes to +. lim = Thus, the right end of the graph heads down. There is no EB on the left. 7. Sketch: 2 y y = With more complicated radical functions, it might be a good idea to look at just the sketching part before putting a whole problem together. 460

38 EX 4 Find the traits and sketch y =. Domain: ( + 2) 0 y DNE + 0 DNE Zero: ( 0, 0) 3. y-int: ( 0, 0) The domain is ( 2, 0 ( 2, ). 4. Etreme Points: dy d = = ( 2 4) = ( 2 4) ( 2 4) 2 2 So ( 0, 0) dy d = = 0 = no solution; dy = does not eist = 0, 2, 2 d y( 0) = 0 y ( ±2) DNE is the only etreme point. 46

39 5. Range: y (, 0 6. The domain of this function is ( 2, 0 ( 2, ). EB on the left: EB on the right: None lim = 0 Therefore, there is a horizontal asymptote at y = 0 on the right. 7. VA: = ±2 8. POE: None 9. Sketch: y y =

40 EX 5 Find the traits and sketch y = on 2, 7. As we saw in EX 3, here are the traits and sketch without the arbitrary domain: Domain: 3, 3 4, ) Zeros: ( ±3, 0) and ( 4, 0) y-int: ( 0, 6) Etreme Points: ( 0.852, 6.336), ( 3, 0), ( 3, 0), and ( 4, 0) Range: y (, 0 End Behavior: Left end none, right end down With the arbitrary domain, the traits become Domain: 2, 3 4, 7 ) Zeros: ( 3, 0) and ( 4, 0) 463

41 y-int: ( 0, 6) Etreme Points: ( 0.852, 6.336), ( 2, 5.477), ( 3, 0), 4, 0 and 7, Range: y 0.954, 0, End Behavior: None And the sketch becomes 464

42 8-4 Free Response Homework. Given the traits below, sketch a graph of the function. Domain:.2, 2 ( 3, 5 ) 5, Zeros: (.2, 0), ( 2, 0) VA: = 3 POE: 5, 6 y-int: 0, 2 Etreme Points: (.2, 0), (, 2.5), ( 2, 0), 4, 5.8 End Behavior: Left: None Right: y = Range: y 0, , ) 2. Given the traits below, sketch a graph of the function. Domain: 5, 3 (, ) 3, 5 Zeros: ( 5, 0), ( 3, 0), ( 5, 0), 3, 0 VA: = ± POE: None y-int: 0, 3.5 Etreme Points: ( 5, 0), ( 4, 2), ( 3, 0), ( 0, 3.5), ( 3, 0), ( 4, 2), 5, 0 End Behavior: Left: None Right: None Range: y 0, 2 3.5, ) Find the traits and sketch. 3. y = y = y =

43 6. y = 5 on 0, 5 7. y = on 4, 4 8. y = y = y = y = on 5, 3 2. y = Multiple Choice Homework. Find the absolute maimum value of y = 8 2 on the interval 3, 3. (a) 0 (b) 0 (c) 9 (d) 8 (e) 2. The graphs of the functions f and g are shown in the figures. If h( ) = g( f ( ) ), which of the following statements are true about the function h? I. h( 0) = 4 II. h is increasing at = 2 III. The graph of h has a horizontal tangent at = 4. (a) I only (b) II only (c) I and II only 466

44 (d) II and III only (e) I, II, and III 3. The table below gives the values of at selected values of f, f, g, and g. If h( ) = f ( g( ) ), then h ( ) = f ( ) g( ) g ( ) f (a) 5 (b) 6 (c) 9 (d) 0 (e) 2 467

45 8-5: Implicit Differentiation In the Chain Rule section, composite functions were denoted as f ( g( ) ). But if y is a function of, epressions like y 3 or 4y 3 +8y 5 are also composite functions, 3. because y = g( ). For eample, if y = + 2 +, then y 3 = LEARNING OUTCOMES Take derivatives of relations implicitly. Use implicit differentiation to find higher order derivatives. EX Find d d y5. Note that in d d dy y5 But for d d d y5 d y5, the derivative in terms of. If it were in terms of y, or, the derivative would be simple: d y5 d dy y5 = 5y4, it is implied that y is a function of. By the chain rule, should equal 5y4 times the derivative of y. The actual function that y equals is unknown, but its derivative is dy d. So, d d y5 = dy 5y4 d In terms of functions, this may not be very interesting or important. But consider a non-function, like this circle. 468

46 EX 2 Find dy d if 2 + y 2 = 25. d d 2 + y 2 = 25 d + d d y dy d = 0 = d d y2 d 25 Now isolate dy d 2y dy d = 2 dy d = y EX 3 Find dy d for the hyperbola 2 3y y 2 = 0. d d 2 3y y 2 = 0 2 6y dy dy d d = = 6y dy dy +2 d d dy = 6y +2 dy d = y +2 = + 2 3y + 6 d 469

47 8-5 Free Response Homework Find dy d for each of these relations y y 5= 0 2. y y + 4 = y y + 4 = y y + 44 = 0 5. y 2 = = y + y Find the line tangent to each of these conics at the given point y y +58 = 0 at ( 0, 8) y + 25= 0 at ( 5, 0) y y + 9 = 0 at ( 3, 3.6) 0. 2 y 2 6y 3= 0 at ( 3, 0) y y + 4 = 0 at ( 0, 2) y y + 44 = 0 at (, 3) 470

48 8-5 Multiple Choice Homework. If y 2 3 = 7, then dy d = (a) 6 7 y 3 (b) 3 y 3 (c) 3 (d) 3 2y (e) 9 4y 3 2. For what values of does the curve y = 8 have horizontal tangent lines? (a) = 0 only (b) = 0 only (c) = 0 only (d) = 0 and = 0 (e) = 0, = 0, and = 0 3. An equation of the tangent line to the curve 2 + y 2 = 69 at the point ( 5, 2) is (a) 5y 2 = 20 (b) 5 2y = 9 (c) 5 2y = 69 (d) 2 + 5y = 0 (e) 2 + 5y = 69 47

49 8-6: Related Rates Derivatives have primarily been interpreted as the slope of the tangent line. But, as with rectilinear motion, there are other contets for the derivative. One overarching concept is that a derivative is a Rate of Change. The tendency is to think of rates as distance per time unit, like miles/hour or feet/second, but even slope is a rate of change it is just that rise and run are both measured as distances. LEARNING OUTCOME Solve related rates problems. The idea behind related rates is two-fold. First, change is occurring in two or more measurements that are related to each other by the geometry (or algebra) of the situation. Second, an implicit chain rule situation eists in that the and y-values are functions of time, which may or may not be a variable in the problem. Therefore, when taking the derivative of an or y, an Implicit Rate Term d dt dy or dt occurs. A classic problem is the falling ladder. 472

50 EX A 3 foot tall ladder is leaning against a wall. The bottom of the ladder slides away from the wall at 4 ft/sec. How fast is the top of the ladder moving down the wall when the ladder is 5 feet from the wall? y 3' As can be seen in the picture, the height of the top of the ladder and the distance the bottom of the ladder is from the wall are related by the Pythagorean theorem. Both are variables, because the ladder is moving. Therefore, 2 + y 2 =3 2 4 ft/sec is the rate at which the -value is changing i.e. d dt differentiate 2 + y 2 =3 2 to get 2 d dy + 2y dt dt = 0. To find dy dt, This is essentially an equation in four variables. But and d dt and y = 2 (by the Pythagorean theorem). So, 2( 5) 4 2 d dy + 2y dt dt = 0 + 2( 2) dy dt = 0 are known, dy dt = 5 3 ft / sec It should make sense that dy dt falling. is negative since the top of the ladder is 473

51 Another common related rate problem is where a tank of a particular shape is filling or draining. EX 2 A tank shaped like an inverted cone 8 feet in height and with a base diameter of 8 feet is filling at a rate of 0 ft 3 /minute. How fast is the height changing when the water is 6 feet deep? 4 r 8 h The units on the 0 tell us that it is the change in volume, or dv dt. The volume of a cone is V = 3 πr 2 h. But this equation has too many variables for us to differentiate it as it stands. Since the rate of change of the height i.e., dh was what the question is, eliminate the r from the equation. By dt similar triangles, r h = 4 8 and r = h. Substitution gives a volume equation 2 in terms of height only: V = 3 π 2 h = π 2 h3 2 h 474

52 Differentiate and plug into to solve for dh dt. V = π 2 h3 dv dt = π dh 4 h2 dt 0 = π 4 6 dh dt = 0 9π 2 dh dt ft/ min 475

53 8-6 Free Response Homework. A spherical balloon is being inflated so that its volume is increasing at a rate of 6 ft 3 /min. How fast is the radius changing when r = 0 ft? 2. A 25-foot tall ladder is leaning against a wall. The bottom of the ladder is pushed toward the wall at 5 ft/sec. How fast is the top of the ladder moving up the wall when it is 7 feet up? 3. A circular ink stain is spreading (i.e. the radius is changing) at half an inch per minute. How fast is the area changing when the stain has a -inch diameter? 4. A cylindrical tank of height 30' and radius 0' is leaking at a rate of 300 ft 3 /min. How fast is the oil level dropping? 5. Two cars approach an intersection, one traveling south at 20 mph and the other traveling west at 30 mph. How fast is the direct distance between them decreasing when the westbound car is 0.6 miles and the southbound car is 0.8 miles from the intersection? 6. Sand is dumped onto a pile at 30 ft 3 /min. The pile forms a cone with the height always equal to the base diameter. How fast is the base area changing when the pile is 0 feet high? 7. According to Boyle s Law, gas pressure varies directly with temperature and inversely with volume or P = kt V. Suppose that the temperature is held constant while the pressure increases at 20 kpa/min. What is the rate of change of the volume when the volume is 600 in 3 and the pressure is 50 kpa? 8. Water is siphoned out of an equilateral triangular prism of length 5' and triangular base edges 2' at a rate of 2 ft 3 /min. What is the rate of change of the height of the water when the trough is 25% full? 2' 2' 2' 5' 476

54 8-6 Multiple Choice Homework. Gravel is being dumped from a conveyor belt at a rate of 35 ft 3 /min and its coarseness is such that it forms a pile in the shape of a cone whose base diameter and height are always equal. How fast is the height of the pile increasing when the pile is 5ft high? (a) 0.27 ft/min (b).24 ft/min (c) 0.4 ft/min (d) 0.2 ft/min (e) 0.6 ft/min 2. Two cars start moving from the same point. One travels south at 28 mi/h and the other travels west at 70 mi/h. At what rate is the distance between the cards increasing 5 hours later? (a) mi/h (b) mi/h (c) 76.4 mi/h (d) mi/h (e) mi/h 3. The radius of a sphere is decreasing at a rate of 2 centimeters per second. At the instant when the radius of the sphere is 3 cm, what is the rate of change, in square centimeters per second, of the surface area of the sphere? (The surface area S of a sphere with radius r is S = 4πr 2.) (a) 08π (b) 72π (c) 48π (d) 24π (e) 6π 477

55 4. Water is flowing into a spherical tank with 6-foot radius at the constant rate of 30π cu ft per hour. When the water is h feet deep, the volume of the water in the tank is given by V = π h2 ( 3 8 h ). What is the rate at which the depth of the water in the tank is increasing the moment when the water is 2 feet deep? (a) 0.5 ft/hour (b).0 ft/hour (c).5 ft/hour (d) 2.0 ft/hour (e) 2.5 ft/hour 5. The radius of a circle is in increasing at a constant rate of 0.2 meters per second. What is the rate of increase in the area of the circle at the instant when the circumference of the circle is 20π meters? (a) 0.04π m 2 /sec (b) 0.4π m 2 /sec (c) 4π m 2 /sec (d) 20π m 2 /sec (e) 00π m 2 /sec 6. If the rate of change of a number with respect to time t, is, what is the rate of change of the reciprocal of the number when = 4? (a) 6 (b) 4 (c) 48 (d) 48 (e) 4 478

56 7. For what values of does the curve y = 0 have horizontal tangents? (a) = 0 only (b) = 0 only (c) = 0 only (d) = 0 and = 0 (e) = 0, = 0, and = 0 Bonus (this one is tough!) 8. Let f be a differentiable function such that f 3 and f 6 = 5, f ( 6) = 3, f ( 3) = 8, = 2. The function g is differentiable and g( ) = f ( ) for all.? What is the value of g 3 (a) 2 (b) 8 (c) (d) 6 3 (e) 3 479

57 Radical Functions Practice Test Part : CALCULATOR REQUIRED Round to 3 decimal places. Show all work. Multiple Choice (3 points each). Use the linear approimation of the function f = 9 at a = 0 to approimate the number (a) 3.02 (b) 0.5 (c) 7.44 (d) 7.40 (e) Find the tangent to the semicircle y = 4 2 at the point (, 3). (a) y = (b) y = (c) y = (d) y = (e) None of these Free Response (0 pts. each). Find the zeros and domain of y = Find the zeros and domain of y =

58 3. Make a sign pattern, find the intervals of increasing, and find the etreme 3 points for y = List all traits and sketch y = Domain: Zeros: y-int: VAs: EB: POEs: Etreme Points: Range: 48

59 Radical Functions Practice Test Part 2: NO CALCULATOR ALLOWED Round to 3 decimal places. Show all work. Multiple Choice (3 pts. each) 3. lim (a) 4 (b) (c) 4 (d) 0 (e) 4. If d d f = g (a) 4 2 f ( 3 2 )+ 2g 2 (b) f 3 2 and g( ) (c) f 4 (d) 2 f ( 3 2 )+ 2g 2 (e) 2 f 3 2 = f 3, then d d f 2 is 482

60 Free Response (0 pts. each) 5. List all traits and sketch y = Domain: Zeros: y-int: VAs: POEs: EB: Etreme Points: Range: 483

61 Radical Functions Homework Answer Key 8- Multiple Choice Homework. D 2. B 3. B 4. C 5. E 6. G 7. E 8- Free Response Homework. Zeros: ( 3, 0), ( 3, 0) Domain: 3, 3 2. Zeros: ( 3, 0), ( 7, 0) Domain: (, 3 7, ) 3. Zeros: ± 6, 0 Domain: (, 6 + 6, ) 4. Zeros: ( 5, 0), ( 2, 0), 4, 0 Domain: 5, 2 4, 5. Zeros: (, 0), 3, 0 Domain: { } 3, ) ) 6. Zeros: (, 0), ( ± 5, 0), 2, 0 Domain: (, 5 2, 5, ) Domain: 2, 0 7. Zeros: 0, 0 ( 2, 8. Zeros: ( 0, 0), ( 3, 0) Domain: (, 3 (, 0 ( 3, ) 8-2 Multiple Choice Homework. B 2. E 3. D 4. B 5. B 6. B 7. E 8. E 8-2 Free Response Homework. 2. ( ) 2 ( 3 + 2)

62 ( ) ( 4 2 ) ( 5 3) ( 3 2 ) ( 25 2 ) ( 5 + 6) ( 2 2 ) ( 2 9) f ( 5) = ( 2 +) ( ) 4 ( 8 5) ( 3 ) Multiple Choice Homework. B 2. A 3. B 4. E 5. D 6. E 7. D 8. C 485

63 8-3 Free Response Homework. Domain:, 5 3 c.v.: = Domain: (, 5, c.v.s: =, 5 3. Domain: 3, 4 9 c.v.s: = 4, ± 3, ) 3, ) 4. Domain: 3, 2 2, 3 c.v.s: = ± 3 2, ± 3, ± 2 5. Domain: (, 9 c.v.: = 9 6. c.v.s: = 0,, Domain: (, 2), 2, 7. c.v.s: = 0, ± 2 Domain: (, 2 (, ) 2, 8. Domain: (, 6 2, 2 6, ) c.v.s: = 0, ± 2, ± 6 ) DNE DNE + 0 DNE DNE + y' Min Min Ma Min Min 9. Domain:.5,.5 c.v.s: = 0, ± 3 2 DNE + 0 DNE y' Min Ma Min 0, 3 0. Domain:, 3 c.v.s: = 0, ± 3. v( 2) = ) + DNE DNE + DNE y' VA Min VA = = miles Cost = $ , 0 486

64 7. ( 0.86,.02), ( 0.86,.02) y 0 8. ( 0, 6), ( ±4, 0) None ( 2, 0) 2. (, 0) y 22. ( ±3, 0), ( 0, 2) 23a. 2, 3 4 b. 0.77, a. t ( 5, 0) b. Never at rest 8-4 Multiple Choice Homework. C 2. D 3. D 8-4 Free Response Homework Domain: (, 5, Range: y 0, ) Zeros: (, 0), 5, 0 2 ) y-int: None EB: Both ends up VA: None POE: None Etreme Points: (, 0), ( 5, 0) y

65 4. Domain:.5,.5 Range: y 3, 0 Zeros: (.5, 0),.5, 0 y-int: 0, 3 EB: None VA: None POE: None Etreme Points: ( 0, 3), (.5, 0), (.5, 0) y Domain: 0, 5 Range: y 0, 5 Zero: 5, 0 y-int: 0, 5 EB: None VA: None POE: None Etreme Points: 5, y, 0, 5 5. Domain: 5, Range: y 0, 3 Zeros: ( 5, 0),, 0 y-int: 0, 5 EB: None VA: None POE: None Etreme Points: 2, 3 ( 5, 0), (, 0), 488

66 7. Domain: 4, 2, 4 Range: y 0, Zeros: ( 4, 0), (, 0), 2, 0 y-int: None EB: None VA: None POE: None Etreme Points: 4, 0 ( 2.732, 3.224), (, 0), ( 2, 0), ( 4, 8.994) 8. Domain: (, 3 0, 2 ( Range: y, 0 Zeros: ( 3, 0), ( 0, 0), 2, 0 y-int: 0, 0 EB: Left end down, right end none VA: None POE: None Etreme Points: ( 3, 0), ( 0, 0), (.20, 2.05), 2, Domain: (, 2), ( 2, ) Range: y (, ) 2, 0 Zeros: (, 0), (, 0) y-int: 0, 2 VA: = 2, = 2 POE: None EB: HA y = Etreme Points: 0, 2 (, 0), (, 0), y y 489

67 0. Domain:, 2, 2 Range: y 0, Zeros: ) 2, 0 y-int: 0, 2 VA: =, = 2 POE: None EB: HA y = 0 on right, left end none Etreme Points: 2, 0. Domain: 5, y 5, 3 Range: y , 0 Zeros: ± 5, 0 y-int: None VA: None POE: None EB: None, Etreme Points: ± 5, 0 ( 5, ), 3, Domain: 3, 2 2, 3 Range: y 0, 2.5 Zeros: ( 3, 0), ( 2, 0), ( 2, 0), ( 3, 0) y-int: None VA: None POE: None EB: None Etreme Points: ( 3, 0), ( 2, 0), ( 2, 0), ( 3, 0), 3 2, 2.5, 3 2, y 490

68 8-5 Multiple Choice Homework. D 2. D 3. C 8-5 Free Response Homework y + 2 y y 3( + 3) 2 y 2 y + ( + y) 2 y 7. y 8 = 3( 0) 8. y = 0 9. y 3.6 = 3 ( 20 3 ) 0. y = 3 3 ( 3 ) = 3 3. y + 2 = 3 ( 2 0 ) 2. y + 3= 6 ( 5 + ) 8-6 Multiple Choice Homework. D 2. D 3. C 4. C 5. C 6. E 7. D 8. A 8-6 Free Response Homework π ft/min 20 7 ft/sec π 2 in2 /min 3 π ft/min mph 6. 6 ft 2 /min in 3 /min ft/min 49

69 Radical Functions Practice Test Answer Key Multiple Choice. A 2. B 3. C 4. E Free Response. Zeros: ( 2, 0), (, 0), 2, 0 Domain: 2, 2, 2. Zeros: None Domain: 3, 3. y DNE 3 No etreme points No intervals of increasing 4. Domain: 2, 2, Range: y (, 0 Zeros: ( 2, 0), (, 0), 2, 0 y-int: None EB: Left end none, right end down VA: None POE: None Etreme Points: ( 2, 0), (.560,.008), (, 0), 2, Range: y ( 0, ) 5. Domain: 3, Zeros: None y-int: None EB: Left end none, HA y = 0 on right VA: = 3 POE: None Etreme Points: None y y 492