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Mdeling the Dynamics f Life: Calculus and Prbability fr Life Scientists Frederick R. Adler cfrederick R. Adler, Department f Mathematics and Department f Bilgy, University f Utah, Salt Lake City, Utah 84

5 3. Supplementary Prblems fr Part 3 Fr the functins shwn: a. Sketch the derivative. b. Label lcal and glbal maima. c. Label lcal and glbal minima. d. Subsets f the dmain with psitive secnd derivative. EXERCISE 3. 4.5 4 3.5 y 3.5.5.5.5.5 3 EXERCISE 3..5 y.5..4.6.8. Use the tangent line and the quadratic Taylr plynmial t nd apprimate values f the fllwing. Make sure t write dwn the functin r functins yu use and the equatin f the tangent line. Check with yur calculatr. EXERCISE 3.3 =(3 + : ). EXERCISE 3.4 e 3(:) +(:). Write dwn the tangent line apprimatin fr the fllwing functins and estimate the requested values. EXERCISE 3.5 f() = + + e 3. Estimate f( :3). EXERCISE 3.6 g(y) = ( + y) 4 ln(y). Estimate g(:). Sketch graphs f the fllwing functins. Find all critical pints, and state whether they are minima r maima. Find the limit f the functin as!.

3.. SUPPLEMENTARY PROBLEMS FOR PART 3 53 EXERCISE 3.7 ( + )e fr psitive. EXERCISE 3.8 ln()=( + ) fr psitive. D nt slve fr the maimum, just shw that there must be ne. Find the Taylr plynmial f degree apprimating each f the fllwing. EXERCISE 3.9 f() = + + fr near. EXERCISE 3. f() = + + fr near. EXERCISE 3. g() = + + e fr near. EXERCISE 3. h() = e fr near. Cmbine the Taylr plynmials frm the previus set f prblems with the leading behavir f the functins fr large t sketch graphs EXERCISE 3.3 g(). EXERCISE 3.4 h(). EXERCISE 3.5 Between days and 5 (measured frm Nvember ), the snw at a certain ski resrt is given by S(t) = 4 t4 + 6t 3 4t + 96t where S is measured in micrns (ne micrn is 4 centimeters). a. A yetti tells yu that this functin has critical pints at t =, t = 4 and t =. Cnrm this assertin. b. Find the glbal maimum and glbal minimum amunts f snw in feet. Remember that inch =.54 centimeters. c. Use the secnd derivative test t identify the critical pints as lcal maima and minima. d. Sketch the functin. EXERCISE 3.6 Let r() be the functin giving the per capita prductin as as a functin f ppulatin size. per capita prductin = 4 + 3 a. Find the ppulatin size that prduces the highest per capita prductin. b. Find the highest per capita prductin. c. Check with the secnd derivative test. EXERCISE 3.7 An rganism is replacing 5% f the air in its lung each breath and the eternal cncentratin f a chemical is = 5: 4 mles/liter. Suppse the bdy uses a fractin f the chemical just after breathing. That is, the chemical fllws c t! ( )c t! mi f 75% used air and 5% ambient air

54 a. Write the discrete-time dynamical system fr this prcess. b. Find the equilibrium level in the lung as a functin f. c. Find the amunt absrbed by the bdy each breath at equilibrium as a functin f. d. Find the value f that maimizes the amunt f chemical absrbed at equilibrium. e. Eplain yur result in wrds. EXERCISE 3.8 An rganism is replacing a fractin q f the air in its lung each breath and the eternal cncentratin f a chemical is = 5: 4 mles/liter. Suppse the bdy uses a fractin = q f the chemical just after breathing. The chemical fllws c t absrptin! ( )c t breathing! mi f used and ambient air a. Write the discrete-time dynamical system fr this prcess. b. Find the equilibrium level in the lung as a functin f q. Is it stable? c. Find the amunt absrbed by the bdy each breath at equilibrium as a functin f q. d. Find the value f q that maimizes the amunt f chemical absrbed at equilibrium. e. Eplain yur result in wrds. EXERCISE 3.9 Suppse the vlume f a plant cell fllws V (t) = ( e t )m 3 fr t measured in days. Suppse the fractin f cell in a vacule (a water-lled prtin f the cell) is H(t) = e t =( + e t ). a. Sketch a graph f the ttal size f the cell as a functin f time. b. Find the vlume f cell utside the vacule. c. Find and interpret the derivative f this functin. Dn't frget the units. d. Find when the vlume f the cell utside the vacule reaches a maimum. EXERCISE 3. Cnsider the functin a. What is lim t! F (t)? b. What is lim t! F (t)? c. What is lim t! F (t)? d. Sketch a graph f this functin. F (t) = ln( + t) t + t : EXERCISE 3. During Thanksgiving dinner, the table is replenished with fd every 5 minutes. Let F t represent the fractin f the table laden with fd. Suppse that and that a = : and b = :5. F t+ = F t amunt eaten + amunt replenished : amunt eaten = bf t + F t amunt replenished = a( F t ) a. Eplain the terms describing amunt eaten and amunt replenished. b. If the table starts ut empty, hw much fd is there after 5 minutes? Hw much is there after minutes? c. Use the quadratic frmula t nd the equilibria. d. Hw much fd will there be ve minutes after the table is 6% full? Sketch the slutin.

3.. SUPPLEMENTARY PROBLEMS FOR PART 3 55 EXERCISE 3. Let N t represent the dierence between the sdium cncentratin inside and utside a cell at sme time. After ne secnd, the value f N t+ is 8 < : N t+ = :5N t if N t < N t+ = 4:N t 7 if < N t < 4 N t+ = :5N t + if 4 < N t : a. Graph the updating functin and shw that it is cntinuus. b. Find the equilibria and their stability. c. Find all initial cnditins which end up at N =. EXERCISE 3.3 Cnsider lking fr a psitive slutin f the equatin e = + : a. Draw a graph and pick a reasnable starting value. b. Write dwn the Newtn's methd iteratin fr this equatin. c. Find yur net guess. d. Shw eplicitly that the slpe f the updating functin fr this iteratin is zer at the slutin. EXERCISE 3.4 Cnsider trying t slve the equatin ln() = 3 : a. Cnvince yurself there is indeed a slutin and nd a reasnable guess. b. Use Newtn's methd t update yur guess twice. c. What wuld be a bad chice f an initial guess? EXERCISE 3.5 Suppse a bee gains an amunt f energy F (t) = 3t + t after it has been n a wer fr time t, but that it uses t energy units in that time (it has t struggle with the wer). a. Find the net energy gain as a functin f t. b. Find when the net energy gain per wer is maimum. c. Suppse the travel time between wers is =. Find the time spent n the wer that maimizes the rate f energy gain. d. Draw a diagram illustrating the results f parts b and c. Why is the answer t c smaller? EXERCISE 3.6 Cnsider the functin fr net energy gain frm the previus prblem. a. Use the Etreme Value Therem t shw that there must be a maimum. b. Use the Intermediate Value Therem t shw that there must be a residence time t that maimizes the rate f energy gain. EXERCISE 3.7 A peculiar variety f bacteria enhances its wn per capita prductin. In particular, the number f spring per bacteria increases accrding t the functin Suppse that r = :5 and that K = 6. per capita prductin = r( + b t K ):

56 a. Graph per capita prductin as a functin f ppulatin size. b. Find the discrete-time dynamical system fr this ppulatin and graph the updating functin. c. Find the equilibria. d. Find their stability. EXERCISE 3.8 A type f buttery has tw mrphs, a and b. Each type reprduces annually after predatin. % f type a are eaten, and % f type b are eaten. Each type dubles its ppulatin when it reprduces. Hwever, the types d nt breed true. Only 9% f the spring f type a are f type a, the rest being f type b. Only 8% f the spring f type b are f type b, the rest being f type a. a. Suppse there are, f each type befre predatin and reprductin. Find the number f each type after predatin and reprductin. b. Find the discrete-time dynamical systems fr types a and b. c. Find the discrete-time dynamical system fr the fractin p f type a. d. Find the equilibria. e. Find their stability. EXERCISE 3.9 A ppulatin f size t fllws the rule per capita prductin = 4 t + 3 : t a. Find the updating functin fr this ppulatin. b. Find the equilibrium r equilibria. c. What is the stability f each equilibrium? d. Find the equatin f the tangent line at each equilibrium. e. What is the behavir f the apprimate dynamical system dened by the tangent line at the middle equilibrium? EXERCISE 3.3 Cnsider a ppulatin fllwing the discrete-time dynamical system N t+ = rn t + Nt : a. What is the per capita prductin? b. Find the equilibrium as a functin f r. c. Find the stability f the equilibrium as a functin f r. d. Des this psitive equilibrium becme unstable as r becmes large?

Chapter 9 Answers 3.. y Derivative 3.5.5.5 -.5 - -.5.5.5.5 3 y 4.5 4 3.5 Minima, maima and curvature 3 secnd secnd derivative derivative.5 psitive negative lcal maimum lcal maimum.5 lcal minimum glbal minimum.5.5.5 3 glbal maimum secnd derivative psitive 3.. y Derivative 3.5.5.5 -.5 -..4.6.8. y.5 secnd derivative psitive Minima, maima and curvature secnd derivative negative glbal minimum lcal maimum.5..4.6.8. secnd derivative psitive glbal maimum lcal minimum 3.3. Let f() = =(3 + ). Then f () = =(3 + ). Substituting =, we nd f() = =4 and f () = =8, s ^f() = =4 =8( ), and ^f(:) = =4 =8(: ) = :4875. 3.4. Let f() = e 3 +. Then f () = (6 + )e 3 +. Substituting =, we nd f() = e 5 and f () = 8e 5, s ^f() = e 5 + 8e 5 ( ) and ^f(:) = e 5 + 8e 5 (: ) = :6e 5. 3.5. ^f() = :5 :5. ^f( :3) = :575. In this case, f( :3) :568. 3.6. ^g(y) = 8(y ). ^g(:) = :6. 3.7. f () = ( )e which is at = p. Because f() = and lim! f() = (epnential declines faster than quadratic increases), this must be a maimum. 79

8 CHAPTER 9. ANSWERS.5 f().5 3 4 5 3.8. This functin is negative fr <, zer at = and psitive fr >. Als, its limit at innity is zer. It must indeed have a psitive maimum between and innity. f() - 3.9. We have that f () = - 4 6 8 ( + ) ; f () = ( 3 + 3 + 3 ) ( + ) 3 : Therefre, f () = and f () =, and ^f() = +. 3.. Using the derivative in the previus prblem, f () = = and f () =, s ^f() = ( ). 3.. g () = e ( + e ) ; g () = e ( 3 + e e ) ( + e ) 3 : Then g() = =, g () = =4 and g () = = and ^g() = + 4 4. 3.. h () = e + e ( + e ) ; h () = e ( 4 e + e ) ( + e ) 3 : Then h() =, h () = and h () = and ^h() = +. 3.3..7.6 g().5.4.3.. slpe = /4 3 4 5 appraches

8 3.4. h() 5 4 blws up 3 slpe = appraches - - -3-4 -5 3 4 5 3.5. a. S (t) = t 3 + 8t 8t + 96, which is at t =, t = 4 and t =. b. Substituting the endpints (t = and t = 5) and the critical pints int the functin S(t), we nd a maimum f 576, at t =. c. S (t) = 3t 3 + 36t 8. Then S () =, S (4) = 6, and S () = 8. The rst and last are maima and the middle ne is a minimum. 6e+6 5e+6 4e+6 S(t) 3e+6 e+6 e+6 d. 4 6 8 4 t 3.6. a. If r() = 4 +3, then r () = 4(3 ) ( + 3 ) : The critical pints are then at = p =3. Only psitive values make sense, s we need nly cnsider :577. Because r() = and lim! r() =, this must be a maimum. b. The value is = p 3 :55. c. The secnd derivative is r () = 7( ) ( + 3 ) which is negative at = p =3. 3.7. a. c t+ = :75( )c t + :5. b. c = :5=(:5 + :75). c. c = :5=(:5 + :75).

8 CHAPTER 9. ANSWERS d. This has its maimum at =. e. Sure, absrb as much as yu can as fast as yu can. 3.8. a. c t+ = ( q)( )c t + q. b. c = q=(q + ( q). Denitely stable because the slpe f the updating functin is ( q)( ) <. c. c = q=(q + ( q). d. Taking the derivative with respect t q, we nd the derivative is ( + q q) which is always psitive. The maimum is at q =. e. Sure, absrb as much as yu can by breathing as deeply as yu can. 3.9. 8 V(t) 6 4 a. 3 4 5 b. The fractin utside is H(t) = =( + e t ), s the ttal vlume utside is ( e t ) + e t : t c. Call this functin V (t). V (t) = ( + e t e t ) ( + e t ) : d. This is a bit tricky. The maimum is the critical pint where V (t) =, r where +e t e t =. Letting = e t, this is + = =. Multiplying bth sides by, we get the quadratic + =, which can be slved with the quadratic frmula t give = + p, s that t = ln( + p ) :88 days. 3.. a. lim t! F (t) =. b. lim t! F (t) = 3. c. lim t! F (t) = 3.

83.5 F(t).5 - -.8 -.6 -.4 -...4.6.8 t 3.. a. The mre fd there is, the mre is eaten, up t a limit f b. The replenishment is enugh t rell the table. b. The system is F t+ = F t :5F t + F t + F t = :5F t + F t : If F =, then F = and F = :5 = :5. c. Slving fr the equilibrium F with F = :5F + F F ( + F ) = + F :5F F + (F ) = :5F :5F + (F ) = (F :5)(F + ) = : The nly psitive equilibrium is at F = :5. d. G(:6) = :4375..8.6 F t.4. 3 4 5 t 3.. a. Each piece is cntinuus, being linear. And at the jump pints, they match. That is, at N t =, :5N t = 4N t 7 =, and at N t = 4, 4N t 7 = :5N t + = 9.

84 CHAPTER 9. ANSWERS 8 N t+ 6 4 4 6 8 N t b. Alng the rst piece, there is an equilibrium at N =, which is stable because the slpe is.5. The diagnal intersects the secnd piece at N = :333, anther equilibrium. This ne is unstable. The third piece hits the diagnal at N = 8, a stable equilibrium with negative slpe f -.5. c. Anything belw.333 denitely des, because it just decreases. And anything that drps belw.333 n the secnd step, r if :5N t + < :333 r N t > 3:67. All the values between.333 and 3.67 end up at the 8. 3.3. a. It lks like = is a gd guess. 5 5 e + crssing pint b. We need t slve g() = e =, s c. = :39. d. The derivative f the updating functin is which is when e =. 3 t+ = t g( t) g ( t ) = t e e : e (e ) (e ) ; 3.4. a. At =, the left hand side is smaller. At = e, the left hand side is larger. And at = e, the left hand side is smaller again. There must be at least tw slutins, ne between and e and anther between e and e. T nd the lwer ne, I bet that is a gd guess.

85 b. Our functin is f() = ln() =3. The Newtn's methd discrete-time dynamical system is t+ = t f( t) f ( t ) = t ln( t) t =3 = t =3 : Substituting =, we nd = :84, and then that = :857. T ve decimal places, the eact answer is.857. c. f (3) =. This wuld be a rather bad guess. 3.5. a. Let N(t) be net energy gain. Then N(t) = F (t) t. b. When N (t) = r F (t) = 3 p ( + t) =. This has slutin t = 3= :5. c. We need t maimize N(t) + t, which ccurs when t = :. d. The answer t c is smaller because the bee has ther ptins besides sucking as much nectar as pssible ut f the wer...8 N(t).6.4. - -.8 -.6 -.4 -...4 t 3.6. a. N() = as des N(:5). Because the values are psitive in between, there must be a maimum. b. The derivative f N(t) + t is 5t 3.7. ( + t) which is psitive fr t = and negative fr t =. There must be critical pint in between, which must be a maimum because the functin switches frm increasing t decreasing. 3 per capita prductin.5.5 b t+.5.5.5 a..5.5 b.5.5 b t

86 CHAPTER 9. ANSWERS b. The discrete-time dynamical system is b t+ = rb t ( + bt K ). c. Equilibria at b = and b = 6. d. Equilibrium at b = is stable and equilibrium at b = 6 is unstable. 3.8. a. Fr type a: there are 8, after predatin, they reprduce t make 6,, 9% r 4,4 f which are a's and % r 6 f which are b's. Fr type b: there are 9, after predatin, they reprduce t make 8,, 8% r 4,4 f which are b's and % r 36 f which are a's. b. a t+ = :9 :8a t + : :9b t = :44a t + :36b t. b t+ = : :8a t + :8 :9b t = :6a t + :44b t. c. d. p = :6. e. f (p ) = :74 and the equilibrium is stable. p t+ = :44p t + :36( p t ) :6p t + :8( p t ) : 3.9. a. The updating functin f is f() = 4 =( + 3 ), multiplying the per capita prductin by, the number f individuals. b. First, nd that is a slutin. The rest is a quadratic, which has slutins at = =3 and =. c. We nd that f () = 8=( + 3 ). Then f () =, f (=3) = 3= and f () = =. Therefre, the equilibria at and are stable and the ne at /3 is unstable. d. The tangent line at is ^f() =. The tangent line at = =3 is ^f() = =3 + 3=( =3). The tangent line at = is ^f() = + =( ). e. This dynamical system shts t innity fr starting pints greater than /3, and t negative innity fr starting pints less than /3. This is dierent frm the behavir f the riginal system, in which such slutins apprach = and = respectively. 3.3. a. It is r. + Nt b. N = p r. c. The derivative at the equilibrium is r r. d. N, this is always greater than -.