HOMEWORK SET 4 SOLUTIONS. 1. Problem (8ex each) Prove or disprove the following identities. (a)2 3 n n = n

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1 HOMEWORK SET 4 SOLUTIONS MATH Problem (8ex each) Prove or disprove the following identities. (a)2 3 n n = n (b)4 n+1 5 n 4 n 5 n+1 = 4 n 5 n (c)(e t ) 2 = e 2t (e) em+t e = e 2t m t (f)e (t µ)=eµ e t (g) 1 1+e (t µ) = e µ e µ +e t (h)(e t + e t ) 2 (e t e t ) 2 = 4 Definition to analyze the graph of a function means to find its maxima, minima, to find its zeros, to describe its behavior (increasing, decreasing), find its vertical and horizontal asymptotes, and to find its behavior at and. 2. Problem (a) (3ex) Sketch the graph of y = t e t. (b) (8ex) Analyze the graph. (c) (3ex) Sketch the graph of y = t 2 e t. (d) (8ex) Analyze the graph. (e) (10ex) Sketch the graphs of y = t k e t, What is it? k = 1, 2, 3,.... Do you see a pattern? 3. Problem (a) (3ex) Sketch the graph of y = t e t2. (b) (8ex) Analyze the graph. (c) (3ex) Sketch the graph of y = t 2 e t2. (d) (8ex) Analyze the graph. 1

2 2 MATH 1113 (e) (10ex) Sketch the graphs of y = t k e t2, k = 1, 2, 3,.... Do you see a pattern? What is it? 4. Problem Let P (t) = 1 1+e, t > 0. t a) (3ex) Calculate P (0). P (0) = 1 2 b) (3ex) Graph the function. c) (8ex) Analyze the graph. d) (5ex) Solve the equation P (t) = 1. Solution:N osolution. e) (5ex) Solve the equation P (t) = 2. Solution:t = 0 f) (10ex) Let Q(t) = 2 2+e. Graph it. Notice that the graph of Q(t) is just a t translation of the graph of P (t). Recall that if you transform the function P (t) into P(t-2)the effect is a translation of the graph for the two units to the right. Using this fact and the identity (f) in Problem 1. find for how much we have to translate the graph of P (t) to obtain the graph of Q(t), i.e. find m such that P (t m) = Q(t). Solution:m = ln(2) 5. Problem (12ex) Recall the Euler formula for exponential function: e x = 1 + x 1! + +x2 2! + x3 3! + x4 4! + x5 5! + Calculate 3 e 2 using the Euler formula to the precision of 5 decimals. Solution: Plug 2 3 for x and calculate sums adding term by term till first 5 decimals are fixed. 6. Problem (10ex)Recall the Euler formula for the exponential function: e x = 1 + x 1! + +x2 2! + x3 3! + x4 4! + x5 5! + Use this to find similar formula, i.e., the series for e x2. Solution: Plug x 2 for x in the Euler formula. 7. Problem (8ex each)prove the identities (a) cosh(x) + sinh(x) = e x (b) cosh(x) sinh(x) = e x (c) sinh(x + y) = sinh(x)cosh(y) + sinh(y)cosh(x) (d) cosh(x + y) = cosh(x)cosh(y) + sinh(x)sinh(y) (e) sinh(2x) = 2sinh(x)cosh(x)

3 HOMEWORK SET 4 3 (f) cosh(2x) = cosh 2 (x) + sinh 2 (x) 8. Problem (10ex) Two telephone poles are erected 60ft apart. Each pole is 30ft height. The wire connecting the poles is 27ft from the ground at the lowest point. Model the shape of the wire using the function Y (x) = b cosh(ax) where a and b are parameters to be determined. Find a and b and sketch the graph. Solution: b = 27 and a = 1 30 ln ( 10± 19 9 ) 9. Problem (20ex)Do the previous problem but in general case. Let the poles be l ft apart, their height h ft and the wire( hanging d ) ft from the ground at the lowest point. Solution: b = d and a = 2 l ln h± h 2 d 2 d (a)(8ex) Let 10. Problem P (t) = A 1 + e λt. Find A and λ if P (0) = 0.25 and P (0.21) = are data obtained by some kind of measurement. Solution:A = 1 2 and λ = (b)(8ex) For which values of t the function P (t) > 0.45? Solution:t > Problem (a)(3ex)for which value of t the equation e Y = 4 t has a Y-solution. Solution: t < 4. (b)(8ex)for which value of t the equation has a Y-solution. Solution:t ( 2, 2) e Y = 4 t 2

4 4 MATH Problem Let P λ (t) = e λt for t 0). (a)(5ex) Given that P λ is at t=2, find λ. Solution:λ = 1 2 ln(0.165) (b)(5ex)using λ from (a)find for which t, P λ is smaller than Solve it graphically and algebraically. Solution:t > ln(0.05 ln(0.165) 13. Problem Let P A (t) = Ae t for t 0. (a)(5ex) Graph P A (t) for A = 1, 2, 3. What do you observe? (b)(3ex) Evaluate P A (0) for A = 1, 2, 3. Solution:P 1 (0) = 1, P 2 (0) = 2, P 3 (0) = 3 (c)(3ex) Evaluate P A (0) for general A. Solution: P A (0) = A (d)(5ex) Given that P A is 1.1 at t = 2, find A. Solution:A = 1.1e Problem Let P λ,a (t) = Ae λt for t 0. (a)(8ex) Find λ and A given that P λ,a is 0.5 at t = 1, and 0.25 at t = 2. Solution:A = 1 and λ = ln(2) (b)(8ex) For P λ,a with λ and A from (a) find its maximal value and where it is attained. Solution: maximum v alue = 1 at t = 0. Let 15. Problem p(t) = e t (1 + e t ) 2. (a)(3ex)find p(0). Solution:p(0) = 1 4 (b)(8ex)find the time t when p is 0.6. Solution:Never. (c)(8ex)what is the maximal value of p(t) and where is it attained? Solution:Graph it.t = 0, maximal value is 1 4 Prove the following ) identities: ( (a)(5ex) ln( Y1 Y 2 = ln Y2 Y 1 ). (b)(5ex) e λt e µt = e (λ+µ)t (c)(5ex) e λt 1 e = λt e λ(t1 t2) Problem

5 HOMEWORK SET 4 5 Solve the following equations. (a)(5ex) e t = 0.16 Solution:t = ln(0.16) (b)(5ex) e (t 2) = 2 Solution:t = 2 ln(2) 1 (c)(5ex) 1+e = 0.05 t Solution:t = ln(19) (d)(5ex) e t2 = 0.05 Solution:t = ± ln(0.05) (e)(5ex) cosh(x) = 4 Solution:x = ln(4 ± 15) (f)(5ex) sinh(2x 3) = 2 Solution:x = 3+ln(2+ 5) Problem 18. Problem Let R a (t) = ln(at) t 0 and a > 0. (a)(5ex) Graph R a for a = 1, 2, 3. What do you observe? (b)(5ex) Find all t for which R a has value 0. Do it for a = 1, 2, 3. (c)(8ex) For general a find all t for which R a has value Problem What is the natural domain of the following functions, i.e., find all t for which the following functions can be computed. (a)(5ex) ln(t) Solution:for all t > 0 (b)(5ex) ln(ln(t)) Solution:for all t > 1 (c)(5ex) ln( t) Solution:for all t < 0 (d)(5ex) ln(1 + t) Solution:for all t > 1 (e)(5ex) ln(2 + t) Solution:for all t > 2 (f)(5ex) ln(3 + t) Solution:for all t > 3 (g)(5ex) ln(a + t) where a is any real number Solution:for all t > a (h)(5ex) ln(t 2 ) Solution:R \ {0} (i)(5ex) ln(1 + t 2 ) Solution:R

6 6 MATH 1113 Let us define 20. Problem L 1 (x) = ln(x) L 2 (x) = ln(l 1 (x)) = ln(ln(x)) L 3 (x) = ln(l 2 (x)) = ln(ln(ln(x))) and in general L n (x) = ln(l n 1 (x)) = ln(ln(... ln(x)). }{{} n times (a)(8ex)solve the equations L n (x) = 0 for n = 1, 2, 3. Solution:for n = 1 x = 1, for n = 2 x = e, for n = 3 x = e 2 (b)(10ex)solve the equation L n (x) = 0 for a general n. Solution:x = exp(exp(exp( exp(e)))) }{{} n times 21. Problem Let N µ = e (x µ)2 2. (a)(8ex) Let µ = 6. For which values of x N 6 < 0.025? Solution:for x (, 6 2ln(0.025)) (6 + 2ln(0.025), ) (b)(8ex) Same as in(a) except for general µ,i.e., for which values of x N µ (x) < 0.025? Solution:for x (, µ 2ln(0.025)) (µ + 2ln(0.025), ) Show the following identities are true. (a)(5ex) ln(e t1 e t1 ) = t 1 + t 2 (b)(5ex) ln(e t e 1 t ) = Problem 23. Problem Prove the following properties of the logarithms. (a)(10ex) log(x k ) = k log(x) for x > 0. (b)(10ex) log( x y ) = log(x) log(y) for x > 0 and y > 0. (a)(8ex) Solve the equation 24. Problem e λt1 e λt2 = 1 2 t 1 + t 2 = 1. Solution:λ = 1 2 (b)(10ex) Solve the equation e λt1 e λt2... e λtn = 1 2

7 HOMEWORK SET 4 7 t 1 + t t n = 1. Solution:λ = 1 2 (a)(8ex) Solve the equation Solution:λ = ± e (b)(10ex) Solve the equation 25. Problem ln(λt 1 ) + ln(λt 2 ) = 1 t 1 t 2 = 1. ln(λt 1 ) + ln(λt 2 ) ln(λt n ) = 1 t 1 t 2... t n = 1. Does the solution depend on n? Solution:Yes. λ = n e