Full Solutions to the Sample Exam for Multivariable Calculus but with material from Calc1 and Calc2 for someone whose RUID is, 123456789 NAME: Doron ZEILBERGER RUID: 123456789 EMAIL: DrZ@gmail.com BELOW WRITE THE LIST OF THE ANSWERS Answer[1]= 13/7 Answer[2]= -1909/216 Answer[3]= 128 Answer[ 4]= exp(3)*exp(exp(3))*(exp(3)+1) Answer[ 5]=1214/405 Answer[ 6]= 143/12 Answer[7]=ln(2)+ln(3) Answer[8]= -9/23 Answer[9]= 36 Answer[10]= 1/10 --------------------------------------------- Problem 1: Let P1=(a[1],a[3]), P2=(a[5],a[7]), P3=(a[9],a[1]), P4=(a[4],a[3]) Find the x-coordinate of the point of intersection of the line joining P1 and P2 and the line joining P3 and P4 With my RUID data the question is --------------- Let P1 = [1, 3], P2 = [5, 7], P3 = [9, 1], P4 = [4, 3] Find the x-coordinate of the point of intersection of the line joining P1 and P2 and the line joining P3 and P4 --------------- Here is how I do it (Explain everything) Let's first find the equation of the line P1P2 P2[2] - P1[2] The slope,m, is ------------- P2[1] - P1[1] that equals, 1 So the equation of the line P1P2 is y - 3 = x - 1 That simplifies to y = 2 + x We now find the equation of the line P3P4 P4[2] - P3[2] The slope is, ------------- P4[1] - P3[1] that equals -2/5 So the equation of the line P3P4 is 2 x y - 1 = - --- + 18/5 5 That simplifies to 2 x y = 23/5 - --- 5 In order to find the point of intersection of the two lines we set them equal to each other 2 x 2 + x = 23/5 - --- 5 Getting that the x-coordinate is 13/7 Ans.: 13/7 --------------------------------------------- Problem 2: With my RUID data the question is Find the y-coordinate of the inflection point of the function f(x):= 3 2 4 x - x + 2 x - 9 Here is how I do it (Explain everything) The first derivative is 2 12 x - 2 x + 2 The second derivative is 24 x - 2 Setting it equal to 0, and solving for x we get x = 1/12 3 2 To get the y-coordinate, we plug into the function, 4 x - x + 2 x - 9 Getting -1909 y = ----- 216 -1909 Ans.: ----- 216 --------------------------------------------- Problem 3: With my RUID data the question is Find the absolute maximum value of f(x):= 2 7 x + 5 x - 4 In the closed interval 0<=x<=4 Here is how I do it (Explain everything) We first find the critical point of the function The derivative is, 14 x + 5 Setting it equal to 0 14 x + 5 = 0 we get the root -5 x = -- 14 It is outside the interval so we ignore it The finalists are, 0, 4 Plugging them into the function we get the VALUES [-4, 128] The maximum value is, 128 Ans.: 128 --------------------------------------------- Problem 4: With my RUID data the question is Find f''(3) if f(x)= exp(exp(x)) Here is how I do it (Explain everything) Using the chain rule, The first derivative is, exp(x) exp(exp(x)) 2 Using the product rule and the chain rule, The second derivative is, exp(x) exp(exp(x)) + exp(x) exp(exp(x)) plugging in x= 3 we get Ans.: exp(3) exp(exp(3)) (exp(3) + 1) --------------------------------------------- Problem 5: With my RUID data the question is Use Newton's rule to approximate (1/5) 242 Here is how I do it (Explain everything) We have to solve the equation 5 x - 242 = 0 5 Let f(x)= x - 242 A crude approximation is x0= 3 We get a better approximation using x1=x0-f(x0)/f'(x0) The derivative is 4 5 x f(, 3, )= 1 f'(, 3, )= 405 Plugging into Newton's equation we get 3 minus 1/405 1214 ---- 405 1214 Ans.: ---- 405 --------------------------------------------- Problem 6: Find the area under the curve y = a[1]*x^3 + a[5]*x^2+ a[2]*x+a[9] between the vertical lines x=0 and x=1 With my RUID data the question is Find the area under the curve 3 2 y = x + 5 x + 2 x + 9 between the vertical lines x=0 and x=1 Here is how I do it (Explain everything) The indefinite integral of the function is 4 3 2 1/4 x + 5/3 x + x + 9 x Since this is a definite integral, we don't care about the arbitrary constant+C 143 At x=1 is equals --- 12 At x=0 is equals 0 subtracting we get 143 Ans.: --- 12 --------------------------------------------- Problem 7: With my RUID data the question is Find the area under the curve 3 x - 4 y = ------------ 2 x - 3 x + 2 between the vertical lines x=3 and x=4 Leave your answer in terms of log (or ln), i.e. do not convert if to decimals Here is how I do it (Explain everything) 3 x - 4 Since the function is always positive (above the x-axis), the area is the integral of ------------ between x=3 and x=4 2 x - 3 x + 2 Factoring the denominator we get that the function is 3 x - 4 --------------- (x - 1) (x - 2) Doing partial-fraction decomosition, we get 2 1 ----- + ----- x - 2 x - 1 Integrating we get ln(x - 1) + 2 ln(x - 2) Plugging into x=4, and x=3, and subtracting we get ln(3) + 2 ln(2) minus ln(2) + 2 ln(1) Ans.: ln(2) + ln(3) --------------------------------------------- Problem 8: With my RUID data the question is Find the slope of the tangent line to the curve 2 2 x + 7 x y + 8 y = 16 at the point (1,1) Here is how I do it (Explain everything) 2 2 First we make sure that the point (1,1) lies on the curve, x + 7 x y + 8 y = 16 it does Using IMPLICIT differentiation we get, let y'=y1 7 x y1 + 16 y y1 + 2 x + 7 y = 0 Plugging-in x=1, y=1 we get 9 + 23 y1 = 0 Solving for y' we get -9 Ans.: -- 23 --------------------------------------------- Problem 9 With my RUID data the question is Find the coefficient of x^2 in the Taylor series around x=0 or the function f(x)= 9 (1 + x) Here is how I do it (Explain everything) 9 Our function f(x) is (1 + x) The coefficient of x^2 in the Taylor series around x=0 is f''(0)/2 8 The first derivative is 9 (1 + x) 7 The sedond derivative is 72 (1 + x) Plugging x=0 and dividing by 2 we get 36 Ans.: 36 --------------------------------------------- Problem 10 With my RUID data the question is Find the coefficient of x in the Taylor series around x=0 or the function f(x)= (1/10) (1 + x) Here is how I do it (Explain everything) (1/10) Our function f(x) is (1 + x) The coefficient of x in the Taylor series around x=0 is f'(0) 1 The first derivative is ---------------- (9/10) 10 (1 + x) Plugging x=0 we get 1/10 Ans.: 1/10