166 Calculus, Practice Exam 3. (* The actual Exam 3 is Friday, April 4, 2003 7:30 to 9:00 AM *) Sections 8.8-8.9, 9.3-9.4, 11.1-11.4 Instructions: You are welcome to check your answers with your calculators. But you will not be given full credit unless your work is clear, complete, and correct. 1. Write the equation of the circle given that the segment between (2,4,8) and (1,2,3) is a diameter. ans: Center = (3/2,3,11/2) (x-3/2)^2 + (y-3)^2 + (z-11/2)^2 = 1/4 (1^2 + 2^2 + 5^2) You do not have to expand the equation out. But if you did it would look like this. 2 2 2 0 = 34 - 3 x + x - 6 y + y - 11 z + z 2. Write the equation of the plane containing (2,3,4), (1,1,1) and (5,0,5). | i j k | ans: Normal: | 1 2 3 | = (11,8,-9) | 4 -1 4 | 11 x + 8 y -9 z = 10 <==== answer (2,3,4) 22 24 -36 10 (1,1,1) 11 8 -9 10 (5,0,5) 55 0 -45 10 3. Write the equation of the line containing (1,2,3) and (5,6,3). x-1 y-2 ans: ---- = ------ z=3 4 4 or x = 1 + 4 t y = 2 + 4 t z = 3 4. The points A = (1,3,9), B = (2,1,3), C = (0,9,9) are a triangle. Find the angle at A. ans: AB = ( 1,-2,-6) AC = (-1, 6, 0) -1-12+0 -13 Cos[A] = ----------------- = ------------------ = -0.333773 Sqrt[41] Sqrt[37] Sqrt[41] Sqrt[37] A = 109.498 degrees 4. P is the point (2,3,4). W is the plane 2x + 3y + z = 7. (a) Find the distance from P to the plane W. (b) Find the point on the plane W closest to P. | 4+9+4-7 | 10 ans(a) ------------- = ----------- Sqrt[4+9+1] Sqrt[14] ans(b) x = 2 + 2 t y = 3 + 3 t z = 4 + 1 t 2(2+2t) + 3(3+3t) + (4+t) = 7 17 + 14 t = 7 14 t = -10 t = -5/7 x = 2-10/7 = 4/7 y = 3-15/7 = 6/7 z = 4-5/7 = 23/7 check 8/7 + 18/7 + 23/7 = 7 on the plane Sqrt[(2-4/7)^2+(3-6/7)^2+(4-23/7)^2] = 5 Sqrt[2/7] correct distance 5. P is the point (3,1,1). L is the line x = 1 + t y = 5 + 3 t z = 7 - 4 t (a) Find the distance from P to L. (b) Find the point on L closest to P. ans(a) P (3,1,1) /| / | ---------> / | (2,-4,-6) / |d / | /_____|____________________________ line (1,5,7) -----------> (1,3,-4) d = |(2,-4,-6)x(1,3,-4)| | i j k | ------------------- = 1/Sqrt[26] | 2 -4 -6 | Sqrt[26] | 1 3 -4 | 1/Sqrt[26] ( 34, 2, 10) = Sqrt[1260]/Sqrt[26] = 3 Sqrt[70/13] = 6.96143 ans(b). x+3y-4z = 2 x = 1 + t y = 5 + 3 t z = 7 - 4 t (1+t)+3(5+3t)-4(7-4t) = 2 -12 + 26 t = 2 26 t = 14 t = 14/26 = 7/13. x = 1+7/13 y = 5 + 21/13 z = 7 -28/13 Point is (20/13, 86/13, 63/13) <=======answer. It it on the line and the distance from P is Sqrt[ (3-20/13)^2 + (1-86/13)^2 + (1-63/13)^2 ] = 3 Sqrt[70/13] 6. L1 is the line x = 1 + t L2 is the line x = 5 + t y = 2 - t y = 6 - 3 t z = 3 + 2 t z = 4 + 2 t (a) Find the distance between L1 and L2. | i j k | Normal to the planes | 1 -1 2 | = (4,0,-2) | 1 -3 2 | plane 1 is 4x-2z = -2 20-8+2 14 Distance from (5,6,4) to plane 1 is --------- = --------- = 7/Sqrt[5] Sqrt[20] 2 Sqrt[5] (b) Find the point on L1 and the point on L2 which are closest together. ans(b) (1+t -5-s,2-t-6+3s,3+2t-4-2s)o(1,-1,2) = 0 (1+t -5-s,2-t-6+3s,3+2t-4-2s)o(1,-3,2) = 0 (-4 - s + t, -4 + 3 s - t, -1 - 2 s + 2 t)o(1,-1,2) = 0 (-4 - s + t, -4 + 3 s - t, -1 - 2 s + 2 t)o(1,-3,2) = 0 -2 - 8 s + 6 t = 0 6 - 14 s + 8 t = 0 -8 s + 6 t = 2 -14 s + 8 t = -6 -4 s + 3 t = 1 -7 s + 4 t = -3 | 1 3 | |-3 4 | 13 s = --------- = --------- |-4 3 | 5 |-7 4 | |-4 1 | |-7 -3 | 19 t = ----------- = ----- |-4 3 | 5 |-7 4 | L1 is the line x = 1 + 19/5 L2 is the line x = 5 + 13/5 y = 2 - 19/5 y = 6 - 39/5 z = 3 + 38/5 z = 4 + 26/5 (24/5,-9/5,53/5) (38/5, -9/5, 46/5) Distance = Sqrt[ (14/5)^2 + 0 + (-7/5)^2 ] = 7/Sqrt[5] 8. Find n so that the error from Simpson's rule for x=10 INT Sin[x^2] dx is less than or equal to 0.0001 . x=0 5 K(b-a) (4) | Es | <= ----------- | f | <= K 4 180 n 3 K(b-a) (2) | Et | <= ----------- | f | <= K 2 12 n Solution: f(x) = Sin[x^2] 2 f'(x) = 2 x Cos[x ] 2 2 2 f'(x) = 2 Cos[x ] - 4 x Sin[x ] 3 2 2 f'"(x)= -8 x Cos[x ] - 12 x Sin[x ] 2 2 2 4 2 f""(x) = -48 x Cos[x ] - 12 Sin[x ] + 16 x Sin[x ] |f""(x)| <= 4800 + 12 + 160000 = 164812 5 K(b-a) (4) | Es | <= ----------- | f | <= K 180 n^4 164812 10^5 | Es | <= -------------- < 0.0001 180 n^4 164812 10^5 ------------ < n^4 180 0.0001 (164812 10^5)/(180 0.0001) 978.203 < n Let n = 980. 9. Use the Trapezoid rule and Simpson's Rule to compute the area of this lake. . . ' | ' . . . | | |'. . ' | | | ' . . ' | | | | |'. . ' | | | | | | ' . .' | | | | | | |'. .' 30 35 45 50 45 35 10 '. :.......|.....|......|......|......|......|......|.....:| |<-10->|<-10->|<-10->|<-10->|<-10->|<-10->|<-10->|<-10->| width Trap Simp 0 0 0 30 60 120 35 70 70 45 90 180 50 100 100 45 90 180 35 70 70 10 20 40 0 0 0 ------ ----- 500 760 *10/2 *10/3 ----- ------ 2500 2533 1/3 Trapezoid Simpson <=================== Answer 10. Find the area of the surface obtained by rotating the 3 2 curve (t , t ) 0 <= t <= 1 about the x axis. INT 2 Pi y ds t=1 2 Pi INT t^2 Sqrt[9 t^4 + 4 t^2] dt t=0 t=1 2 Pi INT t^3 Sqrt[9 t^2 + 4 ] dt t=0 2 Pi Integrate[t^3 Sqrt[9 t^2 + 4],{t,0,1}] 64 247 Sqrt[13] = 2 (---- + ------------) Pi = 4.93642 1215 1215