ANSWERS TO QUIZ#1 ----------------- Warren Smith Math 75 quiz #1. Show all work. name WD Smith, prof ID# --------------------------------- ------------------------------- 1. decide if these 3 points A,B,C are the vertices of a right triangle: A=(6, -9) B=(12, -7) and C=(18, -14) ? 2 2 2 2 2 2 2 2 2 AB = 6 + 2 = 40 BC = 6 + 7 = 36+49 = 85 AC = 12 + 5 = 144+25 = 169 (where "AB" here means "the distance from A to B") 40+85 is not 169 so Pythagoras fails so NOT a right triangle. [Another method: one student computed lines thru each pair of points, found their slopes, and realized no two slopes S1 and S2 were perpendicular since no pair of slopes multiply out to -1. Another way: compute the VECTORS C-B=(6,-7) and A-B=(-6,-2) and their dot product=-6*6+2*7=14-36 is not zero so not a right angle.] 10000 2. What is log 2 accurate to 3% error (or better)? 10 10 Hint: 2 = 1024 is all you need to know, don't need a calculator. 10000 10 log 2 = 1000 log 2 = 1000 log 1024 = 3010 since log 1000 = 3. 10 10 10 10 This used the multiplicative property of the logarithm: log(XY) = log(X)+log(Y) i.e. multiplying inside the argument of the log causes the log to add. k log(x ) = k log(x) for the same reason. Here k=1000 and x=1024. Or you could use k=10000 and x=2, works too, get 10000 log 2. 10 3. What is 5 + 8 + 11 + 14 + 17 + ... + 3002 + 3005 ? Re-express this problem using the sum-symbol notation, and also find the answer as an exact number. 1001 ---- > (3k+2) = (3010)(1001)/2 = (PairSum) (#pairs) ---- k=1 = (1505)(1001) = 1506505 [The "3k" is because I want steps of 3. The "+2" is because I want k=1 to give 5.] Another answer: 1000 ---- 5005 + 3 > k ---- k=0 2 -8 x + 7 x + 3 4. lim --------------- = -8 get rid of 7x+3 and 54 x --> infinity 2 as comparatively negligible x + 54 when x huge 2 2 2 (x+5) - 25 x + 10 x + 25-25 x + 10x (x+10)x 5. lim ------------ = lim ----------------- = lim ------- = ------- x --> 0 x x x x 10x Now get rid of the x since it is comparatively = --- = 10 negligible versus the 10 it is added to. x NOTE: You can often figure out the answer by trying x=.01, x=.001, etc with your calculator and "seeing" what we are approaching. For example x=.01 causes us to get 10.01 which is near to the limit value 10. Just using the calculator in this way makes it almost impossible to get a very-wrong answer. In the previous problem try x=1000, x=10000, etc to "see" what we get, for example x=1000 gives -7.993, near to the limit value 8.) 3 6. lim log x = log 27 = 3 since 27 = 3 . x --> 27 3 3 [Since nothing funny happens in this limit, lim F(x)=F(v) x-->v Boring!] 2 2 7. (x-3) + (y-4) = 25 represents a circle with radius=5=sqrt(25) and center=(3,4) This equation is just saying the DISTANCE from any point (x,y) on the circle-curve to its center (3,4), is 5. It is just Pythagoras. 8. The line thru (5,4) and (7,8) has equation y=2x-6 Find the equation of another line parallel to this one? y=2x Find the equation of another line perpendicular to this one? 2y = -x (many other answers also valid). 9. The line x = 2y intersects the circle in question 7 at what point (or points, or say none if there's no intersection; I want to know both how many points, and what their coordinates are)? x = 2y 2 2 (x-3) + (y-4) = 25 now substitute 1st equation into 2nd 2 2 2 2 so (2y-3) + (y-4) = 25 = 4y - 12y + 9 + y - 8y + 16 2 so 5y - 20y + 25 = 25 2 so 5y - 20y = 0 so y = 4 or y=0 (roots of this quadratic) so (since x=2y) x = 8 or x=0 respectively. (2 points; if line had not intersected circle at all, the quadratic would have had no real roots.) Note: you could also have found this out, by, instead of solving the quadratic like I did, just draw a picture of the circle and the line. The two points are near (8,4) and (0,0). So then just try (8,4) and (0,0) in the two equations and see if they work - and they do. Also a good SANITY CHECK is trying your (x,y) in the two equations to see if they really work. 10. suppose 0 < a < pi and 0 < b < pi radians. Re-express the sin addition law sin(a+b) = sin(a)cos(b) + sin(b)cos(a) so that sin(a+b) is expressed as a function of sin(a) and sin(b) ONLY. 2 2 sin(a+b) = sin(a) sqrt( 1-sin(b) ) + sin(b) sqrt( 1-sin(a) ) 2 2 by using pythagoras sin x + cos x = 1 identity to get rid of (i.e re-express in terms of sin's only) the cos's. 2 11. If F(x) = tan(7 x + x - 3), then find the function G(y) so F and G are inverse functions. (If there is more than 1 branch, then restrict to one.) Say what the range and domain of your G(y) is. [Hint: tan(A) and arctan(B) are inverse functions. The domain of arctan(b) is B=all real numbers, and its range is -pi/2 < arctan(B) < pi/2 radians.] Let y = F(x). We solve this equation for x to get x=G(y) which gives us a formula for the inverse function G(y). Remember, y=F(x) inputs x and spits out y; now the inverse function G(y) inputs that y and gets x back out again. To solve: Let a = arctan(y). Solve 2 a = 7 x + x - 3 for x to get -1 +or- sqrt(1 + (4)(7)(3+a)) x = ----------------------------- 14 so the inverse function is -1 +or- sqrt(1 + 28 (3+arctan(y))) G(y) = ---------------------------------- 14 Also, valid (equivalent formula) is -1 +or- sqrt(85 + 28 arctan(y)) G(y) = -------------------------------- 14 If we use the + sign only, restricting to one branch, -1 + sqrt(1+28(3-pi/2)) -1 + sqrt(1+28(3+pi/2)) then the RANGE is ( -----------------------, ----------------------- ) 14 14 since the arctan ranges from -pi/2 to +pi/2, and the DOMAIN is y = all real numbers. [Fortunately pi/2 < 3; if pi/2 had been 4.5, say, then it would have been trickier because then there would be risk of taking the square root of a negative number, and we might have had to use a smaller domain to prevent that risk.] [total time consumed by me (WDS) in finding all solutions and typing them up neatly was: 23 minutes.]