Math 127(3) - Warren D. Smith - Homework #1 - due Tues 28 Jan 2003 ------------------------------------------------------------------ ANSWERS 1. For the real-valued function F(x)=sqrt(sin(x)) on the real numbers: a. What is the range of F? F's output value set: [0,1] (note, kind of parens is important) b. What is the domain of F? F's allowed inputs: reals x so that sin(x)>=0, namely x must be in [0,pi] U [2pi,3pi] U ... generally [2n pi, (2n+1) pi] for integer n=0,1,2,... and -1,-2,-3,... c. Find a maximum of F? x=pi/2, F(x)=1. 2. Find the derivatives (d/du) and integrals (du from 0 to x) of these functions: 2 a. F(u) = u sin(u ) 2 2 2 F'(u) = sin(u ) + 2 u cos(u ) by product & chain rules 2 integral = (1/2) [ 1 - cos(x ) ] chain rule in reverse direction gives indefinite integral is -(1/2) cos(uu). Sanity checks: get 0 if x=0; derivative is F(x). 2 b. F(u) = u sin(u). 2 F'(u) = 2u sin(u) + u cos(u) by product rule 2 integral = -x cos(x) + 2 cos(x) + 2x sin(x) - 2 Integral can be done by method of undetermined coefficients: You guess that answer is of form 2 2 A x cos(x) + B cos(x) + C + D x sin(x) + E x sin(x) then solve for A,B,C,D,E so that its derivative matches 2 x sin(x). A more direct way (no inspired guess needed!) to do the integral is by INTEGRATION BY PARTS. Remember the integration by parts formula integral A'(x) B(x) dx = A(x) B(x) - integral A(x) B'(x) dx. 2 So here use A'(x)=sin(x) and B(x)=x. Then A(x) = -cos(x), B'(x)=2x. 2 2 integral sin(x) x dx = -cos(x) x + 2 integral x cos(x) dx. That reduced the problem to a simpler integral. We can now do integration by parts another time to do this integral: This time use A'(x) = cos(x), B(x)=x, A(x) = sin(x), B'(x)=1. integral x cos(x) dx = x sin(x) - integral sin(x) dx = x sin(x) + cos(x). So... the indefinite integral is 2 2 integral sin(u) u du = -cos(u) u + 2 u sin(u) + 2 cos(u). If we now take this at u=x minus this at u=0 we get at last 2 integral(from 0 to x) sin(u) u du = 2 -cos(x) x + 2 x sin(x) + 2 cos(x) - 2. Finally, a third way to do this integral is you can just look it up a a table of integrals in the library. Table of integrals, series, and products by I. S. Gradshteyn and I. M. Ryzhik QA55.G6613, QA55.R943 1965 (TU math lib. ref, TU physics lib., Paley lib.) CRC handbook of math QA47.H324, QA47.H32 (Many TU libs) Useful SANITY CHECKS: get 0 if x=0; take derivative and really do get F(x). 3. (Complex numbers.) Let i be the square root of -1. What is (1+7i) / (3+4i) ? 2 2 (31+17i)/25 using the formula for reciprocal of A+iB is (A-iB)/(A + B ) (here with A=3, B=4) then multiply by 1+7i. (SANITY CHECK: multiply ans by 3+4i and should get 1+7i.) ANOTHER WAY: multiply both top and bottom by 3-4i. (Do not multiply by different things on top & bottom!) 4. Solve the following system of 3 equations for the 3 unknowns x,y, & z (get exact rationals as answers): 3x-7y+z = 8 x+y = 9 5x+2y-z = 1 Answer (x,y,z) = (54, 63, 383)/13 = (4.15, 4.85, 29.46). exact approx One way is to solve middle eqn for x getting x=9-y. Substitute 9-y instead of x in the other 2 eqns getting -10y+z = -19 -3y -z = -44 add them getting -13y = -63 so y = 63/13. Now use this to see z = 44-3y = 44-189/13 = 383/13 from -3y -z = -44, and also use it to see x = 9-63/13 = 54/13 from x+y=9. (SANITY CHECK: resubstitute alleged answer into equations.) This is a specific example of "elimination" technique where you successively reduce a system of N eqns in N unknowns to one with N-1 eqns in N-1 unknowns... until you have only 1 eqn in 1 unknown. You solve it, then you back up successively finding all the other unknowns. Another way to solve is to use Gaussian elimination, another way is to use Cramer's rule. But I like my elimination method the best since it work for nonlinear equations too. 5. [vector algebra problems] page 803 sec 12.3 of my (7th ed) book #2. (4,2,-1) . (-2,2,1) = -8 +4 -1 = -5. Some people returned answers which were VECTORS. Wrong. The dot product of two vectors is a SCALAR, i.e. a plain old number, like -5. The CROSS product aXb would have been a vector. 2 2 #8. a.(a-b) + b.(b+a) = a.a + b.b = |a| + |b| since -a.b + b.a = 0 cancels (dot product is commutative & distributive). Note, if these had been cross products instead of dot products we would have had a X (a-b) + b X (b+a) = a X a + b X b - a X b + b X a = 0 + 0 - 2 a X b using the ANTIcommutativity of cross products. 2 #30a if a.a+b.b = |a+b| then 2a.b=0 so a,b perpendicular. 2 #30b if a.a+b.b = |a-b| then -2a.b=0 so a,b perpendicular. These can also be seen (draw picture) to be just statements of Pythagoras thm for right triangle. In fact, you've just produced your own proof of Pythagoras's theorem thru vectors! 5. [THE WRONG vector algebra problems] Since there was a book-edition screwup, a few people solved other problems... I'll award credit this time... #2 length of (4-2, -2-4) is sqrt(40) #8 (1,-2,3) + 3(3,0,-1) - 2(-4,2,1) = (1+9+8, -2+0-4, 3-3-2) = (18,-6,-2). #30 if (3,1,-1) and (a,-4,4) are parallel then a is -12 since the second vector is -4 times the first. Parallel (same direction) vectors happen only if one vector is a constant multiple of the other, and here the only possible constant multiple is -4 since only that makes ther last 2 coordinates work.