1 2 1 2 2 4 6 8 10 1 2 Figure 1.11: Plots of f(x) = p(x)/q(x). 1.6 Hints Hint 1.1 area = constant ×diameter 2 . Hint 1.2 A pair (x, y) is a solution of the equation if it make the equation an identity. Hint 1.3 The domain is the subset of R on which the function is defined. Hint 1.4 Find the slope and x-intercept of the line. Hint 1.5 The inverse of the function is the reflection of the function across the line y = x. Hint 1.6 The formu la for geometric growth/decay is x(t) = x 0 r t , where r is the rate. 14 Hint 1.7 Note that p(x) and q(x) appear as a ratio, they are determined only up to a multiplicative constant. We may take the leading coefficient of q(x) to be unity. f(x) = p(x) q(x) = ax 2 + bx + c x 2 + βx + χ Use the properties of the function to solve for the unknown parameters. Hint 1.8 Write the polynomial in factored form. 15 1.7 Solutions Solution 1.1 area = π ×radius 2 area = π 4 × diameter 2 The constant of proportionality is π 4 . Solution 1.2 1. If we multiply the equation by y 2 − 4 and divide by x + 1, we obtain the equation of a line. y + 2 = x − 1 2. We f actor the quadratics on the right side of the equation. x + 1 y −2 = (x + 1)(x − 1) (y −2)(y + 2) . We note that one or both sides of the equation are undefined at y = ±2 because of division by zero. There are no solutions for these two values of y and we assume from this point that y = ±2. We multiply by (y −2)(y +2). (x + 1)(y + 2) = (x + 1)(x − 1) For x = −1, the equ ation becomes the identity 0 = 0. Now we consider x = −1. We divide by x + 1 to obtain the equation of a line. y + 2 = x − 1 y = x −3 Now we collect the solutions we have found. {(−1, y) : y = ±2} ∪ {(x, x −3) : x = 1, 5} The solutions are depicted in Figure /reffig not a line. 16 -6 -4 -2 2 4 6 -6 -4 -2 2 4 6 Figure 1.12: The solutions of x+1 y−2 = x 2 −1 y 2 −4 . Solution 1.3 The denominator is nonzero for all x ∈ R. Since we don’t have any division by zero problem s, the domain of the function is R. For x ∈ R, 0 < 1 x 2 + 2 ≤ 2. Consider y = 1 x 2 + 2 . (1.1) For any y ∈ (0 . . . 1/2], there is at least one value of x that satisfies Equation 1.1. x 2 + 2 = 1 y x = ±  1 y − 2 Thus the range of the function is (0 . . . 1/2] 17 Solution 1.4 Let c denote degrees Celsius and f denote degrees Fahrenheit. The line passes through the points (f, c) = (32, 0) and (f, c) = (212, 100). The x-intercept is f = 32. We calculate the slope of the line. slope = 100 − 0 212 − 32 = 100 180 = 5 9 The relationship between fahrenheit and celcius is c = 5 9 (f − 32). Solution 1.5 We p lot the various transformations of f(x). Solution 1.6 The formula for geometric growth/decay is x(t) = x 0 r t , where r is the rate. Let t = 0 coincide with 6:00 pm. We determine x 0 . x(0) = 10 9 = x 0  11 10  0 = x 0 x 0 = 10 9 At 7:00 pm the n umber of bacteria is 10 9  11 10  60 = 11 60 10 51 ≈ 3.04 × 10 11 At 3:00 pm the n umber of bacteria was 10 9  11 10  −180 = 10 189 11 180 ≈ 35.4 18 Figure 1.13: Graphs of f(−x), f(x + 3), f(3 − x) + 2, and f −1 (x). Solution 1.7 We write p(x) and q(x) as general quadratic polynomials. f(x) = p(x) q(x) = ax 2 + bx + c αx 2 + βx + χ We wil l u se the properties of the function to solve for the unknown parameters. 19 Note that p(x) and q(x) appear as a ratio, they are determined only up to a multiplicative constant. We may take the leading coefficient of q(x) to be unity. f(x) = p(x) q(x) = ax 2 + bx + c x 2 + βx + χ f(x) has a second order zero at x = 0. This means that p(x) has a second order zero there and that χ = 0. f(x) = ax 2 x 2 + βx + χ We n ote that f (x) → 2 as x → ∞. This d etermine s the parameter a. lim x→∞ f(x) = lim x→∞ ax 2 x 2 + βx + χ = lim x→∞ 2ax 2x + β = lim x→∞ 2a 2 = a f(x) = 2x 2 x 2 + βx + χ Now we use the fact that f(x) is even to conclude that q(x) is even and thus β = 0. f(x) = 2x 2 x 2 + χ Finally, we use that f(1) = 1 to determine χ. f(x) = 2x 2 x 2 + 1 20 Solution 1.8 Consider the polynomial p(x) = (x + 2) 40 (x − 1) 30 (x − π) 30 . It is of degree 100. Since the factors only vanish at x = −2, 1, π, p(x) only vanishes there. Since factors are non- negative, the polynomial is non-negative. 21 Chapter 2 Vectors 2.1 Vectors 2.1.1 Scalars and Vectors A vector is a quantity having both a magnitude and a direction. Examples of vector quantities are velocity, force and position. One can represent a vector in n-dimensi onal space with an arrow whose initial point is at the origin, (Figure 2.1). The magnitude is the length of the vector. Typographically, variables representing vectors are often written in capital letters, bold face or with a vector over-line, A, a,a. The magnitude of a vector is denoted |a|. A scalar has only a magnitude. Examples of scalar quantities are mass, time and speed. Vector Algebra. Two vectors are equal if they have the same magnitude and direction. The negative of a vector, denoted −a, is a vector of the same magnitude as a but in the opposite direction. We add two vectors a and b by placing the tail of b at the head of a and defining a + b to be the vector with tail at the origin and head at the head of b. (See Figure 2.2.) The difference, a − b, is defined as the sum of a and the negative of b, a + (−b). The result of multiplying a by a scalar α is a vector of magnitude |α||a| with the same/opposite direction if α is positive/negative. (See Figure 2.2.) 22 [...]... expand the cross product a × b = (a1 i + a2 j + a3 k) × (b1 i + b2 j + b3 k) = a1 i × (b1 i + b2 j + b3 k) + a2 j × (b1 i + b2 j + b3 k) + a3 k × (b1 i + b2 j + b3 k) = a1 b2 k + a1 b3 (−j) + a2 b1 (−k) + a2 b3 i + a3 b1 j + a3 b2 (−i) = (a2 b3 − a3 b2 )i − (a1 b3 − a3 b1 )j + (a1 b2 − a2 b1 )k Next we evaluate the determinant i j k a a a a a a a1 a2 a3 = i 2 3 − j 1 3 + k 1 2 b2 b3 b1 b3 b1 b2 b1 b2... 2 .15 .) The area of a triangle defined by the two vectors a and b is 1 |a · b| 2 44 The area of the quadrilateral is then, 1 1 1 1 |(3i + j) · (i + 2j)| + |(i − 5j) · (−i − 4j)| = (5) + (19 ) = 12 2 2 2 2 y (3,7) (2, 3) (4 ,2) (1, 1) x Figure 2 .15 : Quadrilateral Solution 2. 7 The tetrahedron is determined by the three vectors with tail at (1, 1, 0) and heads at (3, 2, 1) , (2, 4, 1) and (1, 2, 5) These are 2, ... tetrahedron is 6 1 1 7 | 2, 1, 1 · 1, 3, 1 × 1, 2, 5 | = | 2, 1, 1 · 13 , −4, 1 | = 6 6 2 45 Solution 2. 8 The two vectors with tails at (1, 2, 3) and heads at (2, 3, 1) and (3, 1, 2) are parallel to the plane Taking the cross product of these two vectors gives us a vector that is orthogonal to the plane 1, 1, 2 × 2, 1, 1 = −3, −3, −3 We see that the plane is orthogonal to the vector 1, 1, 1 and passes... through the point (1, 2, 3) The equation of the plane is 1, 1, 1 · x, y, z = 1, 1, 1 · 1, 2, 3 , x + y + z = 6 Consider the vector with tail at (1, 2, 3) and head at (2, 3, 5) The magnitude of the dot product of this vector with the unit normal vector gives the distance from the plane √ 4 4 3 1, 1, 1 =√ = 1, 1, 2 · | 1, 1, 1 | 3 3 46 Part II Calculus 47 Chapter 3 Differential Calculus 3 .1 Limits of Functions... (a2 b3 − a3 b2 )i − (a1 b3 − a3 b1 )j + (a1 b2 − a2 b1 )k Thus we see that, i j k a × b = a1 a2 a3 b1 b2 b3 Solution 2. 6 The area area of the quadrilateral is the area of two triangles The first triangle is defined by the vector from (1, 1) to (4, 2) and the vector from (1, 1) to (2, 3) The second triangle is defined by the vector from (3, 7) to (4, 2) and the vector from (3, 7) to (2, 3) (See Figure 2 .15 .)... What is the volume of the tetrahedron with vertices at (1, 1, 0), (3, 2, 1) , (2, 4, 1) and (1, 2, 5)? Hint, Solution Exercise 2. 8 What is the equation of the plane that passes through the points (1, 2, 3), (2, 3, 1) and (3, 1, 2) ? What is the distance from the point (2, 3, 5) to the plane? Hint, Solution 37 2. 4 Hints The Dot and Cross Product Hint 2 .1 First prove the distributive law when the first vector... i = j × j = k × k = 0 31 • i × j = k, j × k = i, k × i = j • i j k a × b = (a2 b3 − a3 b2 )i + (a3 b1 − a1 b3 )j + (a1 b2 − a2 b1 )k = a1 a2 a3 , b1 b2 b3 cross product in terms of rectangular components • If a · b = 0 then either a and b are parallel or one of a or b is zero Scalar Triple Product Consider the volume of the parallelopiped defined by three vectors (See Figure 2 .10 .) The area of the base... Solution Exercise 2. 3 What is the angle between the vectors i + j and i + 3j? Hint, Solution Exercise 2. 4 Prove the distributive law for the cross product, a × (b + c) = a × b + a × b Hint, Solution Exercise 2. 5 Show that i j k a × b = a1 a2 a3 b1 b2 b3 Hint, Solution 36 Exercise 2. 6 What is the area of the quadrilateral with vertices at (1, 1) , (4, 2) , (3, 7) and (2, 3)? Hint, Solution Exercise 2. 7 What is... y · xm = ci xi · xm i =1 n ci xi · xm = i =1 = cm xm · xm y · xm cm = xm 2 Thus y has the expansion n y= i =1 If in addition the set is orthonormal, then y · xi xi xi 2 n (y · xi )xi y= i =1 35 2. 3 Exercises The Dot and Cross Product Exercise 2 .1 Prove the distributive law for the dot product, a · (b + c) = a · b + a · c Hint, Solution Exercise 2. 2 Prove that a · b = ai bi ≡ a1 b1 + · · · + an bn Hint,... the vector a is |a| = a2 + · · · + a2 1 n z a a3 k a1 i y a2 j x Figure 2. 4: Components of a vector 2 .1. 2 The Kronecker Delta and Einstein Summation Convention The Kronecker Delta tensor is defined δij = 1 if i = j, 0 if i = j This notation will be useful in our work with vectors Consider writing a vector in terms of its rectangular components Instead of using ellipses: a = a1 e1 + · · · + an en , we . = 10 9 = x 0  11 10  0 = x 0 x 0 = 10 9 At 7:00 pm the n umber of bacteria is 10 9  11 10  60 = 11 60 10 51 ≈ 3.04 × 10 11 At 3:00 pm the n umber of bacteria was 10 9  11 10  18 0 = 10 18 9 11 18 0 ≈. is R. For x ∈ R, 0 < 1 x 2 + 2 ≤ 2. Consider y = 1 x 2 + 2 . (1. 1) For any y ∈ (0 . . . 1/ 2] , there is at least one value of x that satisfies Equation 1. 1. x 2 + 2 = 1 y x = ±  1 y − 2 Thus. 1 2 1 2 2 4 6 8 10 1 2 Figure 1. 11: Plots of f(x) = p(x)/q(x). 1. 6 Hints Hint 1. 1 area = constant ×diameter 2 . Hint 1. 2 A pair (x, y) is a solution of the