Mathematics 140

Chapter 2

• Review Worksheet for Chapter 2

  1. 1. Let f be a function defined on an open interval containing a, except possibly at a. Then L is the limit of f(x) as x approaches a if for every positive number epsilon, there is a positive number delta (depending on epsilon) such that when |x - a| < delta and x is not equal to a, |f(x) - L| < epsilon.
     2. If f(x) = x² + 2x + 3, a = 10 (so that L = 123), and epsilon = 0.1, then delta = 0.004 works in the above definition.
  2. The function f is continuous at x = 0 and continuous from the left (but not from the right) at x = 1.
  3. 1. lim as u -> 0 [e^3u - e⁰]/[2u] = [3/2] [lim as u -> 0 [e^3u - e⁰]/[3u]] =
        (substitution rule) [3/2][lim as x -> 0 [e^x - e⁰]/x]
        = (3/2)*(1) = 3/2.
     2. lim as t -> 0 [sqrt t + tan (2 sqrt t)]/(sqrt t) = (substitution rule) lim as s -> 0 [s + tan (2s)]/s = 1 + 2 = 3.
     3. 2-sided limit doesn't exist since function undefined for z negative. The one-sided limit from the right is 0.
  4. 1. Let f(x) = x³ - x. Then f(x) is a polynomial function and so is continuous in x (everywhere). But f(0) = 0 > -2 and f(-2) = -8 - (-2) = -6 < -2. So by the intermediate value theorem, there is a value of x between -2 and 0 with f(x) = -2.
     2. Inspection of the graph shows that there is only one solution of the equation, located at about x = -1.52.

• Math 140 First Hourly Exam, September 19, 1997

  1. a) Exponentiating gives (x²-1) / (x+1)² = 2, so x² - 1 = 2(x² + 2x + 1), and x² + 4x + 3 = 0. Factoring gives (x + 3)(x + 1) = 0, so x = -1 or -3. In fact, all of these are spurious solutions since ln (x + 1) is undefined at these points, so in fact there are no solutions of the original equation.
     b) Note that l has slope -1/2, so any line perpendicular to l has slope 2. Thus the desired line is y = 2x + 4.
     c) sin [pi/3 + pi/4] = sin [7pi/12] = sin[pi/12 + pi/2]. But sin[x + pi/2] = cos x. So we need to compute cos[pi/12]. By the double angle formula, cos[pi/6] = 2 cos²[pi/12] - 1. But cos[pi/6] = (sqrt 3) / 2. So 2 cos²[pi/12] = (2 + sqrt 3) / 2, cos²[pi/12] = (2 + sqrt 3) / 4, and cos[pi/12] = sqrt(2 + sqrt 3) / 2.
  4. a) A function f is said to be continuous on the closed interval [a, b], if the function is defined on this interval, and if the limit of f(x) as x approaches c is f(c), for any c satisfying a < c < b, and if the one-sided limit of f(x) as x approaches a (respectively, b) from the right (resp., left) is equal to f(a) (resp., f(b)).
     b) The Intermediate Value Theorem doesn't apply to [-1, 1], since f is not defined at 0. It does guarantee existence of a zero of f in the interval [1, 3], since f(1) = -1, f(3) = 1/3, and f is continuous on [1, 3]. On the interval [-3, -1], the function is everywhere negative.

Chapter 3

• Math 140 Second Hourly Exam, October 10, 1995

  1. (a) Suppose f is a function differentiable at a value a. The equation for the tangent line to the graph of f at (a, f(a)) is: y = f(a) + f'(a)(x - a). (b) Let f(x) = cube root of x, a = 27. Then f(a) = 3 and f'(a) = (1/3)(27^-2/3) = 1/27. So the cube root of 28 is roughly 3 + (1/27)*(28-27) = 3 + (1/27) = 3.037.
     
  2. (a) 1/(3x^2/3) - x sin x + cos x
     (b)
     (c) [(1 + 3^x)(ln 2)(2^x) - (2^x)(ln 3)(3^x)] / (1 + 3^x)².
  3. (10 points) Implicit differentiation applied to e^xy = y gives e^xy(y + xy') = y'. Solving for y' gives y' = e^xyy / (1 - e^xyx).
     b) (10 points) The Newton-Raphson Method applied to the function f(x) = x³ - 28 replaces an approximation x to the root by x - f(x) / f(x) = x - (x³ - 28) / (3x²). So if we start with x = 3, successive approximations are 3, 3.037037037, 3.036589038, 3.036588972, 3.036588972. (From then on all approximations agree, to calculator accuracy.)
  4. (30 points) Let x be the distance from Joan to the kite, i.e., the length of the string that has been let out, and let y be the vertical height of the kite. (We measure all of these in feet.) Note that x is the hypotenuse of a right triangle with legs y and 15. We are given that dx/dt = 10 (in units of feet per second), and we want dy/dt when y = 20. So differentiate the Pythagorean theorem x² = y² + 15², getting 2x dx/dt = 2y dy/dt. Now substitute y = 20, dx/dt = 10, and x = sqrt(20² + 15²) = 25. We get dy/dt = (25)(10)/(20) = 12.5 feet per second.

• Review Worksheet for Chapter 3

  1. a.
     b. You can use x = 0 as an initial value, and you find the solution 0.6367326508.
     c.
  3. b. Let x be a leg of the triangle, and let h = (sqrt 2)x be the hypotenuse. Then the area is A = x² / 2. Let's measure x and h in miles and time t in years. We are given that dA/dt = 100. But dA/dx = x, so dividing gives dx/dt = (dA/dt) / (dA/dx) = 100 / x, or dh/dt = (sqrt 2)(100) / x. When A = sqrt 2, we have x² = 2 sqrt 2, or x = 2^3/4, so dh/dt = 100 / 2^1/4.
     c.

Chapter 4

Chapter 5

The Whole Course

• Final Exam, May 1997

  1. 1. 3x² + 3 sec² 3x
     2. (1 + 2x) cot (1 + x + x²)
  2. Differentiate implicitly: (d / dx) (x⁴ + y⁴) = 4 x³ + 4y³(dy / dx) = 0. So dy / dx = - x³ / y³ = 8 at (2,1), and the slope of the tangent line is 8. Thus the desired equation is y = 8(x-2) + 1 = 8x - 15.
  4. The number of grams in the sample decays exponentially with time, hence must be of the form f (x) = 100 e^-kt, where k is a positive constant, and where t is time measured in years. Since the half-life is 590 years, we know that e^-1590k = .5, or 1590k = ln 2. We need to find t for which e^-kt = .01, or kt = ln 100 = 2 ln 10. So t = 2 ln 10 / k = (1590)(2 ln 10 / ln 2) = 10564 years.
  5. 1. The function x⁵ - x² - 3 is a polynomial function and thus is continuous (for all real x). Since 0⁵ - 0² - 3 = -3 < 0 and 2⁵ - 2² - 3 = 32 - 4 - 3 > 0, the Intermediate Value Theorem implies there must be some x between 0 and 2 for which x⁵ - x² - 3 = 0, or x⁵ + 1 = x² + 4.
     2. The derivative of x⁵ - x² - 3 is 5x⁴ - 2x, which is 0 when x = 0 or 5x³ = 2, x = (2/5)^1/3. So for x between 0 and (2/5)^1/3 (= .737), one can see that x⁵ - x² - 3 is decreasing, and for x > (2/5)^1/3, x⁵ - x² - 3 is increasing. Thus x⁵ - x² - 3 is less than or equal to -3 for x < .737, and for x > .737, x⁵ - x² - 3 is increasing and therefore can only assume the value 0 once.
     3. Let's take x = 1 as an initial guess of a solution. Our iteration scheme is
        x - (x^5 - x^2 - 3) / (5*x^4 - 2*x) -> x.
        So we get the iterates 1, 2, 1.671..., ..., 1.373391379.
  6. Let r be the radius of the base and let h be the height of the barrel, both measured in meters. Then pi h r² = 32, while the cost of the materials for the side is 2 pi h r and the cost of the materials for either the top or the base is 2 pi r². Since h = 32 / (pi r²), the total cost is C(r) = (64 / r) + 4 pi r², with r > 0. We want to find the minimum value of this function. Since C'(r) = 8 pi r - (64 / r²), which is negative for small r, positive for large r, and 0 only when pi r³ = 8, the minimum must occur when r = 2(pi^-1/3). Then h = 32 / (pi r²) = 8(pi^-1/3).
  10. The curves y = 3x and y = x³ - x intersect when 3x = x³ - x, or 4x = x³, or x(x² - 4) = 0, or x = -2, 0, and 2. The desired region thus consists of two parts, one with x between -2 and 0, and one with x between 0 and 2. By symmetry, the two parts have the same area, so it's enough to consider the part with x between 0 and 2 and double the area. This part has an area obtained by integrating 3x - (x³ - x) = 4x - x³ from 0 to 2. This gives 2x² - x⁴/4 evaluated from 0 to 2, or 2*(2²) - (2⁴/4) = 4. The total area is double this, or 8.
  
  

• Final Exam, December 1997

  1. 1. We can rewrite (sqrt (1 + 9x²) / x) as sqrt (9 + x^-2). As x goes to infinity, x^-2 goes to 0, so the limit is sqrt (9) = 3.
     2. This is the definition of the derivative of sin x, computed at x = pi/6. So the limit is cos (pi/6) = (sqrt 3) / 2.
     3. The limit from the right is -2² = -4, while the limit from the left is 4 - 6 = -2. Since these do not agree, there is no limit.
  4. Differentiate implicitly. This gives 2x + 2xy² + x²2yy' + 3y²y' = 0. Substitute x = 1, y = 2. One gets 2 + 8 + 4y' + 12y' = 0, or 16y' = -10. So the slope of the tangent line is -5/8 and the equation of the tangent line is 8(y - 2) = -5(x - 1).
  5. 1. The function f(x) = x³ + x - 7 is a polynomial function and thus is continuous. Also, f(2) = 8 + 2 - 7 = 3 > 0, whereas f(1) = 1 + 1 - 7 = -5 < 0. So by the Internediate Value Theorem, there is a value of x between 1 and 2 for which f(x) = 0.
     2. Note that f'(x) = 3x² + 1 is everywhere positive. So f is strictly increasing and thus the graph of f cannot cross any horizontal line more than once.
     3. Bisect [1,2] at 1.5. We find f(1.5) = -2.125 < 0. So the root lies in [1.5,2]. Bisect [1.5,2] at 1.75. We find f(1.75) = 0.109375 > 0. So the root lies in [1.5, 1.75]. Bisect [1.5, 1.75] at 1.625. We find f(1.625) = -1.08398. So the root lies in [1.625, 1.75]. By Newton's Method, the actual numerical value of the root is about 1.7392.
  6. Let z be the square of the distance from the point to the ellipse. (This is to avoid radicals.) Then z = (x - 1)² + y². Since we are only interested in this for points on the ellipse, we can eliminate y² using the equation (x² / 20) + (y² / 15) = 1, or y² = 15 - (15/20)x². Substituting, we get z as a function of x: z (x) = (x - 1)² + 15 - 3x²/4. We need the minimum of this function, but not over arbitrary values of x, only over the interval from -20^1/2 to 20^1/2, or from -2 sqrt(5) to 2 sqrt(5). Note that z' (x) = 2(x - 1) - 3x / 2 = (1/2)x - 2, which vanishes when x = 4. And z(4) = 3² + 15 - (3)(4²) / 4 = 9 + 15 - 12 = 12. We have to compare this with the values z(-2sqrt(5)) and z(2sqrt(5)). z(-2sqrt(5)) is approximately 30 and is much bigger. z(2sqrt(5)) is 21 - 4sqrt(5), approximately 12.056 and is also bigger, though not by much. So the mimumum distance is sqrt(12), achieved when x = 4 and y = sqrt(3) or -sqrt(3).
  7. 1. Note that f (x) only involves x² and not any odd powers of x. Thus the graph has even symmetry around the y axis.
     2. Note that f' vanishes when x = 0 (this is automatic for any even function) and also when x² + 3 = 8, or x = 5^1/2 or - 5^1/2.
  9. The kite and the person are at two of the vertices of a right triangle with hypotenuse 500 feet and "vertical" leg 300 feet. So the distance from the person to the point just below the kite is 400 feet (by the Pythagorean Theorem). The wind velocity is equal to the kite velocity, and is the same as the rate of change with respect to time of the horizontal distance from the person holding the kite to a point just below the kite. Call this distance x and the time t. Let l be the length of string that has been let out. Then l² + (300)² = x². Differentiating, 2l(dl / dt) = 2x(dx / dt). So when dl / dt = 20 (feet per second) and x = 400 (feet), l = 500 (feet), we have dx / dt = (500/400)(20) = 25 feet per second.
  10. Let u = sqrt x, so du = dx / (2 sqrt x). The integral becomes the integral of 2 e^u du, or 2 e^u, from u = 1 to u = 2^1/2. So the answer is 2(e^{sqrt 2} - e).
  11. The curves intersect when x³ + x² - 2x - 1 = x³ - 1, or x² - 2x = 0, or x = 0 or 2. For x between 0 and 2, x² - 2x < 0, so the area is the integral from 0 to 2 of 2x - x², which is x² - (x³ / 3) evaluated from 0 to 2, or 2² - (2³ / 3) = 4 - (8/3) = 4/3.