Calculus Problems - Assignment Example

Problems in Calculus Solve for t. Assume a and b are positive constants and k is nonzero: Solution: In order to solve for t, we have appliedlogarithm function to both side of the equation as (Mathwords, 2008). Rewritten the above expression, we solve for t: 2. Draw the angle using a ray through the origin, and determine whether the sine, cosine, and tangent of that angle are positive, negative , zero, or undefined Solution: The angle is equal to -240 degree respect to the x-axis positive as shown in figure 1. The angle lies within the second quadrant. Note that the sine function is always positive within the second quadrant. Both cosine and tangent function are negative in the second quadrant (Spector 2010). Figure 1: It shows the angle. The sine function is positive within first and second quadrant, while within third and fourth quadrant is negative. The cosine function is positive within first and fourth quadrant, while within second and third quadrant is negative. The tangent function is positive within the first and third quadrant, while is negative within the second and fourth quadrant. 3. a) Match the functions w = f(t), w = g(t), w = h(t), w = k(t), whose values are in the table, with the functions with formulas: (i) w = 1.5 + sin(t) (ii) w = 0.5 + sin(t) (iii) w = -0.5 + sin(t) (iv) w = -1.5 + sin(t) t f(t) t g(t) t h(t) t k(t) 6 -0.78 3 1.64 5 -2.46 3 0.64 6.5 -0.28 3.5 1.15 5.1 -2.43 3.5 0.15 7 0.16 4 0.74 5.2 -2.38 4 -0.26 7.5 0.44 4.5 0.52 5.3 -2.33 4.5 -0.48 8 0.49 5 0.54 5.4 -2.27 5 -0.46 Solution: f(t) = -0.5 + sin(t) g(t) = 1.5 + sin(t Figure 2: Left) It shows the f(t) function and discrete values. Right) It shows the g(t) function and discrete values. h(t) = -1.5 + sin(t) k(t) = 0.5 + sin(t) Figure 3:Left) It shows the h(t) function and discrete values. Right) It shows the k(t) function and discrete values. b) Based on the table, what is the relationship between the values of g (t) and k (t)? Explain this relationship using the formulas you chose for g(t) and k(t). Solution: Using the table, we release that the relationship between the values of g(t) and k(t) is given by: g(t) – k(t) = 1.0. We assume that g(t) = 1.5 + sin(t) and k(t) = 0.5 + sin(t). Thus, g(t) – k(t) = 1.5 + sin(t) - ( 0.5 + sin(t) ) = 1. c) Using the formulas you chose for g(t) and h(t), explain why all the values of g(t) are positive, whereas all the values of h(t) are negative. Solution: We assume that g(t) = 1.5 + sin(t) and h(t) = -1.5 + sin(t). Since g(t) = 1.5 + sin(t), we know that 0 < 1.5 and -1 < sin(t) < 1. Thus: 1.5 - 1 < 1.5 + sin(t) < 1 + 1.5 0.5 < 1.5 + sin(t) < 2.5 0.5 < g(t) = 1.5 + sin(t) < 2.5 g(t) is always positive. Since h(t) = -1.5 + sin(t), we release that -1.5 < 0 and -1 < sin(t) < 1. Thus: -1.5 - 1 < -1.5 + sin(t) < 1 - 1.5 -2.5 < -1.5 + sin(t) < -0.5 -2.5 < h(t) = -1.5 + sin(t) < -0.5 h(t) is always negative. d) Are the functions continuous on the given intervals? on [-1,1] Solution: Yes because the functions f(x) = x, g(x) = -2 and h(x) = 1/x are continuous within the interval [-1, -1]. The original function can be written as: F(x) = h(x) ( f(x)+g(x) ) which is a composition of functions (Spivak, 1996). on [0, π ] Solution: NO because cos (π/2) = 0, thus 1/cos(π/2) is undefined (Spivak, 1996). e) Find k so that the following function is continuous on any interval: Solution: In order to find out the k value, we must to show that . We want that F(3) = 3k = . If the limit exists, we must calculate: = = In this way: is equal to 5, because F(x) is defined equal to 5 for 3 < x. On the other hand: is equal to 3k, because F(x) is defined equal to k x for x ≤ 3. Thus, == iif, 5 = 3k or k = 5/3. Then the function F(x) is written as: F(x) is continuous. 6. Sketch the graphs of three different functions that are continuous on 0 ≤ x ≤ 1 and that have the values given in the table. The first function is to have exactly one zero in [0, 1 ], the second is to have at least two zeros in the interval [0.6, 0.8]. and the third is to have at least two zeros in the interval [0,0.6]. x 0 0.2 0.4 0.6 0.8 1.0 F(x) 1.0 0.9 0.60 0.11 -0.58 -1.46 Solution: Figure 4: It shows the plot of the function F(x). Figure 5: It shows a sketch when the function has one zero within the interval [0,1]. Figure 6: It shows a sketch when the function has three zeros in the interval [0.6,0.8]. Figure 7: It shows a sketch when the function has 6 zeros in the interval [0.6,0.8]. 7. f (x) = sin 3 x, for the functions in 7) do the following: a. Make a table of values of f (x) for x = 0.1, 0.01, 0.001, 0.0001, -0.1, -0.01, -0.001, and -0.0001. Solution: b. Make a conjecture about the value of Solution: The limit seems to be zero. Because x values near to zero by left hand-side and right hand-side is close to zero as we can see in the table above. c. Graph the function to see if it is consistent with your answers to parts (a) and (b) Figure 8: It shows the plot of the function sin(3x) (red line) as well as the blue dots (table values). d. Find an interval for x near 0 such that the difference between your conjectured limit and the value of the function is less than 0.01. ( in other words, find a window of height 0.02 such that the graph exits the sides of the window and not the top or bottom of the window. Figure 9: It shows a small window around zero. 8. Graphs of Exponential Functions To see how the value of b affects the graph of an equation in the form y = bx, graph the equations y = 3x, y = 4x, y = 7x, and y = .8x for x ≥ 0 on the same pair of axes. Describe the similarities and differences in the curves. To help in your exploration, feel free to use a graphing calculator or spreadsheet. Solution: Figure 10: It shows the potential functions F(x), G(x), H(x) and I(x). Note that when the base is large at the same x value, the function grows faster than when the base is small. Figure 11: It shows the same function, but in logarithm scale in the y-axis. Note that the function with large base at the same x value grows faster than when the base is small. 9. Area of a square. Write a paragraph in which you explain to a middle-school student that the area of a square whose sides each have length is equal to 2. Use two different arguments, one based on geometry and one on the ides of approximating the number by rational numbers. Solution: The area of a square is given by the product of its side. If the side of a square is a, then the area is equal to a2. If the side of a square is a number symbolized as . Then the area is given by 2. Because x = 2. There is an interest formula to calculate approximation of roots (O’Beirne 1955). This formula can be written by: n1 = 0.5 x (n0 + 2 / n0) If you give a guess n0 then we have a better value n1. For example root of 2. Initial guess: n0 =1.5 or 3/2. Since (3/2)2 = 2.25 this cannot be too bad for root of 2. In order to get a better for root of 2; after some algebra: n1 = ½ ( 3/2 + 2 / ( 3/2) ) n1 = ½ ( 3/2 + 4/3 ) n1 = ½ (17/6) n1 = 17/12=1.4167 At the end of this step, we can take the last value as our new initial guess and we get a better approximation and so on. A better approximation for root of 2 can be written as: 99/70 = 1.41421 35623 73095 04880 16887 24209 69807 85696 71875 37694 80731 76679 73799 (to 65 decimal places) (O’Beirne 1955). 10. Dan and John are trying to compute the infinite sum: 1 – 1 + 1 – 1 + 1 – 1 + … Dan writes: (1 – 1) + (1 – 1) + (1 – 1) + … = 0 = 0 = 0 and concludes that the sum of the series is 0. John writes: 1+ (– 1 + 1) + (-1 + 1) + (-1 + 1) +… = 0 = 0 = 0 and concludes that sum must be equal to 1. Who do you think is right? Solution: The above conclusions do not make a sense. The question is not at all trivial. This is a famous infinitum series, where units are subtracted and added intermittently. This can be seen as: S is associated with the series: S = 1 - 1 + 1 - 1 + ... We may also write S = 1 - (1 - 1 + ...) = 1 – S. From which 2S = 1 and S = 1/2. The infinite sum is equal to ½. The mathematician, L. Euler in particular, believed this to be true (Bogomolny 2010). Euler relied on the formula for the geometric series: 1/(1 - x) = 1 + x + x2 + x3 + ..., which he was willing to consider as the definition of the infinite sum on the right for any x, for which the left side was defined. In particular, for x = -1, we obtain:   1/2 = 1 - 1 + 1 - 1 + 1 - ... 11. Billy bob has 200 yards of fencing material, and he wishes to build a rectangular kennel with four sections, as shown below. Find the dimensions of the individual pens if the total enclosed area is to be as large as possible? Prove that you have accurately determined the total maximum area. _________________________ [ ] ] ] ] [ Pen ] pen ] pen ] pen ] (______)____ )____)____ ) Solution: The perimeter is given by l = 200 yards. The area of each individual rectangle is given by its width x per height H. Thus, the total enclosed area is given by: 4 x H. In order to find the H value, we release that 2 H + 4x + 4x = l is the perimeter of the rectangular fence. Thus, 2 H = l – (4x + 4x). Figure 12: It shows the final dimension of the Rectangular divisions. Total enclosed area as a function of x, where x is the width of one interior rectangle: Area(x) = 4 x ( 200 – (4x+4x) ) / 2 Figure 13: It shows the total area as a function of width x. In order to find its maximum area, we get it in the following way: There is a critical point at x = 400/32 = 12. Whereas the second derivative is given by: Which is negative, thus there is a maximum area at x = 12. Area(12) = 4 * 12 * ( 200 – 8*12 ) / 2 = 2496 yards2 Thus individual pens have dimensions: Width x = 12 and Height H = 52. On the other hand, perimeter is given by: 2 H + 4x + 4x = l. 2 * 52 + 4*12 + 4*12 = 200 yards, perimeter. Find a possible formula for the graph. Solution: The estimate formula is given by: The estimate formula is plotted over the handwrite graph. The figure 15 below shows the estimate formula f(x). Figure 14: Estimate function f(x) vs handwrite graph. Figure 15: It shows the estimate function f(x). References: Mathwords, 2008. “Terms and Formulas from Algebra I to Calculus” http://www.mathwords.com/l/logarithm_rules.htm, Accessed 22 October 2010. Lawrence Spector, 2010. “The math page”, http://www.themathpage.com/atrig/graphs-trig.htm, Accessed 22 October 2010. Michael Spivak 1996 “Calculus: Infinitesimal calculus”. Reverte ISBN 8429151362, 9788429151367 Thos H. O’Beirne 1955 “Approximations to Roots” The Mathematical Gazette Vol 39, No 330 pp 303-305. A Bogomoly, 0.9999999…=1? From Interactive Mathematics Miscellany and Puzzles” http://www.cut-the-knot.org/arithmetic/999999.shtml, Accessed 22 October 2010. Read More