Calculus Problems and Tasks - Math Problem Example

1. A REGION TO BE DIVIDED INTO SUITABLE SUBREGIONS Compute the area of the region bounded by the graph of the parabola, the line, the , and the Solution: In order to find out the area of the region bounded, this is given by the following definite integral (Varberg 2007): The lower limit is set to a = 0, as we can see in the figure. The upper limit is calculated when: , i.e. b = 3. The definite integral is divided in two parts: I) using the parabola equation and ii) using the straight line. The intersection between both curves is at: . The solutions are given by: 2. WRITE a paragraph in which you explain to an elementary-school student that the area of a rectangle whose length is 1/2 and whose width is 1/3 is equal to 1/6. Solution: The area is defined as the size of a surface, i.e. the amount of space inside the boundary of a flat (2-dimensional) object such as square (Rod 2010). The area of a rectangle whose length is ½ and whose width is 1/3 may be calculated as follows: The area of a rectangle is given by the product of its length x width. In this case, we have two rational numbers L = ½ and W = 1/3 (Baldor 1992). Thus: Area = L x W = ½ x 1/3 = 1/6 3. Answer question LOTS OF SQUARES 1. IS IT POSSIBLE TO CUT A SQUARE INTO EXACTLY FOUR SQUARE PIECES? IF SO, HOW? Solution: Yes, it is possible. 2. IS IT POSSIBLE TO CUT A SQUARE INTO EXACTLY SEVEN SQUARE PIECES? IF SO, HOW? Solution: Yes, it is possible. 3. IS IT POSSIBLE TO CUT A SQUARE INTO ANY NUMBER OF SQUARE PIECES? IF NOT, WHICH NUMBERS ARE NOT POSSIBLE? HOW DO YOU KNOW? Solution: Not is possible to cut a square into any number of square pieces. However, it is possible to divide the square into 1, 4, 7, 10, 13, 16, 19, 22, 25, 28,… parts. i.e. it obeys the series given by the n-th term: 3n+1 for n = 0, 1, 2, 3,… Divisions Total of Squares 1 1 (original square) 1-1 + 22 4 (see figure above) 4-1 + 22 7 (see figure above) 4-2 + 2 * 22 10 4-3 + 3 * 22 13 4-4 + 4 * 22 16 7-1 + 22 10 7-2 + 2 * 22 13 7-3 + 3 * 22 16 7-4 + 4 * 22 19 7-5 + 5 * 22 22 7-6 + 6 * 22 25 7-7 + 7 * 22 28 10-1 + 22 13 10-2 + 2 * 22 16 And so on… And so on… Note: Squares may be of different sizes, but they may not overlap 4. Background info: Denote by the position function of an object moving along a straight line, then consider the following definitions: The velocity of the object at time t, denoted by , which is the instantaneous rate of change of h with respect to time, i.e. ; The speed of the object at time t, denoted by , which is the absolute value of the velocity , i.e. ; The acceleration of the object at time t, denoted by , which is the instantaneous rate of change of the velocity function with respect to time, i.e. Adrian received a model rocket for his birthday. The launcher is powered by an air pump, and the height attained by the rocket depends on the number of times Adrian pumps air. Adrian knows that five pumps will project the rocket upward according to the formula: , where is the rocket’s height, measured in feet from the ground, t seconds after take-off. 1. Sketch the graph of for . Is this the trajectory of the rocket? Explain. Solution: Yes, it is the trajectory because describe its position h(t) for any given time t (Wilson 1990). 2. Compute the velocity of the rocket. Solution: The velocity is given by the derivative of the position respect to the time, i.e. 3. Sketch the graph of and interpret it based on the rocket’s actual motion. Solution: The initial velocity at t = 0 is 128 feet/sec and the direction of movement is upward (Wilson 1990). 4. Compute the rocket’s velocity at 2, 3, and 5 seconds. How do you interpret the difference in signs of the values you obtained? Solution: The rocket’s velocity at 2 seconds is: -32 (2) + 128 = -64 + 128 = 64 feet/sec. The rocket’s velocity at 3 seconds is: -32 (3) + 128 = -96 + 128 = 32 feet/sec. The rocket’s velocity at 5 seconds is: -32 (5) + 128 = -160 + 128 = -32 feet/sec. At 2 seconds the velocity is 64 feet/sec and the direction of movement is upward. At 3 seconds the velocity is 32 feet/sec and the direction of movement is upward. At 5 seconds the velocity is -32 feet/sec and the direction of movement is downward. 5. Sketch the graph of . What is the speed of the rocket at times 2, 3, and 5 seconds? Solution: It shows the graph of speed of the rocket, s(t) vs t. The rocket’s speed at 2 seconds is: │-32 (2) + 128 │= │-64 + 128│ = 64 feet/sec. The rocket’s speed at 3 seconds is: │-32 (3) + 128│ = │-96 + 128│ = 32 feet/sec. The rocket’s speed at 5 seconds is: │-32 (5) + 128│ = │-160 + 128│ = 32 feet/sec. At 2 seconds the speed is 64 feet/sec, at 3 seconds the speed is 32 feet/sec and at 5 seconds the speed is 32 feet/sec. 6. After how many seconds will the rocket start falling? What will be the maximum height attained by the rocket? Solution: The maximum height is given when the rocket’s velocity is zero. This occurs when, i.e. . sec At 4 seconds the rocket has maximum height. The height at this time is: 7. After how many seconds will the rocket hit the ground? Solution: In order to calculate how many seconds will the rocket hit the ground, we must to solve the trajectory equation by setting, i.e. . After some algebra, we release that: The first solution is, i.e. and the second solution is ,i.e. The rocket hit the ground at 8 seconds after. 8. Determine the rocket’s acceleration. Solution: The acceleration is given by the derivative of the velocity respect to the time, i.e. Fundamental Theorem of Calculus 5. Problems: 1. Compute the area of the plane region bounded by the graph of , the x-axis, and the vertical line first using the Fundamental Theorem of Calculus, then using an area formula. Solution: The area under the curve is given by the defined integral (Verberg 2007): Where a = 0 and b = 2. Thus, On the other hand, the area of the triangle is given by: Base = 2 Height = 4 2. Compute the area of the plane region bounded by the graph of , the x-axis, and the vertical lines , and , first using the Fundamental Theorem of Calculus, and then using an area formula. Solution: The area under the curve is given by the defined integral (Verberg 2007): Where a = 2 and b = 4. 3. Compute the area of the plane region bounded by the graph of , the x-axis, and the vertical lines and using the Fundamental Theorem of Calculus. Solution: The area under the curve is given by the defined integral (Verberg 2007): Where a = 0 and b = 2. 4. Compute the area of the plane region bounded by the graph of , the x-axis, and the vertical lines and using the Fundamental Theorem of Calculus. Remember, can be rewritten as . Solution: The area under the curve is given by the defined integral (Verberg 2007): Where a = 1 and b = 5. 6. A) If f(x) is measured in pounds and x is measured in feet, what the units of ? Solution: The units are. B) Explain in words that the integral represents and give units: where s(x) is rate of change of salinity (salt concentration ) in gm/liter per cm in sea water, where x is depth below the surface of the water in cm. Solution: It is the total salinity in sea water from the surface until 5 cm depth below. C) Find the average value of the function over the given interval. F(x) = 2 over (a, b) Solution: D) Water is leaking out of a tank at the rate of R(t) gallons/hour where t is measured in hours. 1. Write a definite integral that expresses the total amount of water that leaks out in the first two hours. Solution: 2. In the figure below, shade the region whose area represents the total amount of water that leaks out in the first two hours. The red dashed region is the amount of water within 2 hours. 3. Give an upper and lower estimate of the total amount of water that leaks out in the first two hours. Solution: Upper limit: Total_upper = 2 (hrs) x 2.15 (gallons/hrs) = 4.3 gallons. Lower limit: Total_lower = 2 (hrs) x 0.75 (gallons/hrs) = 1.5 gallons. References: Aurelio Baldor 1992 “Geometry and Trigonometry”. Publicaciones Cultural. ISBN 968-439-214-1 Pirce, Rod. “Math is Fun” , http://www.mathsisfun.com/definitions/area.html, accessed on 24 October 2010. Dale Varberg, Edwing J: Purcell and Steven E. Rigdon 2007 “Caculus”. Prenntice Hall Inc. ISBN 0131429248 Jerry D. Wilson. 1990. “Technical College Physics”. McGraw Hill. ISBN 0-03-008494-6. Read More