The Equation of the Circle - Assignment Example

Mathematics Assignment - final Question a) The graph below shows the path of the boat at time t and the circle at the limit of the man's visibility. (i) Note that the center of the circle is at the position (200, 100) and its radius is 200 m. Thus, using the standard form of writing the equation of the circle, (ii) To find the intersection of the two curves above, we must first find their equations. In (i), the equation of the circle is given. We wish to find the equation of the line. Taking any two points form the line, say, when t = 0, then the position of the boat is at (600, -300). When t = 300, then the position of the boat is (-300 +600, 2(300)-300) = (300, 300). Hence, the equation of the line is Thus, solving the system of two equations, we will have Substituting these values to any of the equation, we will have the two intersections to be at points (400, 100) and (320, 260). (iii) We want to find the time when the boat is in the two positions given in (ii). When the position is at (400, 100), t = 200. This can be solved by equating either the y or the x coordinate to its correspondent coordinate of the ordered pair (-t + 600, 2t -300). Also, by this method, at (320, 260), t = 280. Thus, the man can see the boat from time 200 sec to 280 sec or for the duration equal to 80 sec. To find the distance D, using the formula for the distance given two points in the Cartesian plane, we shall have Hence, for the span of 80 seconds, the boat has traveled about 179 meters. (b) The graph on the next page shows the position representation of the three vectors. The resultant vector has been solved graphically through the parallelogram method. (i) The graph above shows vb in blue, vc in yellow and the resultant vector v in black. Their magnitudes are given for a one-second span, i.e., vb has magnitude 2, vc has magnitude 0.5. Note that the direction of the line of the boat is N 40 degrees W, or equivalently, it is 130 degrees from the positive x-axis. We want to find the coordinate of the head of the vector, so Thus, vb = -1.3856i + 1.5321j. Also, note that the direction of the vector vc is S 20 degrees W, or equivalently, it is 250 degrees from the positive x-axis. Hence, its components are given by (ii). Note that the resultant vector is 145 degrees from the positive x-axis, thus using the Sine Law, After finding the magnitude of the resultant vector, r, then Thus, v = -1.3705i + 0.9596j. (iii). Fore a period of one second, the boat moves 0.9596m upward and 1.3705 westward. Thus, starting from (600, -300). The position of the boat at time t will be (600-1.3705t, -300 + 0.9596t). (iv). When the boat is directly south of the position (200, 100), then its x coordinate must be equal to 200, so Thus, at time 291.8711, the boat is directly south of (200, 100) and its position is (200, -19.92). Therefore it is about (100-(-20)) m away from that position or it is 120m away. Question 2 (a) (i) The length of the arc is equal to rX where X is the radian measure of the angle that intercepted the arc. Hence, the length of the wire can be expressed as Manipulating the equation and solving for l, we will have (ii) Note that the area A can be expressed as the difference between the area of the two sectors with radius 6 and with radius (6-l) both intercepted by angle x. Thus In terms of x only, substituting the value of l, (iv) Notice that the formula for A includes a factor (1-x) in the numerator, if we select a value of x greater than 1, then this factor will become less than 0, so the area becomes less than 0 which is not sensible. Also, if we set x to be less than 0, we are assuming that the angle that intercept the arc PS is less than 0 which also doesn't make sense. (b) The graph of f follows. (ii). Using the trace facility as shown, As seen, the graph of f has its maximum value at (0.62, 5.30). (iii) The solution of the equation f(x)=3 is 0.25 and 0.89. (c) (i) The terms of the recurrence relation are tabulated below. Term, n A B 0 0.300000 0.300000 1 0.254850 0.381619 2 0.248144 0.527520 3 0.247221 0.691433 4 0.247095 0.803571 5 0.247078 0.857266 6 0.247076 0.878644 Note the A converges more rapidly than B. (ii)Up to the 6th term of the sequence, the solution of the equation is given by 0.247076 and 0.878644. (d) (i) Solving for the derivative of f, (ii) If we set the first derivative to 0, then Thus x=0.621116 or x=1 or x=-7.77, The value of x that will give f the maximum is 0.621116. (iii) At the value stated above, the value of f that is the maximum it can have is 5.299919. Question 3 (a) (i) Differentiating the function (ii) To find for the stationary points, we set the first derivative of the function to be equal to 0, so we'll have (iii). Substituting the value of x to the function, then The interval at the left of (-/3), the first derivatives are increasing, after that, it decreases, thus - /3 is a local maximum. On the other hand, 2 /3 is a local minimum. (iv). Note that the maximum of the function is given by the function value when x is --/3, therefore, the maximum value of f is 3.067. However, when x is equal to 2/3, the value of f is not the minimum. If we graph f, we will notice that at x approaches negative infinity, the value of y decreases without bounds, thus f has a minimum when x is equal to -. (iv). To find the area, we must first found the bounds of the integral to be used. If we graph the function, (see graph below), note that using a software, the area is equal to 3.7985. (b) (i). Differentiating the function, (ii). Solving for the derivative of h, (iii). When x=5/4 and y=8, c=6, thus the initial value problem is Reference Derivatives. 30 August 2007. Read More