#$&* course mth 173 7/25 6 If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.
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Given Solution: ** We start with the rate dT/dt calculated at the beginning of the t interval for some given initial T. We then use this dT/dt to calculate the amount of change in T over the interval. Then T is calculated for the end of the interval. We then calculate a dT/dt for this T. The two values of dT / dt then averaged to obtain a corrected value. This is then used to calculate a new change in T. This change is added to the original T. The process is then continued for another interval, then another, until we reach the desired t value. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q Problem 1.5.13. amplitude, period of 5 + cos(3x) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: y = 5 + cos(3x) A = 1 P = 2pi/3 the 5 shifts the graph upwards 5 units confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: *&*& The function goes through a complete cycle as 3x changes from 0 to 2 pi. This corresponds to a change in x from 0 to 2 pi / 3. So the period of this function is 2 pi /3. The cosine function is multiplied by 1 so the amplitude is 1. The function is then vertically shifted 5 units. *&*& &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q Explain how you determine the amplitude and period of a given sine or cosine function. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: the constant in from of the sin or cos function is the amplitude and the period is 2pi divided by the constant next to the x in the function. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: *&*& GOOD ANSWER FROM STUDENT: Amplitude is the multiplication factor of the cosine function, such as the absolute value of a in y = a cos(x). the period is 2`pi divided by the coefficient of x. *&*& &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `q query Problem 1.5.27 5th edition, 1.5.28 4th edition. trig fn graph given, defined by 5 pts (0,3), (2,6), (4,3), (6,0), (8,3). What is a possible formula for the graph? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The max y value is 6 and the minimum is 0. So 6/3 = 2 and is the amplitude. To find the period I found that y = 3 at x = 0 and y returns to 3 at x = 8 so 2pi/B = 8 B = pi/4 And we see that at x = 0 y is not equal to 0 so we can use the cos function. Also we can see that the max y value does not occur at x = 0 so the graph has been shifted. the max y value occurs at the point (2, 6) so the function has been shifted 2 units to the right. so the formula is y = 3cos(pi/4(x-2)) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** The complete cycle goes from y = 3 to a max of y = 6, back to y = 3 then to a min of y = 0, then back to y = 3. The cycle has to go thru both the max and the min. So the period is 8. The mean y value is 3, and the difference between the mean y value and the max y value is the amplitude 3. The function starts at it mean when x = 0 and moves toward its maximum, which is the behavior of the sine function (the cosine, by contrast, starts at its max and moves toward its mean value). So we have the function 3 sin( 2 `pi / 8 * x), which would have mean value 0, shifted vertically 3 units so the mean value is 3. The function is therefore y = 3 + 3 sin( `pi / 4 * x). ** STUDENT QUESTION If there was a horizontal shift in the graph, would we also add that value to the function? Or would that make a sin function cosine and vice versa???? INSTRUCTOR RESPONSE Excellent question. A horizontal shift of h units results when x is replaced by (x - h). So for example a horizontal shift of 3 units would result if were were to replace our function y = 3 + 3 sin( `pi / 4 * x) with y = 3 + 3 sin( `pi / 4 * (x - 3)). The argument of this function could also be expanded to give us y = 3 + 3 sin( `pi / 4 * x - 3 pi / 4). If we wanted to express the original function a a cosine function, then since sin(theta) = cos(theta - pi / 2) we have y = 3 + 3 sin( `pi / 4 * x) = 3 + 3 cos( pi/4 * x - pi / 2) We can factor the argument to give us the form y = 3 + 3 cos( pi / 4 * (x - 2) ). The graphs of y = 3 + 3 sin( pi / 4 * x) and y = 3 + 3 cos( pi / 4 * (x - 2) ) are therefore identical. STUDENT QUESTION Is having 3 as a coefficient of sine not representative of the mean of the y-values??? INSTRUCTOR RESPONSE The sine function fluctuates by 1 unit above and below its mean value; multiplying the function by 3 causes it to fluctuate by 3 units. However this doesn't affect its mean value. The mean value of sin(t) is 0, and the mean value of 3 sin(t) is 0. Adding 3 to the fluctuating sine function increases its values by 3, so the mean value increases to 3. So the average value of 3 + 3 sin(t) is 3. In general the average value of A sin(t) + b is b, and the amplitude (the fluctuation above and below the mean) is the coefficient A of the sine function. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `q problem 1.5.29 5th; 1.5.30 4th. Solve 1 = 8 cos(2x+1) - 3 for x. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 1 = 8 cos(2x+1) - 3 4 = 8cos(2x+1) .5 = cos(2x+1) cos(pi/3) = .5 2x + 1 = 2pi/6 2x = 2pi/6 - 1 x = .5(pi/3 - 1) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: ** Starting with 1 = 8 cos(2x+1) - 3 : 1 = 8 cos(2x+1) - 3 Add 3 to both sides to get 4 = 8 cos(2x+1). Divide both sides by 8 and reverse sides: cos(2x + 1) = 1/2. We can find exact expressions for the angles theta at which cos(theta) = 1/2, and we will use this fact in our solution. Were it not possible to find exact expressions for such angles, we would begin by taking the inverse cosine of both sides. We begin by finding the angles theta for which cos(theta) = 1/2. It should be common knowledge that cos(pi/3) = 1/2. If not we should certainly know that cos(60 deg) = 1/2, and that 60 deg = pi/3 rad. The unit-circle model will then reveal that cos(theta) = 1/2 for value of theta which are coterminal with theta = pi/3 or with theta = 2 pi - pi/3 = 5 pi / 3. The angles coterminal with pi/3 are those which differ from pi/3 by an integer multiple of 2 pi, corresponding to an integer number of times around the unit circle. Thus the angles coterminal with pi/3 are the angles theta = pi / 3 + 2 pi n, where n can be any integer Similarly the angles coterminal with 5 pi / 3 are the angles theta = 5 pi / 3 + 2 pi n. Now we want the values of x for which cos(2x + 1) = 1/2. It should be clear that these are the values of x such that 2x + 1 = pi/3 + 2 pi n, or 2x + 1 = 5 pi / 3 + 2 pi n, where n is an integer. More specifically: 2x + 1 = `pi/3, or `pi / 3 + 2 `pi, or `pi / 3 + 4 `pi, ... or `pi / 3 + 2n `pi, ... with n = an integer. So x = (`pi / 3 - 1) / 2. Or x = (`pi / 3 + 2 `pi - 1) / 2 = (`pi / 3 - 1) / 2 + `pi. Or x = (`pi / 3 + 4 `pi - 1) / 2 = (`pi / 3 - 1) / 2 + 2 `pi. Or x = (`pi / 3 + 2 n `pi - 1) / 2 = (`pi / 3 - 1) / 2 + n `pi ... Thus x = (`pi / 3 - 1) / 2 + n `pi, n an integer, is the general solution for angles coterminal with pi/3. Solutions coterminal with 5 pi / 3 are found similarly. We solve 2x + 1 = 5 pi / 3 + 2 pi n, obtaining solutions x = (5 pi/3 + 2 pi n - 1), so that x = (5 pi / 3 -1) / 2 + n pi. We find the solutions for x between 0 and 2 pi: x = (`pi / 3 - 1) / 2 + n `pi takes values (pi/3 - 1)/2 - pi for n = -1, pi/3-1 for n = 0, (pi/3 - 1)/2 + pi for n = 1 and (pi/3 - 1)/2 + 2 pi for n = 2. The first solution is clearly negative, and won't be included. Since pi/3-1>0, the last solution is > 2 pi and won't be included. This leaves the two solutions x = (pi/3 - 1)/2 = pi/6 - 1/2, and x = (pi/3 - 1)/2 + pi = 5 pi / 6 - 1. The solutions of the form x = (5 pi / 3 -1) / 2 + n pi which lie between x = 0 and x = 2 pi are the n = 0 solution (5 pi / 3 - 1) / 2 = 5 pi / 6- 1/2, and the n = 1 solution (5 pi / 3 - 1) / 2 + pi = 11 pi / 6 - 1/2. (note that the n = -1 is negative, and n = 2 solution is > 2 pi). ** NOTEs ON UNIT-CIRCLE MODEL The figure below is the unit-circle model, with a vertical line at x = 1/2. It should be clear that the radial lines which intersect the circle at the same points as the vertical line lie at 60 degrees and 300 degrees relative to the positive x axis, so that the cos(theta) = 1/2 for theta = pi / 3 and for theta = 5 pi / 3.  In the preceding the radial lines would be as described below. (In the process of understanding the preceding statement you will have ideally sketched the above figure and drawn the radial lines.)  You should through prerequisite courses be familiar with the 30-60-90 triangle and the 45-45-90 triangle. The explanation below will remind you of the details: The 30-60-90 triangle is half an equilateral triangle. For example if an equilateral triangle with hypotenuse 1 is divided in half by a line through one of its vertices, either half will be a triangle with hypotenuse 1 and shorter leg 1/2. The Pythagorean Theorem then implies that the longer leg has length sqrt(3) / 2. The figures above and below represent unit circles with a line through x = 1/2. In the figure below the 'blue' triangle therefore has hypotenuse 1 and shorter leg 1/2, which makes it a 30-60-90 triangle. It follows that the 'base angle' of this triangle is 60 degrees, so the radial line that coincides with the hypotenuse is 60 degrees from the positive x axis. By symmetry it follows that the radial line in the fourth quadrant it at 300 degrees. This figure and the accompanying explanation demonstrate why cos(theta) = 1/2 for theta = 60 degrees and 300 degrees, or equivalently for theta = pi / 3 or 5 pi / 3.  OVERVIEW OF THE SOLUTION OF THE EQUATION cos(2x + 1) = 1/2. The above shows why cos ( pi / 3 ) = 1/2 and that cos(5 pi / 3) = 1/2. Therefore the equation cos(2x + 1) = 1/2 will be true if 2x + 1 = pi / 3. The equation will also be true if 2x + 1 = 5 pi / 3. Furthermore if we go around the circle, in either the positive or negative direction, starting from theta = pi / 3, we end up in the same position shown in the above figure. This is also true if we go around the circle any whole number of times, in either the positive or negative direction. To go once around the circle we go through an angle of 2 pi radians. To go n times around the circle we go through an angle of 2 pi n radians. Therefore the equation cos(2x + 1) = 1/2 will be true if 2x + 1 takes any of the values pi / 3 + 2 pi n, where n can be any positive or negative integer. Thus the equation is true for any solution to the equation 2x + 1 = pi / 3 + 2 pi n, where n can be any positive or negative integer. A similar argument shows that the equation cos(2x + 1) = 1/2 will also be true provided 2x + 1 = 5 pi / 3 + 2 pi n, where n can be any positive or negative integer. The original solution includes the remaining details. For more on the circular definition of trigonometric functions, and more, see the 'open qa's' on the Assignments Page, under Assignments 1-6, at the site http://vhcc2.vhcc.edu/pc2fall9/ STUDENT QUESTION I still am lost on what to do after I get to .5=cos(pi/3) INSTRUCTOR RESPONSE We need to solve the equation cos(2x + 1) = 1/2. You know that cos(pi/3) = 1/2. So we get one solution from 2x + 1 = pi / 3. Solving we get x = 1/2 * (pi/3 - 1). There are other angles for whose cosines are .5, and we get additional solutions from those angles. Some other angles whose cosines are pi/3 include: any angle which differs from pi/3 by 2 pi radians the angle -pi/3 and any angle which differs from this by 2 pi radians Setting 2x+1 equal to each of these angles, we get additional solutions for x. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q problem 1.5.52 5th; 1.5.47 4th; (was 1.9.34) arccos fn describe the graph of the arccos or cos^-1 function, including a statement of its domain and range, increasing/decreasing behavior, asymptotes if any, and concavity. Explain why the domain and range are as you describe STUDENT QUESTION Although it can be proven by a calculator, how can cos(5 pie /3) = cos( pie /3)??? INSTRUCTOR RESPONSE This follows from a fairly simple picture. Using the unit circle model cos(theta) is the x component of the unit radial vector making angle theta with the positive x axis. A radial vector at angle -theta will have the same x component, so cos(-theta) = cos(theta). (sketch a picture of a unit vector at 30 deg and another at -30 deg; they both have the same x coordinate; you can do the same for any angle, and should make some sketches until you convince yourself that you always get the same x coordinate). Now since -theta is coterminal with -theta + 2 pi, it follows that cos(-theta) = cos(-theta + 2 pi) = cos(5 pi / 3). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Starting from x = 1 the graph decreases at a decreasing rate and then when x = 0 the graph decreases at an increasing rate. The graph is undefined for x < -1 and x>-1 so this is the domain. The range is from 0 to 3.14. The range is from -1 to 1 because sin(x)^2 +cos(x)^2 = 1 so if either sin or cos is greater or less than one than there combined squares will be greater than 1 and this is not possible. confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The table below includes approximate values of cos(theta) vs. theta for theta between 0 and 2 pi. The approximate value .71 corresponds to the exact value sqrt(2) / 2, which comes from the basic right triangle with angle pi / 4. theta cos(theta) 0 1 pi/4 .71 pi/2 0 3 pi/4 -.71 pi -1 5 pi/4 -.71 3 pi/2 0 7 pi/4 .71 2 pi 1 Reversing the columns of the above table we get cos(theta) theta 1 0 .71 pi/4 0 pi/2 -.71 3 pi/4 -1 pi -.71 5 pi/4 0 3 pi/2 .71 7 pi/4 1 2 pi This table could correspond to a function, if the first-column numbers were all unique. However every number in the first column is repeated at least once. So for example if we ask what second-column number corresponds to the first-column number .71, we will get two answers, pi/4 and 7 pi / 4. We can still get a function out of the table if we choose a set of x values that is unique. We do this by choosing the fifth through eighth rows of the table, obtaining the table below. We name this function the arcCosine function. x arcCos(x) -1 pi -.71 5 pi/4 0 3 pi/2 .71 7 pi/4 If we imagine a more extensive table for the cosine function, we can see that the above process will work if we choose x values starting with -1 and going up to, but not including, x = 1. The graph of the cosine function, for 0 <= x < 2 pi, looks like this:  The graph of the cosine function, and the graph of the inverse function (shown in green), and depicted below. The line y = x is also depicted; each point of the inverse function can be found from a point of the cosine function by reversing its coordinates, which 'reflects' the point across the line y = x. These are principles covered in a precalculus course; if you aren't familiar with these concepts then you should review the topic of 'inverse functions'.  ** To get the graph of arccos(x) we take the inverse of the graph of cos(x), restricting ourselves to 0 < = x < = `pi, where y = cos(x) takes each possible y value exactly once. The graph of this portion of y = cos(x) starts at (0, 1), decreases at an increasing rate (i.e., concave down) to (`pi/2, 0) the decreases at a decreasing rate (i.e., concave up) to (`pi, -1). The inverse function reverses the x and y values. Since y values of the cosine function range from -1 to 1, the x values of the inverse function will run from -1 to 1. The graph of the inverse function therefore runs from x = -1 to x = 1. The graph of the inverse function starts at (-1, `pi) and runs to (0, `pi/2), reversing the values of y = cos(x) between (`pi/2, 0) and (`pi, -1). Since the graph of y = cos(x) is decreasing at a decreasing rate (i.e., concave up) between these points, the graph of the inverse function will be decreasing at a decreasing rate (i.e., concave up) over the corresponding region. The graph of the inverse function continues from (0, `pi/2) to (1, 0), where it decreases at an increasing rate. The values of the inverse cosine function range from 0 to `pi; the domain of this function is [-1, 1].** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: 3 " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!