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Graphing circular functions

  1. Sep 14, 2010 #1
    1. For graph like -3cos(1/2(theta- 3π/2) + 2 what's the best method for finding the relevant intercepts and coordinates of turning points and such? domain being equal to [-π, π]. Also if it were different domain would that only affect the end points in my calculation?

    2. For 100cos[π(x-400)/600]+30 for 0<x<1600 I tried finding the x-intercepts of this equation by hand by finding the inverse cos of -3/10 then solving for x but that only gives me the second intercept of 758... how do I get the first intercept at 41.8?

    Also I tried to find the integral of 100cos[π(x-400)/600]+50 for boundaries 800 to 1200 and received the answer of 59937.66 when it should be 13080, what mistake did I make?

    3. How do I differentiate V= π(h+1)[(ln(h+1))² + h]?
  2. jcsd
  3. Sep 15, 2010 #2
    Anyone? I'm really just concerned with number 1 as I need a solution to continue my studies...
  4. Sep 15, 2010 #3


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    The basic "y= cos(x)" has "amplitude 1", period [itex]2\pi[/itex], and goes through (0, 0).
    In y= A cos(mx+ b)+ c, the "A" 'stretches' the amplitude while the "c" 'moves' the graph up or down. Since the period of cos(x) is 0 to [itex]2\pi[/itex] you need to look at mx+ b= 0 and [itex]mx+ b= 2\pi[/itex] to find starting and ending values for a single period.

    In, for example, [itex]y= -3 cos(1/2(\theta- 3\pi/2)+ 2[/itex], the amplitude is 3. [itex]1/2(\theta- 3\pi/2)= 1/2\theta- 3\pi/4= 0[/itex] when [itex]1/2\theta= 3\pi/4[/itex] so [itex]\theta= 3\pi/2[/itex] and [itex]1/2\theta- 3\pi/4= 2\pi[/itex] when [itex]1/2\theta= 5\pi/4[/itex] so [itex]theta= 5\pi/2[/itex].

    I would recommend marking [itex]x= 3\pi/2[/itex] and [itex]x= 5\pi/2[/itex] on the x-axis. Also mark halfway between them: [itex](3\pi/2+ 5\pi/2)/2= 8\pi/4= 2\pi[/itex] as well as the "quarter points" [itex](3\pi/2+ 2\pi)/2= 7\pi/4[/itex] and [itex](2\pi+ 5\pi/4)/2= 9\pi/4[/itex].

    Now, the lowest value of cos(x) is -1 so the highest value of -3 cos(x) is 3 and the highest value -3 cos(x)+ 2 is 5. The highest value of cos(x) is 1 so the lowest value of -3 cos(x) is -3 and the lowest value of -3cos(x)+ 2 is -1. Draw light erasable horizontal lines at y= -1 and y= 5.

    You should have an idea of what the graph of the base function, cos(x), looks like. It starts each period at y= 1, drops down to 0 1/4 of the way through the period, then -1 half way, back up to 0 3/4 of the way, ending the period at y= 1 again. Since the coefficient of cosine is negative for this function, that pattern is reversed. The graph will start at y=-1, go up to 0 and then y= 5, drop down through 0 to -1 again.

    When [itex]\theta= 3\pi/2[/itex], calculated above as the starting value for one period, the value inside the cosine is x= 0 so the 'beginning point", is at [itex](3\pi/2, -1)[/itex]. When [itex]\theta= 5\pi/2[/itex], calculated above as the ending value for one period, the inside the cosine is [itex]2\pi[/itex] so the 'ending point" is at [itex](5\pi/2, -1)[/itex]. At the midpoint, the graph is at [itex](2\pi, 5)[/itex]. At the "quarter points", y is half way between -1 and 5, (-1+ 5)/2= 2. Mark those five points and draw the "cosine shape" between them. You can repeat that due to the periodicity of the cosine.

    First, the highest y value for this graph is 100+ 30= 130 and the lowest is -100+ 30= -70. Draw light horizontal lines at y= -70 and y= 130 as guides.

    [itex]\pi(x-400)/600= 0[/itex] when [itex])\pi(x- 400)= 0[/itex] or [itex]x= 400[/itex]. One period will start at [itex](400, 100+ 30)= (400, 130)[/itex]. [itex]\pi(x- 400)/600= 2\pi[/itex] when [itex]x- 400= 2/600= 1/300[/itex] or [itex]x= 400+ 1/300[/itex]. On period will end at (400+1/300, 100+ 30)= (400+ 1/300, 130). Midway between 400 and 400+ 1/300 is 400+ 1/600. At that point the cosine will be -1 so the midpoint of the graph is at (400+ 1/300,-100+ 30)= (400+ 1/300, -70). The two "quarter points" are at 400+ 1/1200 and 400+ 1/600+ 1/1200= 400+ 1/400. cosine there is 0 so the quarter points are (400+ 1/1200, 30) and (400+ 1/400, 30).

    How can anyone possibly tell you want mistake you made if you don't tell us what you did!?

    Well, using the product rule and the chain rule, of course,
    [tex]V'= \pi((h+1)'[(ln(h+1))^2+ h]+ (h+1)[2(ln(h+1))\frac{1}{h+1}+ 1][/tex]
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