ScienceNewb said:
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?
The basic "y= cos(x)" has "amplitude 1", period 2\pi, 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 2\pi you need to look at mx+ b= 0 and mx+ b= 2\pi to find starting and ending values for a single period.
In, for example, y= -3 cos(1/2(\theta- 3\pi/2)+ 2, the amplitude is 3. 1/2(\theta- 3\pi/2)= 1/2\theta- 3\pi/4= 0 when 1/2\theta= 3\pi/4 so \theta= 3\pi/2 and 1/2\theta- 3\pi/4= 2\pi when 1/2\theta= 5\pi/4 so theta= 5\pi/2.
I would recommend marking x= 3\pi/2 and x= 5\pi/2 on the x-axis. Also mark halfway between them: (3\pi/2+ 5\pi/2)/2= 8\pi/4= 2\pi as well as the "quarter points" (3\pi/2+ 2\pi)/2= 7\pi/4 and (2\pi+ 5\pi/4)/2= 9\pi/4.
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 \theta= 3\pi/2, calculated above as the starting value for one period, the value inside the cosine is x= 0 so the 'beginning point", is at (3\pi/2, -1). When \theta= 5\pi/2, calculated above as the ending value for one period, the inside the cosine is 2\pi so the 'ending point" is at (5\pi/2, -1). At the midpoint, the graph is at (2\pi, 5). 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.
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?
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.
\pi(x-400)/600= 0 when )\pi(x- 400)= 0 or x= 400. One period will start at (400, 100+ 30)= (400, 130). \pi(x- 400)/600= 2\pi when x- 400= 2/600= 1/300 or x= 400+ 1/300. 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).
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?
How can anyone possibly tell you want mistake you made if you don't tell us what you did!?
3. How do I differentiate V= π(h+1)[(ln(h+1))² + h]?
Well, using the product rule and the chain rule, of course,
V'= \pi((h+1)'[(ln(h+1))^2+ h]+ (h+1)[2(ln(h+1))\frac{1}{h+1}+ 1]
