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Topic: Calling all Calculus Genius'!!!! (Read 688 times) |
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killacal
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Calling all Calculus Genius'!!!!
« on: May 24th, 2006, 1:13pm » |
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Solve this one!!!
Graph y= 3+2cos (x/2 - 3.14/4)
I've been trying since yesterday!! I can't stand this stuff!
HELP!!! 
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SMQ
wu::riddles Moderator
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Re: Calling all Calculus Genius'!!!!
« Reply #1 on: May 24th, 2006, 2:01pm » |
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Well, let's see: first off, we need to know some values of the cosine function by itself:
cos(0) = cos(2pi) = 1
cos(pi/6) = cos(11pi/6) = sqrt(3)/2
cos(pi/4) = cos(7pi/4) = sqrt(2)/2
cos(pi/3) = cos(5pi/3) = 1/2
cos(pi/2) = cos(3pi/2) = 0
cos(2pi/3) = cos(4pi/3) = -1/2
cos(3pi/4) = cos(5pi/4) = -sqrt(2)/2
cos(5pi/6) = cos(7pi/6) = -sqrt(3)/2
cos(pi) = -1
Hopefully with those you could graph y = cos(x) over 0 <= x <= 2pi. If x < 0 or x > 2pi, the pattern just repeats. And if you can do that, you can graph y = 3 + 2cos(x) just as easily -- put in the known points and connect the dots with a smooth curve.
Now, for the problem at hand, we need to find some points where x/2 - pi/4 is one of our known values, right? So for any of our known points, p, we'd like to find the value of x so that x/2 - pi/4 = p. A little algebra, and we have x = 2(p + pi/4). Looking at each of our known points, we get
p = 0:
x = 2(p+pi/4) = 2(0+pi/4) = 2(pi/4) = pi/2
y = 3+2cos(x/2-pi/4) = 3+2cos(p) = 3+2cos(0) = 3+2(1) = 5
p = pi/6:
x = 2(p+pi/4) = 2(pi/6+pi/4) = 2(5pi/12) = 5pi/6
y = 3+2cos(x/2-pi/4) = 3+2cos(p) = 3+2cos(pi/6) = 3+2(sqrt(3)/2) = 3+sqrt(3)
...and so on. Plot the points and connect the dots.
--SMQ
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--SMQ
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killacal
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Posts: 2
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Re: Calling all Calculus Genius'!!!!
« Reply #2 on: May 24th, 2006, 2:14pm » |
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WOW!! Thanks SMQ!
You really broke it down. I'm going to connect those dots, see what I come up with and get back at you.
I appreciate the help to the fullest. Bless your heart!! 
on May 24th, 2006, 2:01pm, SMQ wrote:Well, let's see: first off, we need to know some values of the cosine function by itself:
cos(0) = cos(2pi) = 1
cos(pi/6) = cos(11pi/6) = sqrt(3)/2
cos(pi/4) = cos(7pi/4) = sqrt(2)/2
cos(pi/3) = cos(5pi/3) = 1/2
cos(pi/2) = cos(3pi/2) = 0
cos(2pi/3) = cos(4pi/3) = -1/2
cos(3pi/4) = cos(5pi/4) = -sqrt(2)/2
cos(5pi/6) = cos(7pi/6) = -sqrt(3)/2
cos(pi) = -1
Hopefully with those you could graph y = cos(x) over 0 <= x <= 2pi. If x < 0 or x > 2pi, the pattern just repeats. And if you can do that, you can graph y = 3 + 2cos(x) just as easily -- put in the known points and connect the dots with a smooth curve.
Now, for the problem at hand, we need to find some points where x/2 - pi/4 is one of our known values, right? So for any of our known points, p, we'd like to find the value of x so that x/2 - pi/4 = p. A little algebra, and we have x = 2(p + pi/4). Looking at each of our known points, we get
p = 0:
x = 2(p+pi/4) = 2(0+pi/4) = 2(pi/4) = pi/2
y = 3+2cos(x/2-pi/4) = 3+2cos(p) = 3+2cos(0) = 3+2(1) = 5
p = pi/6:
x = 2(p+pi/4) = 2(pi/6+pi/4) = 2(5pi/12) = 5pi/6
y = 3+2cos(x/2-pi/4) = 3+2cos(p) = 3+2cos(pi/6) = 3+2(sqrt(3)/2) = 3+sqrt(3)
...and so on. Plot the points and connect the dots.
--SMQ |
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rmsgrey
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Re: Calling all Calculus Genius'!!!!
« Reply #3 on: May 24th, 2006, 5:05pm » |
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on May 24th, 2006, 1:13pm, killacal wrote:Solve this one!!!
Graph y= 3+2cos (x/2 - 3.14/4)
I've been trying since yesterday!! I can't stand this stuff!
HELP!!! |
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Another approach (rather than SMQ's perfectly viable find-and-plot-points approach) is to look at the function as made up of a simpler function with transformations applied to it...
Breaking it down, I'd look at it starting from cos(x):
cos(x)
-> cos (x-Pi/2) by shifting the cos graph by Pi/2 to the right.
-> cos [ (x-Pi/2) /2 ] which is effectively cos(u) -> cos (u/2) (where u is a linear function of x) so stretches the graph by a factor of 2 parallel to the x axis.
-> 2{cos [ (x-Pi/2) /2 ]} which is just stretching by 2 parallel to the y-axis.
-> 3+ (2{cos [ (x-Pi/2) /2 ]}) and finally shift everything up by 3.
If you sketch each graph in turn, you'll see that the transformations I describe move you from one to the next...
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