Michael Thomason's write-up for assignment eleven, question one.
Investigate
. First I assign values to a , b
, and k. Setting
them all equal to 1 seems like a good place to start.
> a:=1;b:=1;k:=1;
plot([a+b*cos(k*theta),theta,theta=0..2*Pi],coords=polar,scaling=constrained);
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![[Maple Plot]](images/assignment115.gif)
If
is the
equation for a "k-leaf rose" when a = b and k is an integer, then the above graph is a
1-leaf rose. Here are several graphs with a and b equal to the first few
integers after 1:
> for i from 2 to 6 do k:=i;plot([a+b*cos(k*theta),theta,theta=0..2*Pi],coords=polar,scaling=constrained);od;
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![[Maple Plot]](images/assignment118.gif)
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![[Maple Plot]](images/assignment1110.gif)
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![[Maple Plot]](images/assignment1112.gif)
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![[Maple Plot]](images/assignment1114.gif)
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![[Maple Plot]](images/assignment1116.gif)
One can see that k equals the number of petals in each rose.What if a = 0? Here I compare the above graph, which has a = b = 1 and k = 6, to a graph where b and k are the same as before, but a = 0.
> a:=0;b:=1;k:=6;
plot([a+b*cos(k*theta),theta,theta=0..2*Pi],coords=polar,scaling=constrained);
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![[Maple Plot]](images/assignment1120.gif)
This looks like a 12-leaf rose, but k only equaled 6...
> a:=1;b:=1;k:=12;
plot([a+b*cos(k*theta),theta,theta=0..2*Pi],coords=polar,scaling=constrained);
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![[Maple Plot]](images/assignment1124.gif)
Comparing the a = 0 graph to the standard 12-leaf rose shows that the a = 0 graph is smaller. Here are some other a = 0 graphs for various values of k .
> a:=0;b:=1;k:=2;
plot([a+b*cos(k*theta),theta,theta=0..2*Pi],coords=polar,scaling=constrained);
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![[Maple Plot]](images/assignment1128.gif)
> a:=0;b:=1;k:=3;
plot([a+b*cos(k*theta),theta,theta=0..2*Pi],coords=polar,scaling=constrained);
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![[Maple Plot]](images/assignment1132.gif)
> a:=0;b:=1;k:=4;
plot([a+b*cos(k*theta),theta,theta=0..2*Pi],coords=polar,scaling=constrained);
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![[Maple Plot]](images/assignment1136.gif)
> a:=0;b:=1;k:=5;
plot([a+b*cos(k*theta),theta,theta=0..2*Pi],coords=polar,scaling=constrained);
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![[Maple Plot]](images/assignment1140.gif)
> a:=0;b:=1;k:=6;
plot([a+b*cos(k*theta),theta,theta=0..2*Pi],coords=polar,scaling=constrained);
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![[Maple Plot]](images/assignment1144.gif)
It seems like odd k -values result in a k -leaf rose, but even k s yield 2 k -leaf roses. What if sine replaces cosine?
> a:=1;b:=1;k:=1;
plot([a+b*sin(k*theta),theta,theta=0..2*Pi],coords=polar,scaling=constrained);
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![[Maple Plot]](images/assignment1148.gif)
> a:=1;b:=1;k:=2;
plot([a+b*sin(k*theta),theta,theta=0..2*Pi],coords=polar,scaling=constrained);
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![[Maple Plot]](images/assignment1152.gif)
> a:=1;b:=1;k:=3;
plot([a+b*sin(k*theta),theta,theta=0..2*Pi],coords=polar,scaling=constrained);
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![[Maple Plot]](images/assignment1156.gif)
> a:=1;b:=1;k:=4;
plot([a+b*sin(k*theta),theta,theta=0..2*Pi],coords=polar,scaling=constrained);
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![[Maple Plot]](images/assignment1160.gif)
Changing
cosine to sine rotates the graphs. Click here to open the Maple worksheet from which this page was
created and play with the graphs for different values of a, b, and k.