Team Homework.docx

Next we know that if we plug 2 π into our formula

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Next, we know that if we plug 2 π into our formula for the radius, the radius is once again 3. This means that the total angles that trace out this figure are from 0 to 2 π . So, we can solve our equation to find the area. 3 + sin ( 5 θ ) ¿ 2 dθ≈ 19 π 2 29.8451 1 2 ¿ 0 2 π ¿ b. We know that the smallest value of our radius of the function given for the star will be the largest radius that the circle can be, so we can do this by looking at the graph of the function 3 + sin ( 5 θ ) and find that the minimum value of the graph is 2. We can also know from the graph that the first 5 values of theta at these points where the radius equals 2 are 3 π 10 , 7 π 10 , 11 π 10 , 3 π 2 , 19 π 10 .

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c. We know the formula for the arc length of a polar figure is length = α β . We also know from part a that the values of theta that trace out the figure are form 0 to 2 π . Now, we need to find f ' ( θ ) which from derivative rules we can find out is 5cos ( 5 θ ) . So now we can plug these values into our function and get 0 2 π . d. To evaluate the area of one of these pieces, we can subtract two integrals. First, we need to find our values for alpha and beta by setting the two polar equations equal to each other. 3.5 = 3 + sin ( 5 θ ) , 0.5 = sin ( 5 θ ) ,sin 1 ( 0.5 )= 5 θ 0.52 / 5 = θ . θ = 0.105, π 6 , and we can also verify this by looking at a graph. Now, we can set up our integrals. 3.5 ¿ 2 d θ≈ 0.506 ¿ 1 2 ¿ 3 + sin ( 5 θ ) ¿ 2 0.105 π 6 ¿ 1 2 ¿ 0.105 π 6 ¿ . e. To do this, we first set up our derivative as it is shown in the problem, then we find what values of theta make that derivative equal 0 between 0 and 2 π . So,
we do: y ' ( θ ) x' ( θ ) = 5sin ( θ ) cos ( 5 θ )+ cos ( θ )( sin ( 5 θ )+ 3 ) 5cos ( θ ) cos ( 5 θ )− sin ( θ )( sin ( 5 θ )+ 3 ) = 0 . Now, to solve for the values of theta for which this is true we can graph this and find how many times it crosses the x-axis between 0 and 2 π , which is 8.
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• Fall '07
• Irena

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