Scores and Estimate Grades

Note: E1 and E1Xtr amount to your exam 1 score. E2A+E2B amount to exam 2 score. Est Score is computed as (E1+E1Xtr+E2A+E2B)*0.75/2 + hw%*0.15 +Sum(Q1...q7)/4.5, and Estimated grade based on a straight scale. The curved scale would likely be different. 

13.4#3

I can't seem to setup my bounds properly for this problem. Honestly, setting up my bounds for all 
of these Green's Theorem problems has been a struggle. How should I go about setup up the bounds?



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Well, ok, you've got a piece of pie of radius 4. It's 1/8 of the whole pie. *If* you were going to actually calculate the circulation, that would be the path integral along the edge of the piece of pie. Since it has three pieces, you'd have to break up the path into three pieces: the part going outward from the origin along the x-axis, which can be parameterized as r(t) = <4t,0> for 0≤t≤1, the counterclockwise circular part of radius 4 moving from the x-axis upward and to the left to the point (2√2,2√2), which can be parameterized as r(t) = <4 cos(t), 4 sin(t)> for 0≤t≤π/4, and then the radial part moving inward that can be parameterized as r(t) = (1-t)<2√2, 2√2>. You'd have to substitute x(t) and y(t) for each part and then integrate with respect to t and add them all up.

*BUT* you're not going to do it that way, instead you're going to compute
∂F_2/∂x - ∂F_1/∂y = 5-(-4) = 9,
then use ∫∫_D ∂F_2/∂x - ∂F_1/∂y dA = 9 ∫∫_D dA = 9(Area of circle)/8 = 9(π 4^2)/8
=9*π*16/8 = 18π = 56.55

By the way, I changed the due date to tomorrow night.

12.7#11

I'm struggling with what to do for this problem. My phi bounds are pi/3 to pi/2. My theta bounds are 0 to 2pi. My rho bounds are 0 to 2. My integrand is rho^2 times sin(phi). Can you please give me a clue as to where I'm going wrong? I took this problem to the tutoring center and they couldn't figure it out either.

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 Let's see: z=3r means ρ*cos(φ)=  3ρ*sin(φ) so φ = tan^{-1}(1/3) =0.322 radians or 18.43 degrees. Your only mistake was jumping to the conclusion that webwork would default to a clean and easy lower bound.  

12.6#4

Theta can be found with the following formula, right?

arctan(y/x).   

The coordinate for this problem is (-3, 4, -1), and none of the follow are accepted:

arctan(4/-3) = -0.927295218001612 = -53.1301023542

The previous problem didn't have this issue, and the first entry for this problem accepts the following:

-3/(cos(arctan(4/-3))

Is there a way to fix this, or do I need to do something differently?

*************************************************************It requires a slightly more sophisticated notion of arctan.
The graph of the tangent function, with it's vertical asymptotes looks like this: This function doesn't satisfy the horizontal line test, so it has no inverse function. The red part of the graph is restriction of the domain of tan(x) to (-π/2, π/2), and it does satisfy the horizontal line test and the inverse function of this is what we USUALLY call arctan. These values of theta correspond to (x,y) in the first and fourth quadrants, where x>0.  The point (-3,4) is in the second quadrant though, and if you look at the angle of this point from the positive x-axis its going to have θ>π/2, corresponding to the blue curve on the right, or π-0.927295218001612 = 2.214

Lecture Notes 10/25/19 through 11/1/19

Lecture Notes 10/25/19

Lecture Notes 10/28/19

Lecture Notes 10/30/19

Lecture Notes 11/1/19