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just a little fact about polar coordinates

The radial coordinate is a distance from the origin. Since a distance can never be negative, when you're doing an integral in polar coordinates the lower limit of integration can NEVER be less than zero.

12.3 #10

Hi Dr. Taylor,

I can't see how this is wrong. I know that for polar graphs are integrated in terms of r*dr*dtheta (or r*dtheta*dr) and cartesian graphs are integrated in terms of dydx or dxdy. Am I missing something?

Thank you!

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you left out the integrand f(x,y), respectively f(r cos(t), r sin(t))

12.3 #8

Hello,

Would you be able to help me understand why my bounds are wrong? I listed by bounds as a = 3pi/2, b = pi/2, c = 1, and d = 3 as integral set up from a to b, integral c to d (frdrdt).

Thank you,




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Well, there are a couple ways you got tangled up.  First of all, your outer limits of integration refer to the angular variable on the negative and positive y-axis while your inner variable is θ, so that you either need to have θ on the outside or make c,d refer to the angular variable.  Second, you have a bigger than b. That means when you do an integral that you are doing integral backwards, so that integral would be over the opposite half washer--the one with the negative x-values.  I recommend that you use -π/2 to π/2 instead.  

Lecture Notes 10/3/18 through 10/12/18

Lecture Notes 10/3/18

Lecture Notes 10/5/18

Lecture Notes 10/10/18

Lecture Notes 10/12/18

scores todate and projected final grade

Approximate score computed as 75% Exam, 15% HW and 10% Quiz percentage

11.6#3

I can't figure out how to figure out this equation. What I did first was finding the gradient of the equation, I got gradient of f = <-2xy, -x^2, 3z^2>. Then I solved it at the point, I got gradient of f = <10, -1, 3>. Then I found the unit vector of v. <-3,3,4>/sqrt(34). Finally I dot product to find the directional deriv, which i got -21/sqrt(34). Please help, I don't know what I'm doing wrong.

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As you say: grad(f) = <-2xy, -x^2, 3z^2> . 

This means  grad(f) (5, -1, 1)=<-2(5)(-1), -(5^2), 3(1^2)> = <10, -25, 3> unlike what you say. It looks like you made the mistake of substituting -1 for x in x^2, but really you needed to substitute x=5 into -x^2.

Then since v=<-3,3,4>,  v^ = <-3,3,4>/sqrt(34), as you say, so 

D_v^ f(5, -1, 1)=<10, -25,3>.<-3,3,4>/sqrt(34) =(-30-75+12)/sqrt(34)=-93/sqrt(34)

Lecture Notes 9-24-18 through 10-1-18

Lecture Notes 9-24-18

Lecture Notes 9-26-18

Lecture Notes 9-28-18

Lecture Notes 10-1-18