Lecture Notes 2/12/18
Lecture Notes 2/14/18
Lecture Notes 2/16/18
Lecture Notes 2/19/18
Tuesday, February 20, 2018
Tuesday, February 13, 2018
Friday, February 9, 2018
ASA DataFest 2018 registration now open! March 23-25, 2018
The dates have been set — March 23-25, 2018. Register now for the inaugural ASA DataFest competition at ASU, sponsored by Discover®.
Undergraduate students from any major are invited to participate in this 48-hour data hackathon. Put your team of 2 to 5 students together and register today!
Tuesday, February 6, 2018
Sunday, February 4, 2018
Section 10.9 due date changed to next Friday
I just changed the section 10.9 due date to Friday 2/9/18
Due dates and answers available dates
Good afternoon, wanted to ask about this sections due date since its closed
but not due until tonight.
Thanks for your help.
*************************
No one has every asked me that before but it's a natural confusion. The due date was on Friday. *Your* answers were due then. The *correct* answers for all of the problems for that section will be available later tonight.
but not due until tonight.
Thanks for your help.
*************************
No one has every asked me that before but it's a natural confusion. The due date was on Friday. *Your* answers were due then. The *correct* answers for all of the problems for that section will be available later tonight.
Friday, February 2, 2018
10.7#11
***** The feedback message: *****
I do not understand how to solve this problem. I tried taking the
derivative of each x,y, and z and then solving for t.
*************
Hmmm, you might be trying to represent the tangent line p using the symmetric equations. You could make that work but the parametric form r(t)=u+tv is better for this. And solving for t that way won't work because you are confusing the t that would be the parameter in the parametric equations for the line--that you solve for to get the symmetric equations--for the t that's a variable in the curve. THOSE ARE PARAMETERS FOR DIFFERENT CURVES. You can get the t for the curve from knowing that the point (7, 9, 32) is on the curve, like for example from the equation t^3-1=7 (hence t=2). From the definition of tangent line on the top of page 585 (in my book anyway) you know it has to pass through the point (7,9,32) and be parallel to r'(2)=<12,12,80>,
I do not understand how to solve this problem. I tried taking the
derivative of each x,y, and z and then solving for t.
*************
Hmmm, you might be trying to represent the tangent line p using the symmetric equations. You could make that work but the parametric form r(t)=u+tv is better for this. And solving for t that way won't work because you are confusing the t that would be the parameter in the parametric equations for the line--that you solve for to get the symmetric equations--for the t that's a variable in the curve. THOSE ARE PARAMETERS FOR DIFFERENT CURVES. You can get the t for the curve from knowing that the point (7, 9, 32) is on the curve, like for example from the equation t^3-1=7 (hence t=2). From the definition of tangent line on the top of page 585 (in my book anyway) you know it has to pass through the point (7,9,32) and be parallel to r'(2)=<12,12,80>,
Some Homework Due Dates Changed
I changed the homework due date for section 10.7 to Sunday Feb 4, and the 10.8 due date to Monday Feb 5.
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