#ATTENDANCE QUIZ for Lecture 5 of Math251(Dr. Z.)
#EMAIL  RIGHT AFTER YOU WATCHED THE VIDEO
#BUT NO LATER THAN  Sept. 21, 2020, 8:00PM (Rutgers time)
#THIS .txt FILE (EDITED WITH YOUR ANSWERS)
#TO:
#DrZcalc3@gmail.com
#Subject: aq5
#with an ATTACHMENT CALLED:
#aq5FirstLast.txt
#(e.g. aq4DoronZeilberger.txt)



#THE ATTENDANCE QUESTIONS AND ANSWERS IN LECTURE 5

#LIST ALL THE ATTENDANCE QUESTIONS FOLLOWED BY THEIR ANSWERS

#ATTENDANCE QUESTION 1: 
#Let a be the 7th digit of your RUID.
#Let b be the 9th digit of your RUID.
#Find the curvature of the curve:
#r(t) = <t^a, t^b, t^(2*a)> at the point <2^a, 2^b, 4^a>

#ANSWER TO QUESTION 1:
#My RUID is 190008204. So, a =  2, and b = 4
#The equation for the curve is:
#r(t) = <t^2, t^4, t^4>
#The point that we want to find the curvature for is:
#<2^2, 2^4, 4^2> = <4, 16, 16>
#The formula for the curve's curvature is:
#kappa(t) = (|r'(t) x r''(t)|)/|r'(t)|^3
#Because kappa is in terms of t, we need to find at what t-value does the curve reach the point above:
#t^2 = 4, t^4 = 16, t^4 = 16
#Solution: t = 2
#We then take the derivative of the curve to find r'(t):
#r'(t) = <(t^2)', (t^4)', (t^4)'> = <2*t, 4*t^3, 4*t^3>
#We then take the derivative of r'(t) to find r''(t):
#r''(t) = <(2*t)', (4*t^3)', (4*t^3)'> = <2, 12*t^2, 12*t^2>
#To avoid a really complicated way of taking the cross product of r'(t) and r''(t), we plug in the t-value of 2 into each and then take the cross product of the answers:
#r'(2) = <2*2, 4*2^3, 4*2^3> = <4, 32, 32>
#r''(2) = <2, 12*2^2, 12*2^2> = <2, 48, 48>
#Now we calculate r'(2) x r''(2):
#r'(2) x r''(2) = <4, 32, 32> x <2, 48, 48> = 
# = i*(32*48 - 48*32) - j*(4*48 - 2*32) + k*(4*48 - 2*32) = 
# = i*0 - j*(192-64) + k*(192 - 64) = -128*j + 128*k = <0, -128, 128>
#The magnitude of the cross product is:
#|r'(2) x r''(2)| = sqrt(0^2 + (-128)^2 + 128^2) = sqrt(2*128^2) =
# = 128*sqrt(2)
#Now, we calculate |r'(2)|^3:
#|r'(2)| = |<4, 32, 32|> = sqrt(4^2 + 32^2 + 32^2) = 4*sqrt(129)
#|r'(2)|^3 = 8256*sqrt(129)
#Finally, we calculate the curvature:
#kappa(2) = (128*sqrt(2))/(8256*sqrt(129) = (2/16641)*sqrt(258) ≈ 0.001930458313
#ANSWER: The curvature of the curve at the specified point is (2/16641)*sqrt(258), or approximately 0.001930458313

#ATTENDANCE QUESTION 2:
#Who is your favorite rock band?

#ANSWER TO QUESTION 2:
#My favorite rock band is Santana