#ATTENDANCE QUIZ for Lecture 7 of Math251(Dr. Z.)
#EMAIL  RIGHT AFTER YOU WATCHED THE VIDEO
#BUT NO LATER THAN  Sept. 28, 2020, 8:00PM (Rutgers time)
#THIS .txt FILE (EDITED WITH YOUR ANSWERS)
#TO:
#DrZcalc3@gmail.com
#Subject: aq7
#with an ATTACHMENT CALLED:
#aq7FirstLast.txt
#(e.g. aq7DoronZeilberger.txt)




#LIST ALL THE ATTENDANCE QUESTIONS FOLLOWED BY THEIR ANSWERS

Question 1
Compute the first-order partial derivatives.
24. S=tan^(-1)(w*z)
diff(arctan(w*z),w);
z/(w^2*z^2+1)
diff(arctan(w*z),z);
w/(w^2*z^2+1)
# (arctan(z))'=1/(1+z^2)
#d/dw arctan(w*z)=(1/(1+(w*z)^2))*(w*z)'=         #'=d/dw
#(1/(1+(w*z)^2))*z=z/(1+w^2*z^2)
#The answer is (z/(1+w^2*z^2))

Question 2
Find an equation of the tangent plane of the surface
#z=f(x,y)=x/sqrt(y)
#z-z0=f_x(x0,y0)*(x-x0)+f_y(x0,y0)*(y-y0)
#(x0,y0)=(4,4)
#z0=f(4,4)=4/sqrt(4)=4/2=2
#f_x=(x/sqrt(y))'			# ':=d/dx
#f_y=(x*y^(-1/2))'=x*(-1/2)*y^(-3/2)		# ':=d/dy
#f_x(4,4)=1/sqrt(4)=1/2
#f_y(4,4)=4*(-1/2)*4^(-3/2)=-2*1/(4^(3/2))=-2*(1/4^(1/2))^3
# -2*(1/2)^3=1/4
#(z-2)=1/2*(x-4)-1/4*(y-4)
#z=2+1/2*(x-4)-1/4*(y-2)
#z=2+(1/2)*x-2-(1/4)*y+2
#z=2+x/2-y/4
#The equation is z=2+x/2-y/4

Question 3
Use the linearization of the function f(x,y)=x/sqrt(y) to approximate (4.01)/sqrt(3.98) in other words f(4.01,3.98)
#(x0,y0)=(4,4)
#L(z)=z0+f_x(x0,y0)*(x-x0)+f_y(x0,y0)*(y-y0)
#L(z)=2+1/2*(x-4)-1/4*(y-2)
#L(4.01,3.98)=2+1/2(4.01-4)-1/4*(3.98-4)=2+1/2*0.01-1/4*(-0.02)
#2+1/2*(0.01)+1/2*(0.01)=2+0.01=2.01
(4.01)/sqrt(3.98);
2.010031376
#So the answer is 2.01