Can someone please help me with Q4. Thank you in advance (2) Suppose that z = f(
ID: 2891280 • Letter: C
Question
Can someone please help me with Q4. Thank you in advance
(2) Suppose that z = f(x, y) and that both x and y are functions of a single variable t such that Vf(x, y) = (1°, ) and dit - dy = t2-1. For what values of t in the interval [0, 0) is z increasing? (3) Find the directions in which the directional derivative of f(x,y) = r + sin(xy) at the point (1,0) has the value 1. (4) Find all the points (, y) where Vf(x, y) = 0 for f(x, y) = 13 - 24 - 2y+ + 31°y. (5) Find and classify the critical points of f(x, y) = r +r°y - Y - 4y.Explanation / Answer
4) We have given f(x,y)=x^3-2y^2-2y^4+3x^2y
fx=3x^2+6xy,fy=-4y-8y^3+3x^2
f(x,y)=<fx,fy>=<3x^2+6xy,-4y-8y^3+3x^2>=<0,0>
fx=3x^2+6xy=0
x(3x+6y)=0 implies x=0,3x=-6y,x=-2y
x=0,x=-2y
fy=-4y-8y^3+3x^2=0
plug x=0 into fy
-4y-8y^3=0
y(-4-8y^2)=0 implies y=0,8y^2=-4 implies y^2=-4/8=-1/2 implies y=sqrt(-1/2)=1i/sqrt(2),-1i/sqrt(2)
when x=0,y=0,1i/sqrt(2),-1i/sqrt(2)
plug x=-2y into fy
-4y-8y^3+3(-2y)^2=0
-4y-8y^3+12y^2=0
4(-y-2y^3+3y^2)=0
-y-2y^3+3y^2=0
y(-1-2y^2+3y)=0
y(-2y^2+3y-1)=0
y=0 or (-2y^2+3y-1)=0 implies (-2y^2+2y+y-1)=0 implies -2y(y-1)+1(y-1)=0 implies (y-1)(-2y+1)=0 implies y=1,y=1/2
so we have y=0,1,1/2
plug y=0,1,1/2 into x=-2y
y=0,x=0 ,y=1,x=-2,y=1/2,x=-1
the points are (0,1i/sqrt(2)),(0,-1i/sqrt(2)),(0,0),(-2,1),(-1,1/2) where f(x,y)=0