Implicit Differentiation Calc Questions

Question

Giv‎en a funct‎ion g(x, y) defi‎ned imp‎licitly by the equ‎ation

z³ y^x + sin(x² - y²) = tan^-1(xz)

a) Find ∂z/‎∂x and ∂z‎/∂y . Do not over‎simplify.

(b) Let S be the gra‎ph of z³ y^x + sin(x² - y²) = tan^-1(xz) . Find the sym‎metric f‎orm of the nor‎mal li‎ne to S at the p‎oint (−1, 1, 0).

Answer 1

z³ y^x + sin(x² - y²) = tan^-1(xz)

same as F(x,y,z)=z³ y^x + sin(x² - y²) - tan^-1(xz)=0

to find symmetric form of normal line need F_x, F_y, F_z

which are computed at point of interest on the surface.

What exactly problem wants with dz/dx, and dz/dy, I am not sure

then for normal line it is (x-x₀)/Fx=(y-y₀)/F_y=(z-z₀)/F_z

_where F_x, F_y, F_{z are computed at point}

I find

(x+1)/-2=(y-1)/-2=z

same as x+1=y-1=-2z

does that match your computations?

Yefim and I differ

let's look at simpler example

for

G(x,y)=x²+y²-r²=0,,,at (x,y)=(-1,1)

want normal line,

My method

says it is (x+1)/G_x=(y-1)/G_y,,,,where partial derivatives computed at (-1,1)

(x+1)/-2=(y-1)2,,,,,y=-x as expected

Yefim's method, 2x+2y(dy/dx)=0,,, at (-1,1) dy/dx=1

Yefim method says dy/dy=1

and Yefim method is normal line is (x+1)/(dy/dx)=(y-1)/(dy/dy) ,,,dy/dx computed at (-1,1)

that is normal line is x+1=y-1 which is not correct

the problem is that if z=f(x,y) and F(x,y,z)=f(x,y)-z=0

Then yes F_z=-1 and the normal line is (x-x₀)/f_x=(y-y₀)/f_y=(z-z₀)/-1

when z is isolated to z=f(x,y) ,,,that is explicit, not implicit.

My method is reliable

Answer 2

a)3z²y^x∂z/∂x + z³y^xlny + 2xcos(x² - y²) = 1/(1 + x²z²)(z + x∂z/∂x). From here you can get ∂z/∂x

3z²y^x∂z/∂y + z³xy^x-1 - 2ycos(x² - y²) = 1/(1 + x²z²)x∂z/∂y Find from here ∂z/∂y

b) At point (-1,1,0): - 2 = - ∂z/∂x; ∂z/∂x = 2; - 2 = - ∂z/∂y; ∂z/∂y = 2; ∂z/∂z = 1

Symmetric equation of normal line: (x + 1)/2 = (y - 1)/2 = z/1