Question

y=3z+4x²

Answer 1

$y=3z+4x^2\\$

Making Z the subject of the formula, we get

$Z=\frac{1}{3}y-\frac{4}{3}x^2\\ F(x, y)=\frac{1}{3}y-\frac{4}{3}x^2\\$

The equation to the tangent of the surface is

∂F/∂x(a,b,c)(x−a)+∂F/∂y(a,b,c)(y−b)+∂F/∂z(a,b,c)(z−c)=0

Where

a=0

b=3

c=1

∂F/∂x=$\frac{-8}{3}x$

∂F/∂y=$\frac{1}{3}$

The equation of the tangent plane to the surface z=f(x,y) at the point (a,b,f(a,b)) is

∂f/∂x(a,b)(x−a)+∂f/∂y(a,b)(y−b)−z+f(a,b)=0

So the equation of the tangent plane at the point (0, 3, 1) is

$\frac{-8}{3}(0)(x-0)+\frac{1}{3}(y-3)-z+1=0\\ \frac{-8}{3}(0)+\frac{1}{3}y-1-z+1=0\\ 0+\frac{1}{3}y-z=0\\ y-3z=0\\$