Question

Score: 0 of 1 pt 2 of 10 13.6.11 Find an equation for the plane that is tangent to the given surface at the given point. z = Vy-x (0,1,1) Find the equation for the tangent plane to the surface z = Vy-x at the point (0,1,1). = 0

Answer 1

Let F be a function given by :

$F=\sqrt{y-x}-z$

It's partial derivatives are:

$F_x=-\frac{1}{2\sqrt{y-x}}$

$F_y=\frac{1}{2\sqrt{y-x}}$

and $F_z=-1$

Substituting the given point (0,1,1) in the above relations, we get:

$F_x=-\frac{1}{2}$

$F_y=\frac{1}{2}$

$F_z=-1$

So the gradient vector for the given surface at the point (0,1,1) is :

$\triangledown =\left ( \frac{-1}{2},\frac{1}{2},-1 \right )$

Thus equation of the tangent plane to the surface will be:

$\frac{-1}{2}(x-0)+\frac{1}{2}(y-1)-1( z-1)=0$or, multiplying all terms by 2, we get:

$-x+y-1-2z+2=0$

i.e. $-x+y-2z+1=0$

Or, multiplying all terms by-1, we get:

$x-y+2z-1=0$

Therefore the equation for tangent plane to the surface is $x-y+2z-1=0$ .

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