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Let $f(x,y) = -(1-x^2-y^2)^{1/2}$ for $(x,y)$ such that $x^2+y^2 < 1$. Show that the plane tangent to the graph of $f$ at $(x_0,y_0,f(x_0,y_0))$ is orthogonal to the vector $(x_0,y_0,f(x_0,y_0))$.

I am confused about the overall steps of solving such a problem. The tangent plane is given by $z = f(x_0,y_0) - \frac {x_0} { \sqrt {1-x_0^2-y^2}}(x-x_0) - \frac {y_0}{ \sqrt {1-x_0^2-y_0^2}}(y-y_0)$, which has normal $( \frac {x_0} { \sqrt {1-x_0^2-y^2}} , \frac {y_0} { \sqrt {1-x_0^2-y^2}}, -1)$, and here I am stuck. Perhaps I have gone astray, any hints appreciated.

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Check your calculation of partial derivatives again. The tangent plane equation should be $z = f(x_0,y_0) + \frac {x_0} { \sqrt {1-x_0^2-y_0^2}}(x-x_0) + \frac {y_0}{ \sqrt {1-x_0^2-y_0^2}}(y-y_0)$, so that the normal vector should be $( \frac {x_0} { \sqrt {1-x_0^2-y_0^2}} , \frac {y_0} { \sqrt {1-x_0^2-y_0^2}}, -1)$

The two vectors $(x_0,y_0,f(x_0,y_0))$ and $( \frac {x_0} { \sqrt {1-x_0^2-y_0^2}} , \frac {y_0} { \sqrt {1-x_0^2-y_0^2}}, -1)$ only differ by a scalar.

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  • $\begingroup$ Excellent answer thanks! $\endgroup$ – IntegrateThis May 25 '17 at 3:17
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You are 95% done, except that you have a small sign mistake. Your equation for the curve is $z=-(1-x^2-y^2)^{1/2}$ or $z+(1-x^2-y^2)^{1/2}=0$

The normal direction is given by the gradient, so $$\vec{n}=\left(-\frac{x}{(1-x^2-y^2)^{1/2}},-\frac{y}{(1-x^2-y^2)^{1/2}},1\right)$$

Step 1, just plug in $x_0$ and $y_0$ into the equation for the normal. Now the plane tangent to the graph is orthogonal to your desired vector only if the normal and your vector are parallel. The simplest way to prove this is to use the cross product.$$\vec v \times \vec u=(u_yv_z-u_zv_y,u_zv_x-u_xv_z,u_xv_y-u_yv_z)$$ If the vectors are parallel, the cross product is $0$ Now plugging in $$\left(-\frac{x_0}{(1-x_0^2-y_0^2)^{1/2}},-\frac{y_0}{(1-x_0^2-y_0^2)^{1/2}},1\right)\times(x_0,y_0,-(1-x^2-y^2)^{1/2})=\\ \left(y_0-y_0,x_0-x_0,-\frac{x_0y_0}{(1-x^2-y^2)^{1/2}}+\frac{x_0y_0}{(1-x^2-y^2)^{1/2}}\right)=0$$

You could have noticed in this case that your vector is proportional to the normal direction since the normal is just your vector divided by $f(x_0,y_0)$

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The easiest, quickest, and clearest way that I know to see this is to re-write the equation

$f(x,y) = -(1-x^2-y^2)^{1/2} \tag{1}$

by first squaring it:

$f^2(x,y) = 1-x^2-y^2, \tag{2}$

and then adding $x^2 + y^2$ to each side:

$f^2(x,y) + x^2 + y^2 = 1; \tag{3}$

we recognize (3) as the equation of a sphere of radius $1$ centered at $(0, 0, 0)$. The radial vector to this sphere (which is in fact the unit sphere in $\Bbb R^3$) at $(x_0, y_0, f(x_0, y_0))$ is in fact $(x_0, y_0, f(x_0, y_0))$, hence is normal to the tangent plane at $(x_0, y_0, f(x_0, y_0))$.

One can also see this more analytically by noting that the normal to the tangent plane,

$(\dfrac {x_0} {\sqrt {1-x_0^2-y^2}} , \dfrac {y_0} {\sqrt {1-x_0^2-y^2}}, -1), \tag{4}$

is collinear with the radial vector:

$\sqrt {1-x_0^2-y^2}(\dfrac {x_0} {\sqrt {1-x_0^2-y^2}} , \dfrac {y_0} { \sqrt {1-x_0^2-y^2}}, -1) = (x_0, y_0, -\sqrt {1-x_0^2-y^2})$ $= (x_0, y_0, f(x_0, y_0)). \tag{5}$

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    $\begingroup$ Very interesting, but I'm afraid I'm not sure what you mean by radial vector. $\endgroup$ – IntegrateThis May 25 '17 at 3:17
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    $\begingroup$ The vector joining the origin to a point on the sphere; it points along the radius. $\endgroup$ – Robert Lewis May 25 '17 at 3:19

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