2.5 Tangent Planes and Normal Lines

What are the tangent planes to the two surfaces at $(0,0,0)$?

Yes. The plane $z=0$ is the tangent plane to both surfaces at $(0,0,0)$.

Apply the chain rule to $G\big(\vr(t)\big)=0$.

See the solution.

To find a tangent vector to the curve of intersection of the surfaces $F(x,y,z)=0$ and $G(x,y,z)=0$ at $(x_0,y_0,z_0)$, use Q[2] twice, once for the surface $F(x,y,z)=0$ and once for the surface $G(x,y,z)=0$.

The normal plane is $\vn\cdot\llt x-x_0\,,\,y-y_0\,,\,z-z_0\rgt =0$, where the normal vector $\vn = \vnabla F(x_0,y_0,z_0)\times \vnabla G(x_0,y_0,z_0)$.

To find a tangent vector to the curve of intersection of the surfaces $z=f(x,y)$ and $z=g(x,y)$ at $(x_0,y_0,z_0)$, use Q[2] twice, once for the surface $z=f(x,y)$ and once for the surface $z=g(x,y)$.

Tangent line is

$2x+y+9z=2$

$2x+y+z = 6$

(a) The tangent plane is $4x+2y+z=-3$ and the normal line is $\llt x,y,z\rgt=\llt -2,1,3\rgt+t\llt 4,2,1\rgt$.

(b) The tangent plane is $2y-z=-1$ and the normal line is $\llt x,y,z\rgt=\llt 2,0,1\rgt+t\llt 0,2,-1\rgt$.

$z = -\frac{3}{4} x- \frac{3}{2} y + \frac{11}{4}$

(a) $2ax -2ay +z = -a^2$ (b) $a=\frac{1}{2}$.

The tangent plane is $\frac{8}{25}x-\frac{6}{25}y-z=-\frac{8}{5}$.
The normal line is $\llt x,y,z\rgt = \llt -1,2,\frac{4}{5}\rgt +t \llt \frac{8}{25}\,,\,-\frac{6}{25}\,,\,-1\rgt$.

Let $(x,y,z)$ be a desired point. Then

$\pm(1,0,-2)$

First find a parametric equation for the normal line to $S$ at $(x_0,y_0,z_0)$. Then the requirement that $(0,0,0)$ lies on that normal line gives three equations in the four unknowns $x_0,y_0,z_0$ and $t$. The requirement that $(x_0,y_0,z_0)$ lies on $S$ gives a fourth equation. Solve this system of four equations.

$\big(\frac{1}{\sqrt{2}}\,,\,-1\,,\,-\frac{1}{2}\big)$ and $\big(-\frac{1}{\sqrt{2}}\,,\,-1\,,\,-\frac{1}{2}\big)$

Two (nonzero) vectors $\vv$ and $\vw$ are parallel if and only if there
is a $t$ such that $\vv=t\,\vw$. Don't forget that the point has to be on the hyperboloid.

$\pm \big(\half,-1,-1\big)$

The curve lies in the surface $z^2=4x^2+9y^2$. So the tangent vector to the curve is perpendicular to the normal vector to $z^2=4x^2+9y^2$ at $(2,1,-5)$.

The curve also lies in the surface $6x+3y+2z=5$. So the tangent vector to the curve is also perpendicular to the normal vector to $6x+3y+2z=5$ at $(2,1,-5)$.

$\pm\sqrt{3}\frac{\llt 3,14,-30\rgt}{|\llt 3,14,-30\rgt|} =\pm\sqrt{\frac{3}{1105}}\llt 3,14,-30\rgt$

At the highest and lowest points of the surface, the tangent plane is horizontal.

The horizontal tangent planes are $z=0$, $z=e^{-1}$ and $z=-e^{-1}$. The largest and smallest values of $z$ are $e^{-1}$ and $-e^{-1}$, respectively.

(a) $y+z=3$ $\vr(t)=\llt 0,2,1\rgt+t\llt 0,5,5\rgt$

(b) $z_x(x,y)=\frac{2-2x-y}{y+3z(x,y)^2}$ $z_y(x,y)=-\frac{x+2y+z(x,y)}{y+3z(x,y)^2}$ $z_y(0,2)=-1$

(c) $z_{xy}(x,y)=\frac{1}{y+3z(x,y)^2} -\frac{2-2x-y}{{[y+3z(x,y)^2]}^2} \left(1-6z(x,y)\frac{x+2y+z(x,y)}{y+3z(x,y)^2}\right)$

(b) If $\vv$ is tangent, at a point $P$, to the curve of intersection of the surfaces $S_1$ and $S_2$, then $\vv$

(a) $\llt 1,0,3\rgt$ (b) $\llt 3,3,-1\rgt$ (c) $\vr(t)=\llt 1,1,3\rgt+t\llt 3,3,-1\rgt$

The angle between the curve and the surface at $P$ is $90^\circ$ minus the angle between the curve and the normal vector to the surface at $P$.

$49.11^\circ$ (to two decimal places)

Let $D(x,y)$ be the distance (or the square of the distance) from $(1,1,0)$ to the point $\big(x,y, x^2+y^2)$ on the paraboloid. We wish to minimize $D(x,y)$. That is, to find the lowest point on the graph $z=D(x,y)$. At this lowest point, the tangent plane to $z=D(x,y)$ is horizontal.

$\frac{\sqrt{3}}{2}$