\section{Tangent Planes}
\begin{enumerate}[1.]
\item  {\it Find the equation of the tangent plane to the graph of the function
$f(x,y)=ye^{xy^2}$ at the point $(0,2)$.}

\bigskip

We have
\begin{align*}
f(0,2) &= 2\cdot 1 = 2 
\\
f_1(x,y) &= y \cdot y^2 e^{xy^2} = y^3 e^{xy^2},\ \ f_1(0,2) =8,
\\
f_2(x,y) &= e^{xy^2} + y\cdot 2xy e^{xy^2} = (1+2xy^2) e^{xy^2}, \ \ f_2(0,2) = 1.
\end{align*}
Hence the equation of the tangent plane is 
$$
z=2+ 8(x-0) +(y-2) = 8x + y .
$$

\bigskip

\item {\it Find the equation of the tangent plane to the surface $z=5x^2y-y^2$ at the point
where $x=-1$, $y=3$.}


\bigskip


We have $\displaystyle{\frac{\partial z}{\partial x}=10xy}$ and 
$\displaystyle{\frac{\partial z}{\partial y}=5x^2-2y}$. At $(x,y)=(-1,3)$,
we get
$$z=15-9=6,\ \ \frac{\partial z}{\partial x}=-30,\ \ \frac{\partial z}{\partial y}=5-6=-1.
$$
The equation of the
tangent plane is
$$
z=6-30(x+1)-(y-3)$$
which simplifies to $z=-30x-y-21$.

\bigskip

\item {\it Find all points $(x,y,f(x,y))$ where the tangent plane to the graph of the function
$f(x,y)=x^2+3xy-2y^2$ is parallel to the plane $x+10y-z=3$. For each such point, find the
equation of the tangent plane.}

\bigskip

We have $f_x(x,y)=2x+3y$ and $f_y(x,y)=3x-4y$, so the equation of the tangent plane at
$(x,y)=(a,b)$ is
$$
z=f(a,b)+(2a+3b)(x-a) + (3a-4b)(y-b),
$$
with the normal vector ${\bf N}=\langle -(2a+3b), -(3a-4b), 1)\rangle$.
We want to find all $(a,b)$ where this plane  is parallel to the plane $x+10y-z=3$, with the normal vector ${\bf n}=\langle
1,10,-1\rangle$. This means that ${\bf N}=\lambda {\bf n}$ for some $\lambda\in\rr$, so that
\begin{align*}
-(2a+3b) &= \lambda
\\
-(3a - 4b) &= 10\lambda
\\
1 &= -\lambda.
\end{align*}
The last equation says that $\lambda=-1$. From the first two equations, we then get 
\begin{align*}
2a+3b &= 1
\\
3a - 4b &= 10.
\end{align*}
We now solve this system. Here is one way to do it: 
adding the two equations, we get $5a-b=11$, so that $b=5a-11$. Plugging this into the first equation, we get
\begin{align*}
2a+3b &= 2a+3(5a-11)=2a+15a-33 = 1,
\\
17a & =34,
\end{align*}
so that $a=2$ and $b=5a-11 = 10-11 =-1$. We have $f(2,-1)=4-6-2=-4$. Thus the only point
where the tangent plane  is parallel to the plane $x+10y-z=3$ is $(2,-1,-4)$,
and the equation of that tangent plane is $z=-4+(x-2)+10(y+1) =x+10y+4$.


\bigskip

\item 
{\it Find all points $P=(1,b,c)$ such that $P$ lies on the paraboloid $z=x^2+y^2$ and the tangent plane to this
paraboloid at $P$ passes through the point $(2,4,18)$.}

\bigskip

The partial derivatives of $f(x,y)=x^2+y^2$ are $f_x(x,y)=2x$ and $f_y(x,y)=2y$. At $(x,y)=(a,b)$, the equation of the tangent plane is $z-(a^2+b^2) =2a(x-a)+2b(y-b)=2ax+2by-2a^2-2b^2$. Letting $a=1$, 
we get the equation 
$$
z=2x+2by-1-b^2.
$$
We are looking for all values of $b$ such that 
the point $(2,4,18)$ satisfies that equation: $18=4+8b-1-b^2$, $b^2-8b+15=0$, $b=3$ or 5. The
corresponding points on the paraboloid are $(1,3,10)$ and $(1,5,26)$.

\bigskip


\item {\it
Find all values of $(a,b)$ such that the tangent plane to the surface
$z=4x^2-y^2$ at $(a,b,4a^2-b^2)$ is parallel to the line with parametric
equations $x=t+1$, $y=2t$, $z=-4t+9$.}

\bigskip

Let $f(x,y)=4x^2-y^2$, then $f_x(a,b)=8a$ and $f_y(a,b)=-2b$. Hence the
tangent plane at $(a,b,4a^2-b^2)$ is perpendicular to the vector ${\bf n}=-8a\ii+2b\jj+\kk$.
We want ${\bf n}$ to be perpendicular to the direction vector ${\bf v}$ of the
given line. From
the equations of the line,
${\bf v}=\ii+2\jj-4\kk$. We therefore want ${\bf n}\cdot{\bf v}=0$, so that $-8a+4b-4=0$,
or simplifying, $-2a+b=1$. All values $(a,b)$ with $b=1+2a$
satisfy this condition.
\end{enumerate}
