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\centerline
{
\bf Dr. Z's Math251 Handout \#14.4 (2nd ed.) 
[Differentiability, 
Linear Approximations, and Tangent Planes]
}
By Doron Zeilberger

{\bf Problem Type 14.4a}: 
Find an equation of the tangent
plane to the given surface at the specified point.
$$
z=f(x,y) \quad , \quad (x_0,y_0,z_0) \quad .
$$
{\bf Example Problem 14.4a}: Find an equation of the tangent
plane to the given surface at the specified point.
$$
z=9x^2+y^2+6x-3y+5 \quad , \quad (1,2,18) \quad .
$$
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{\bf Steps} & {\bf Example} \cr  &  \cr
{\bf 1.} First make sure that $z_0=f(x_0,y_0)$ or refuse
to do the problem. Then take
$f_x={{\partial f}  \over {\partial x}}$
and $f_y={{\partial f}  \over {\partial y}}$.
&
{\bf 1.} 
$9\cdot 1^2+2^2+6\cdot 1-3\cdot 2+5=18$, so the
point $(1,2,18)$ indeed lies on the surface.
Now
$$
f_x=
{{\partial} \over {\partial x}} (9x^2+y^2+6x-3y+5)=
18x+6 \quad ,
$$
$$
f_y={{\partial} \over {\partial y}} (9x^2+y^2+6x-3y+5)
=2y-3 \quad .
$$
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{\bf 2.} Plug in $x=x_0,y=y_0$ into $f_x$ and $f_y$ that
you have just found.
&
{\bf 2.} 
$$
f_x(1,2)=18 \cdot 1 +6=24 \quad .
$$
$$
f_y(1,2)=2 \cdot 2 -3=1 \quad .
$$

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{\bf 3.} An equation for the tangent plane for the given
surface at the given point is
$$
z-z_0=f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0) \quad.
$$
Plug-in the $x_0,y_0,z_0$ from the data of the problem
and $f_x(x_0,y_0),f_y(x_0,y_0)$ from step 2.
&
{\bf 3.} $z-18=24(x-1)+(y-2) $. Or, in expanded form: $z=24x+y-8$.
\cr}


{\bf Problem Type 14.4b}: Explain why the function is
differentiable at the given point. Then find
the linearization of that function at the given point.
$$
z=f(x,y) \quad , \quad (a,b) \quad .
$$
{\bf Example Problem 14.4b}: Explain why the function is
differentiable at the given point. Then find
the linearization of that function at the given point.
$$
z=e^x \sin(xy) \quad , \quad (0,\pi/2) \quad .
$$
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{\bf Steps} & {\bf Example} \cr  &  \cr
{\bf 1.} Find the first partial derivatives $f_x$ and
$f_y$.

&
{\bf 1.} 
$$
f_x={{\partial} \over {\partial x}} e^x \sin(xy)=
({{\partial} \over {\partial x}} e^x) \sin(xy)+
e^x({{\partial} \over {\partial x}} \sin(xy)) 
$$
$$
e^x \sin(xy)+e^x y \cos(xy) \quad ,
$$

$$
f_y={{\partial} \over {\partial y}} e^x \sin(xy)=
e^x({{\partial} \over {\partial y}} \sin(xy)) 
=e^x x \cos(xy) \quad .
$$

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{\bf 2.} If both $f_x$ and $f_y$ are {\it continuous} at
the designated point $(a,b)$ (i.e. they are defined and
do not blow up), then the function is {\it differentiable}
at that point, and it is OK to have a linearization.
$$
L(x,y)=
$$
$$
f(a,b)+f_x(a,b)(x-a)+f_y(a,b)(y-b) \quad .
$$
&
{\bf 2.} $f_x=e^x \sin(xy)+e^x y \cos(xy)$ and
$f_y=e^x x \cos(xy)$ are both continuous at the
point $(0,\pi/2)$ (they are continuous everywhere
for that matter), so the function is differentiable
there.
Now
$$
f(0,\pi/2)=e^0 \sin(0)=0 \quad,
$$
$$
f_x(0,\pi/2)=e^0 \sin(0)+e^0 \cdot (\pi/2) \cos(0)=\pi/2 \quad ,
$$
$$
f_y(0,\pi/2)=e^0 \cdot 0 \cdot \cos(0)=0 \quad .
$$ 
The linearization is
$$
L(x,y)=0+(\pi/2)\cdot (x-0)+0\cdot (y-\pi/2)=
(\pi/2)x \quad .
$$
{\bf Ans.:} The linearization of the function 
$e^x \sin(xy)$ at the point $(0,\pi/2)$ is
$(\pi/2)x$.
\cr}

{\bf Problem Type 14.4c}: 
Use the linear apprimation of the function
$f(x,y)$  at $(a,b)$
to approximate $f(a_1,b_1)$, where $(a_1,b_1)$ is
``near'' $(a,b)$.

{\bf Example Problem 14.4b}: 
Use the linear approximation of the function
$f(x,y)=\sqrt{20-x^2-2y^2}$  at $(3,1)$
to approximate $f(3.05,.97)$.

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{\bf Steps} & {\bf Example} \cr  &  \cr
{\bf 1.} The beginning is exactly as before. Just
find the linearization of the function at the designated
point $(a,b)$.

So first find the first partial derivatives $f_x$ and
$f_y$.

&
{\bf 1.} 
$$
f_x={{\partial} \over {\partial x}}(20-x^2-2y^2)^{1/2}=
(1/2)(20-x^2-2y^2)^{-1/2} \cdot (-2x) 
$$
$$
={{-x} \over {\sqrt{20-x^2-2y^2}}} \quad ,
$$
$$
f_y={{\partial} \over {\partial y}}(20-x^2-2y^2)^{1/2}=
(1/2)(20-x^2-2y^2)^{-1/2} \cdot (-4y)
$$
$$
={{-2y} \over {\sqrt{20-x^2-2y^2}}} \quad .
$$
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{\bf 2.} If both $f_x$ and $f_y$ are {\it continuous} at
the designated point $(a,b)$ (i.e. they are defined and
do not blow up), then the function is {\it differentiable}
at that point, and it is OK to have a linearization.
$$
L(x,y)=f(a,b)+f_x(a,b)(x-a)+f_y(a,b)(y-b) \quad .
$$
&
{\bf 2.} $f_x={{-x} \over {\sqrt{20-x^2-2y^2}}}$ and
$f_y={{-2y} \over {\sqrt{20-x^2-2y^2}}}$ are both continuous at the
point $(3,1)$  (the argument of the square-root is $9$ which is positive
and nothing blows up).
Now
$$
f(3,1)= \sqrt{20-3^2-2 \cdot 1^2}=3 \quad .
$$
$$
f_x(3,1)={{-3} \over {\sqrt{20-3^2-2 \cdot 1^2}}}=-1 \quad ,
$$
$$
f_y(3,1)={{-2 \cdot 1} \over {\sqrt{20-3^2-2 \cdot 1^2}}}=-2/3 \quad .
$$ 
and the linearization is
$$
L(x,y)=3-(x-3)-(2/3)(y-1) \quad .
$$
So the {\bf linear approximation}, valid {\bf near} $(3,1)$ is
$$
f(x,y) \approx 3-(x-3)-(2/3)\cdot (y-1) \quad .
$$

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{\bf 3.} Plug-in $(a_1,b_1)$ into this approximation.
&
{\bf 3.}
$$
f(3.05,.97) \approx 3 - (3.05-3)-(2/3)\cdot (.97-1)=
$$
$$
3 -.05+.02=2.97 \quad .
$$
{\bf Ans.:} $f(3.05,.97) \approx 2.97$ .
\cr}

{\bf Problem from a previous Final}

Find an equation of the tangent plane to the surface
$$
z=e^{2x-3y} \quad ,
$$
at the point $(3,2,1)$. Simplify as much as you can!

{\bf Ans.}: $z=2x-3y+1$ (or $2x-3y-z=-1$ ).


\end


linearliz
