Sec2Sess27

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Approximation formula:
If we change:
<math>\mathrm{x\ }</math> -> <math>\mathrm{\ x\ +\ change_x}</math>
<math>\mathrm{y\ }</math> -> <math>\mathrm{\ y\ +\ change_y}</math>
<math>\mathrm{w=f(x,y)\ then\ change_w \approx f_x*change_x+f_y*change_y}</math>

Justify the formula: tangent plane to w=f(x,y)
know: f(x),f(y) are slopes of 2 tangent lines

Combining planes:
<math>pd(x)(x_0,y_0)=a\ =>\ L_1\ =\ {w=w_0+a(x-x_0),\ y=y_0}\ (1st\ plane)</math>
<math>pd(y)(x_0,y_0)=b\ =>\ L_2\ =\ {w=w_0+b(y-y_0),\ x=x_0}\ (2cnd\ plane)</math>

<math>\mathrm{L_1,\ L_2\ are\ both\ tangent\ to\ the\ graph\ w\ =\ f(x,y)}</math>
Together they determine a plane:
<math>\mathrm{w=w_0+a(x-x_0)+b(y-y_0)}</math>

the approximation only works well for small delta x and small delta y

Approximation formula says: the graph of f is close to its tangent plane.

* The most important use for the tangent plane is the give an approximation that is the
basic formula in the study of functions of several variables - almost everything follows
in one way or another from it.

general equation of a plane through <math>(x_0,y_0,w_0)</math> is:
<math>A(x\ -\ x_0)+B(y\ -\ y_0)+C(w\ -\ w_0)\ =\ 0\ .</math>
Assuming plane is not vertical, C <> 0, divide through C to solve for <math>w - w_0</math>
<math>w\ -\ w_0\ =\ a(x-x_0)+b(y-y_0)\ a=A/C,\ b=B/C.</math>

"Linearization" of <math>f(x,y)\ at\ (x_0,\ y_0)</math>
<math>f(x,y)\ \approx\ w_0+pd(x)(x-x_0)+pd(y)(y-y_0)</math>
linear function with best approx. to f(x,y) for (x,y) close to <math>(x_0,y_0)</math>

the change (change over time) notation has an equivalent form:
<math>\mathrm{change_w\ \approx\ pd_0(x)*change_x+pd_0(y)*change_y\ if\ change_x\ \approx\ 0,\ and\ change_y\ \approx\ 0}</math>

for 3 variables:
w = f(x,y,z)
<math>\mathrm{change_w\ \approx\ pd_0(x)*change_x+pd_0(y)*change_y+pd_0(z)*change_z\ if\ change_x,\ change_y,\ change_z\ \approx\ 0}</math>

Beyond 3 variables the approximation formula does not work well.

Example 1
reasonable square at (1,1) over <math>w=x^3y^4</math> does not vary more than +/- .1
<math>pd(x)=3x^2*y^4\ ,\ pd(y)\ =\ 4x^3*y^3</math>
<math>x_0\ =\ 1,\ y_0\ =\ 1</math>
<math>\mathrm{change_w\ \approx\ pd_0(x)*change_x+pd_0(y)*change_y}</math>
apply <math>(x_0,y_0)</math> to pd(x) and pd(y)
<math>change_w\ \approx\ 3*change_x+4*change_y</math>
plug in values to make <math>change_w\ <=\ .1</math>
<math>change_x,\ change_y\ <=\ .01</math> work due to:
<math>.07\ \approx\ 3*.01+4*.01</math>
answer is therefore square at center (1,1) with:
|x-1|<= .01 and |y-1|<= .01

Which variable causes greatest results change:
Sensitivity Principle: The numerical value of w = f(x,y,...) calculated
at some point <math>(x_0,y_0,...)</math>, will be most sensitive to small changes in
that variable for which the corresponding partial derivative <math>w_x,w_y,</math>...
has the largest absolute value at the point.
NOTE: the largest absolute value seem to occur because taking the partial
derivative of the variable that contributes the least to the total leaves
the rest of the greatest contributing variables without their derivative
taken and makes that point contribute the most.

Area and other functions can be found with formula:
<math>\mathrm{f(x,y)\ \approx\ w_0+pd(x)(x-x_0)+pd(y)(y-y_0)}</math>
For example take <math>(X_0,y_0)</math> and use it in the formula with (x,y) to get
area solution. <math>w_0</math> can be found as x*y in area formula (A=x*y)