class: center, middle, inverse, title-slide .title[ # Total Derivative and Tangent Plane ] .subtitle[ ## Session 06 ] .author[ ### Alejandro Ucan ] .date[ ### 2024-02-25 ] --- # Goals for the session: * Approximate a function using a plane. <br/><br/> * Introduce the equation of the plane (and the Taylor polynomial). <br/><br/> * Describe the total derivative using this approximation. <br/><br/> --- # Motivation > The function `\(f(x,y)=\ln(x+y)\)` models the height of a physiological fault in California. Because we are in a field investigation, we do not have tools to calculate the value of the height at the point `\((0.3,-0.3).\)` How would you approximate this height?  --- # Recall > Recall that my partial derivatives `\(f_x\)` and `\(f_y\)` are the slopes of the tangent lines to the function at `\((x_0,y_0).\)` Let's see these tangent lines at the point `\((1,0)\)` which is close to the desired point. <iframe src="https://www.geogebra.org/classic/txd5dnqk?embed" width="1000" height="300" allowfullscreen style="border: 1px solid #e4e4e4;border-radius: 4px;" frameborder="0"></iframe> --- # The tangent plane > Note that if we want to know the equation of the plane `\(T\)` tangent to `\(f\)` at the point `\((x_0,y_0),\)` it necessarily has to contain all the tangent lines to the curves that pass through `\((x_0,y_0,f(x_0,y_0)).\)` <br/><br/> Remember that the equation of a line is: `$$z-z_0 = A(x-x_0)+B(y-y_0),$$` where `\(z_0=f(x_0,y_0).\)` <br/><br/> If we cut the function and the plane with the plane `\(y=y_0,\)` we obtain that `$$z-z_0=A(x-x_0)$$` and this is a tangent line to the curve. --- # The tangent plane > If we cut the function and the plane with the plane `\(x=x_0,\)` we obtain that `$$z-z_0 = B(y-y_0)$$` and this is a tangent line to the curve. <br/><br/> These two tangent lines are known, we know that they have slopes `$$f_x(x_0,y_0)\quad \mbox{and}\quad f_y(x_0,y_0).$$` Therefore, the equation of the tangent plane is `$$z-z_0=f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0).$$` --- ## How we approximate with the tangent plane? > Recall that we want to approximate the value of `\(\ln(0.3,-0.3)\)` and for this we will use the equation of the tangent plane to `\(f\)` at the point `\((1,0).\)` The partial derivatives of `\(f\)` at `\((1,0)\)` are: `$$f_x(1,0)=1\quad \mbox{and}\quad f_y(1,0)=-1.$$` So the tangent plane to `\(f\)` at `\((1,0)\)` is: `$$z-0=1(x-1)+(-1)(y-0).$$` The value of `\(z\)` when `\(x=0.3\)` and `\(y=-0.3.\)` <br/><br/> Substituting in the equation of the plane we obtain: `$$z=(0.3-1)+(-1)(-0.3)=-0.4.$$` So `\(f(0.3,-0.3)\approx -0.4.\)` --- #### Example 1: > Compute the equation of the tangent plane to `\(f(x,y)=2x^2+y^2\)` at the point `\((1,1,3).\)` -- 1. Find the partial derivatives of `\(f\)` at `\((1,1):\)` `$$f_x(1,1)=4(1)=4 \quad \mbox{y} \quad f_y(1,1)=2(1)=2.$$` <br/><br/> 2. Substituting the values of the partial derivatives and `\((x_0,y_0,z_0)=(1,1,3)\)` in the equation of the plane `$$z-3=4(x-1)+2(y-1).$$` --- #### Example 2: > Using the linearization of the function `\(f(x,y)=xe^{xy}\)` approximate the value of the function at `\((1.1,-0.1).\)` -- 1. Note that the point `\((1,0)\)` is close to `\((1.1,-0.1).\)` 2. Compute the partial derivatives of `\(f\)` at `\((1,0):\)` `$$f_x(1,0)=1e^0+(1)(0)e^0=1\,\mbox{y}\, f_y(1,0)=(1)^2e^0=1.$$` 3. Write the tangent plane: `$$z-1=1(x-1)+1(y-0)=x-1+y$$` 4. Aproximate `\(f(1.1,-0.1)\)` `$$z-1=1.1-1-0.1\Leftrightarrow z=0.1-0.1+1=1.$$` --- ### Example 3: > With the following table, approximate the values of the function when they are close to `\((96,70).\)`  --- ## Total Derivative > With the equation of the tangent plane to `\(f\)` at `\((x_0,y_0)\)` we can induce the value of the _total_ change of the function. Remember that the equation of the tangent plane is `$$z-z_0=f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0).$$` -- <br/><br/> If we move closer to `\((x_0,y_0)\)` my increments become infinitesimal and we can rewrite them as `$$dz=f_x dx + f_y dy$$` where `\(dx\)` and `\(dy\)` mean the increment that we have given from `\((x_0,y_0)\)` to `\((x,y).\)` --- #### Ejemplo > Find the total derivative of `\(f(x,y)=x^2+3xy-y^2.\)` Calculate the value of the total derivative when we change from `\((2,3)\)` to `\((2.05,2.95).\)` -- 1. Find the partial derivatives of `\(f:\)` `$$f_x(x,y)=2x+3y\quad\mbox{y}\quad f_y(x,y)=3x-2y.$$` 2. Express the diferencial: `$$dz= f_x dx+ f_y dy =(2x+3y)dx+(3x-2y)dy$$` 3. Compute the value of the total derivative: `$$dz=(2(2)+3(3))(0.05)+(3(2)-2(3))(-0.05)= 0.65$$` --- #### Example 4: > The radious of the base and the height of a circular cone have measures of `\(10cm\)` and `\(25cm,\)` respectively. But an error of `\(0.1cm\)` was detected in the measurement. Use the differential to estimate the maximum possible error in the volume.