# 11. Partial Derivatives and Tangent Planes

## d. Tangent Plane to the Graph of a Function

## 2. Algebraic Formula

## b. Comparison of Tangent Line and Tangent Plane Derivations

Here is a comparison of the derivations of the tangent line to the graph of a function of \(1\) variable and the tangent plane to the graph of a function of \(2\) variables, in two-column format.

### Tangent Line

We want to construct the tangent line at a point, \(x=a\), on the graph of a function, \(y=f(x)\). The most general line has the standard equation: \[ Ax+By=C. \] The line is vertical if \(B=0\) and non-vertical if \(B\ne0\). Assuming it is not vertical, we can solve for \(y\) and put the equation into slope-intercept form: \[ y=mx+b \qquad (1) \] where \(m\) is the slope and \(b\) is the \(y\)-intercept.

### Tangent Plane

We want to construct the tangent plane at a point, \((x,y)=(a,b)\), on the graph of a function, \(z=f(x,y)\). The most general plane has the standard equation: \[ Ax+By+Cz=D. \] The plane is vertical if \(C=0\) and non-vertical if \(C\ne0\). Assuming it is not vertical, we can solve for \(z\) and put the equation into slope-intercept form: \[ z=mx+ny+c \qquad (1) \] where \(m\) is the \(x\)-slope, \(n\) is the \(y\)-slope and \(c\) is the \(z\)-intercept. (Note the \(x\)-slope is the slope of the \(x\)-trace and the \(y\)-slope is the slope of the \(y\)-trace.)

Now suppose we want to find the equation of the line tangent to \(y=f(x)\) at \(x=a\). We know the slope is \(m=f'(a)\). So equation \((1)\) becomes: \[ y=f'(a)x+b \qquad (2) \] We know the line passes through the point \((x,y)=(a,f(a))\). So equation \((2)\) tell us: \[ f(a)=f'(a)a+b \] or \[ b=f(a)-f'(a)a \] Using this formula for \(b\), equation \((2)\) becomes: \[\begin{aligned} y&=f'(a)x+f(a)-f'(a)a\\ &=f(a)+f'(a)(x-a) \end{aligned}\] which is the equation for the tangent line. We define the formula on the right to be the tangent function. \[ f_{\tan}(x)=f(a)+f'(a)(x-a) \] so that the equation of the tangent line is \(y=f_{\tan}(x)\).

Now suppose we want to find the equation of the plane tangent to \(z=f(x,y)\) at \((x,y)=(a,b)\). We know the \(x\)-slope is \(m=f_x(a,b)\) and the \(y\)-slope is \(n=f_y(a,b)\). So equation \((1)\) becomes: \[ z=f_x(a,b)x+f_y(a,b)y+c \qquad(2) \] We know the plane passes through the point \((x,y,z)=(a,b,f(a,b))\). So equation \((2)\) tell us: \[ f(a,b)=f_x(a,b)a+f_y(a,b)b+c \] or \[ c=f(a,b)-f_x(a,b)a-f_y(a,b)b \] Using this formula for \(c\), equation \((2)\) becomes: \[\begin{aligned} z&=f_x(a,b)x+f_y(a,b)y+f(a,b)-f_x(a,b)a-f_y(a,b)b\\ &=f(a,b)+f_x(a,b)(x-a)+f_y(a,b)(y-b) \end{aligned}\] which is the equation for the tangent plane. We define the formula on the right to be the tangent function: \[ f_{\tan}(x,y)=f(a,b)+f_x(a,b)(x-a)+f_y(a,b)(y-b) \] so that the equation of the tangent plane is \(z=f_{\tan}(x,y)\).

The equation of the tangent line to the graph of the function \(y=f(x)\) at \(x=a\) is: \[\begin{aligned} y&=f_{\tan}(x) \\ &\equiv f(a)+f'(a)(x-a) \end{aligned}\] In differential notation, this is: \[\begin{aligned} y&=f_{\tan}(x) \\ &\equiv f(a)+\left.\dfrac{df}{dx}\right|_{x=a}(x-a) \end{aligned}\]

*You have probably already memorized this.*

The equation of the tangent plane to the graph of the function \(z=f(x,y)\) at \((x,y)=(a,b)\) is: \[\begin{aligned} z&=f_{\tan}(x,y) \\ &\equiv f(a,b)+f_x(a,b)(x-a)+f_y(a,b)(y-b) \end{aligned}\] In differential notation, this is: \[\begin{aligned} z&=f_{\tan}(x,y) \\ &\equiv f(a,b) +\left.\dfrac{\partial f}{\partial x}\right|_{(a,b)}(x-a) +\left.\dfrac{\partial f}{\partial y}\right|_{(a,b)}(y-b) \end{aligned}\]

*Memorize this!*