11. Partial Derivatives and Tangent Planes

d. Tangent Plane to the Graph of a Function

2. Algebraic Formula

On the previous page, we defined the tangent plane to the graph of a function to be the plane containing the tangent lines to its \(x\)-trace and \(y\)-trace. On this page, we derive a formula for the tangent plane. As we do this, it would be good to compare the derivation to that for the tangent line to the graph of a function of \(1\) variable as shown here: Tangent Line to the Graph of a Function It would be good to open the two pages side by side in two browser windows to see the comparison. The two derivations are also available in a two-column format for comparison as shown here: Comparison of Tangent Line and Tangent Plane Derivations

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

Memorize this!

Compute the tangent plane to \(z=-x^2y^3\) at \((x,y)=(3,2)\) discussed in the second exercise on the previous page. Then find its \(z\)-intercept.

The solution is straightforward: find the required information and plug into the formula for the equation of the tangent plane. We identify \(a=3\) and \(b=2\). We need to find \(f(3,2)\), \(f_x(3,2)\), and \(f_y(3,2)\). It is good to make a table: \[\begin{aligned} f&=-x^2y^3 \qquad &f(3,2)&=-72 \\ f_x&=-2xy^3 \qquad &f_x(3,2)&=-48 \\ f_y&=-3x^2y^2 \qquad &f_y(3,2)&=-108 \end{aligned}\] So the equation of the tangent plane is: \[\begin{aligned} z&=f(a,b)+f_x(a,b)(x-a)+f_y(a,b)(y-b) \\ &=f(3,2)+f_x(3,2)(x-3)+f_y(3,2)(y-2) \\ &=-72-48(x-3)-108(y-2) \\ &=-48x-108y+288 \end{aligned}\] We identify the \(z\)-intercept as \(c=288\).