14. Directional Derivatives and Gradients

e. Tangent and Normal, Lines and Planes to Level Sets

2. Tangent Planes to Level Surfaces in \(\mathbb R^3\)

A surface in space is frequently defined as the graph of an equation. In that case, the surface can be regarded as the level set of a function \(F(x,y,z)\). To determine the tangent plane to the surface at a point \(P=(a,b,c)\), we need to find the normal vector at \(P\), which in turn, is the gradient of \(F\) at \(P\).

Find the plane tangent to the ellipsoid \(36x^2+9y^2+4z^2=108\) at the point \((x,y,z)=(1,2,3)\).

Recall from the chapter on Lines and Planes that the normal equation for a plane in \(\mathbb R^3\) is given by \[ \vec N\cdot X=\vec N\cdot P \] where \(P\) is a particular point on the plane, \(\vec N\) is a normal vector to the plane and \(X=(x,y,z)\) is a general point on the plane. In the case of a tangent plane to a level surface of a function \(F\), \(P\) is the point of tangency and \(\vec N\) is the normal to the level surface, i.e. the gradient of \(F\) at \(P\).

In this example, the function is \(F(x,y,z)=36x^2+9y^2+4z^2\), its gradient is \[ \vec\nabla F=\langle 72x,18y,8z\rangle \] and the normal at \(P=(1,2,3)\) is \[ \vec N=\left.\vec\nabla F\right|_{(1,2,3)}=\langle 72,36,24\rangle \] So the normal equation of the plane is \[ \langle 72,36,24\rangle\cdot(x,y,z) =\langle 72,36,24\rangle\cdot(1,2,3) \] or \[ 72x+36y+24z=216 \] or \[ 6x+3y+2z=18. \]

It is important to distinguish between three (or four) methods of finding the tangent plane to a surface.

1. If the surface is given as the graph of the a function, \(z=f(x,y)\), then the tangent plane at \((x,y)=(a,b)\) is: \[ z=f_{\tan}(x,y)=f(a,b)+f_x(a,b)(x-a)+f_y(a,b)(y-b) \]
2. If the surface is given as a parametric surface, \(\vec R(s,t)=\langle x(s,t),y(s,t),z(s,t)\rangle\), then there is a parametric equation for the tangent plane which will be given in the chapter on parametric surfaces.
3. If the surface is given as the graph of an equation \(F(x,y,z)=C\), (equivalently, the level surface of the function \(F(x,y,z)\) with value \(C\)), then the normal vector is \(\vec N=\left.\vec\nabla F\right|_{(a,b,c)}\) and the tangent plane at \(P=(a,b,c)\) is: \[ \vec N\cdot X=\vec N\cdot P \]
4. If the surface is given as the graph of an equation \(F(x,y,z)=C\), then an alternate method to find the tangent plane at \((a,b,c)\) is to apply method (1) to the implicitly defined function \(z=f(x,y)\): Since the point is \((a,b,c)\), the function value is \(f(a,b)=c\). Using implicit differentiation the \(x\) and \(y\) slopes are \(f_x(a,b)=\left.\dfrac{\partial z}{\partial x}\right|_{(a,b,c)}\) and \(f_y(a,b)=\left.\dfrac{\partial z}{\partial y}\right|_{(a,b,c)}\). Then the tangent plane at \((a,b,c)\) is: \[ z=f_{\tan}(x,y)=f(a,b)+f_x(a,b)(x-a)+f_y(a,b)(y-b) \] This implicit differentiation method is the least useful of the four: In general, if the surface is a level set \(F(x,y,z)=C\), then it is much easier to compute the gradient, \(\vec\nabla F\), than to use implicit differentiation to compute the slopes, \(\dfrac{\partial z}{\partial x}\) and \(\dfrac{\partial z}{\partial y}\).

Notice that these two exercises were the same problem. Which method did you think was easier? For more complex problems, different methods may have a distinct advantage, depending on the way the problem is stated.