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Directional Derivatives and Gradients · Tutorial 399 of 999

Level Surface Geometry Workshop

Learn to use the gradient of an implicit surface to find and verify its tangent plane, normal line, and tangent directions.

Intermediate 10 min read

What You'll Learn

  • Check whether a point lies on a sphere, ellipsoid, or other implicit surface.
  • Compute a gradient and determine whether the point is regular.
  • Find tangent planes and normal lines from the gradient.
  • Test whether a vector is tangent to a surface at a point.
  • Check geometric answers using substitutions and dot products.

Using Gradients to Read Surface Geometry

A surface described by an equation \(F(x,y,z)=c\) may be difficult to picture from its equation alone. At a point where the gradient is nonzero, however, the gradient gives a reliable local guide: it points normal to the surface, and the plane perpendicular to it is the tangent plane.

This workshop applies that idea to spheres, ellipsoids, and a surface defined by a more general equation. In each case, first check that the point belongs to the surface, then compute the gradient there. A final substitution or dot product checks the resulting geometry.

Definition: Let \(F\) be differentiable near a point \(\mathbf{p}=(x_0,y_0,z_0)\), and consider the level surface \(F(x,y,z)=c\). The point \(\mathbf{p}\) is a regular point if \(F(\mathbf{p})=c\) and \(\nabla F(\mathbf{p})\ne\mathbf{0}\). At a regular point, \(\nabla F(\mathbf{p})\) is a normal vector to the surface. The tangent plane consists of the points \((x,y,z)\) satisfying \(\nabla F(\mathbf{p})\cdot((x,y,z)-\mathbf{p})=0\). A nonzero vector \(\mathbf{v}\) is a tangent direction at \(\mathbf{p}\) precisely when \(\nabla F(\mathbf{p})\cdot\mathbf{v}=0\).

Why the Gradient Is Normal

The geometric rule follows from the chain rule. Suppose a differentiable curve \(\mathbf{r}(t)\) lies on the surface and passes through \(\mathbf{p}\) at \(t=0\). Because the curve stays on the level surface, \(F(\mathbf{r}(t))=c\) for all relevant \(t\). Differentiating this constant with respect to \(t\) gives:

$$ \frac{d}{dt}F(\mathbf{r}(t)) = \nabla F(\mathbf{r}(t))\cdot\mathbf{r}'(t) = 0. $$

At \(t=0\), the curve’s velocity \(\mathbf{r}'(0)\) is therefore perpendicular to \(\nabla F(\mathbf{p})\). Tangent vectors to the surface lie in this perpendicular plane. At a regular point, the gradient is nonzero, so its perpendicular plane is two-dimensional; it is the tangent plane to the smooth level surface there. This also explains why every tangent direction must have zero dot product with the gradient.

Important condition: The gradient method gives a normal and tangent plane at a regular point, where \(\nabla F(\mathbf{p})\ne\mathbf{0}\). If the gradient is zero, this method does not determine a tangent plane; the point may be singular, and the local geometry needs separate investigation.

Worked Example: A Sphere

Worked Example: Tangent Geometry on a Sphere

Find the tangent plane, a normal line, and two tangent directions to the sphere \(x^2+y^2+z^2=9\) at \(\mathbf{p}=(1,2,2)\).

Step 1: Check the point. Substitution gives \(1^2+2^2+2^2=1+4+4=9\), so the point lies on the sphere. Define \(F(x,y,z)=x^2+y^2+z^2\). Its gradient is \(\nabla F=(2x,2y,2z)\), hence \(\nabla F(1,2,2)=(2,4,4)\). This is nonzero, so the point is regular.

Step 2: Find the tangent plane. The point-normal equation is \((2,4,4)\cdot(x-1,y-2,z-2)=0\). Expanding gives \(2(x-1)+4(y-2)+4(z-2)=0\), or \(2x+4y+4z=18\). Dividing by \(2\), the tangent plane is \(x+2y+2z=9\).

Step 3: Find a normal line. Use the gradient as a direction vector. A normal line through \(\mathbf{p}\) is \((x,y,z)=(1,2,2)+t(2,4,4)\). The direction \((2,4,4)\) is parallel to the simpler vector \((1,2,2)\), so either gives the same line.

Step 4: Find tangent directions and check. The vector \(\mathbf{v}_1=(2,-1,0)\) satisfies \((2,4,4)\cdot(2,-1,0)=4-4+0=0\). The vector \(\mathbf{v}_2=(0,1,-1)\) satisfies \((2,4,4)\cdot(0,1,-1)=0+4-4=0\). Both are nonzero and therefore tangent directions. As a plane check, the point gives \(1+2(2)+2(2)=9\), as required.

Worked Example: An Ellipsoid

Worked Example: Tangent Geometry on an Ellipsoid

For the ellipsoid \(\frac{x^2}{4}+\frac{y^2}{9}+\frac{z^2}{16}=1\), find the tangent plane, a normal line, and two tangent directions at \(\mathbf{p}=(0,\frac{3}{\sqrt{2}},2\sqrt{2})\).

Step 1: Check the point. At this point, \(x^2/4=0\), \(y^2/9=(9/2)/9=1/2\), and \(z^2/16=8/16=1/2\). Their sum is \(1\), so the point lies on the ellipsoid. Set \(F(x,y,z)=x^2/4+y^2/9+z^2/16\). Differentiating each term gives \(\nabla F=(x/2,2y/9,z/8)\). Therefore, \(\nabla F(\mathbf{p})=(0,\sqrt{2}/3,\sqrt{2}/4)\), a nonzero vector.

Step 2: Find the tangent plane. The point-normal equation is \((\sqrt{2}/3)(y-3/\sqrt{2})+(\sqrt{2}/4)(z-2\sqrt{2})=0\). Multiplying by \(12/\sqrt{2}\), a nonzero scalar, simplifies this to \(4(y-3/\sqrt{2})+3(z-2\sqrt{2})=0\). In expanded form, the plane is \(4y+3z=12\sqrt{2}\).

Step 3: Find a normal line. The gradient is a scalar multiple of \((0,4,3)\), since \((\sqrt{2}/12)(0,4,3)=(0,\sqrt{2}/3,\sqrt{2}/4)\). Thus a normal line is \((x,y,z)=(0,3/\sqrt{2},2\sqrt{2})+t(0,4,3)\).

Step 4: Find tangent directions and check. The vectors \(\mathbf{v}_1=(1,0,0)\) and \(\mathbf{v}_2=(0,3,-4)\) are nonzero. Their dot products with the gradient are \(0\) and \(3\sqrt{2}/3-4\sqrt{2}/4=\sqrt{2}-\sqrt{2}=0\), respectively. They are tangent directions. Substituting the point into the plane gives \(4(3/\sqrt{2})+3(2\sqrt{2})=6\sqrt{2}+6\sqrt{2}=12\sqrt{2}\), confirming that it lies in the plane.

Worked Example: A General Implicit Surface

Worked Example: Tangent Geometry for a Mixed-Term Surface

Find the tangent plane, a normal line, and a tangent direction to \(xy+yz+zx=3\) at \(\mathbf{p}=(1,1,1)\).

Step 1: Check the point and compute the gradient. Substitution gives \(1\cdot1+1\cdot1+1\cdot1=3\). Define \(F(x,y,z)=xy+yz+zx\). Its partial derivatives are \(F_x=y+z\), \(F_y=x+z\), and \(F_z=x+y\), so \(\nabla F=(y+z,x+z,x+y)\). At \((1,1,1)\), the gradient is \((2,2,2)\), which is nonzero.

Step 2: Find the tangent plane. The point-normal equation is \(2(x-1)+2(y-1)+2(z-1)=0\). Dividing by \(2\) and simplifying gives \(x+y+z=3\).

Step 3: Find a normal line. The gradient \((2,2,2)\) is a normal direction, so a normal line is \((x,y,z)=(1,1,1)+t(2,2,2)\).

Step 4: Find a tangent direction and check. Take \(\mathbf{v}=(1,-1,0)\). The dot product is \((2,2,2)\cdot(1,-1,0)=2-2+0=0\), so this is a tangent direction. The point satisfies the plane equation because \(1+1+1=3\). The normal line passes through the point at \(t=0\), and its direction is the computed gradient.

A Reliable Problem-Solving Routine

The three examples use the same sequence of checks. Keeping the steps separate helps prevent errors, especially when the surface equation contains fractions or mixed products.

1
Verify membership.
Substitute the proposed point into the surface equation. If it does not satisfy the equation, it cannot be the point of tangency on that surface.
2
Differentiate the defining function.
Compute all three partial derivatives, then evaluate them at the point. Check that the resulting gradient is not zero.
3
Use the gradient as a normal.
Write the tangent plane with the point-normal equation. Use the same gradient, or any nonzero scalar multiple of it, as the direction of a normal line through the point.
4
Test tangent directions.
A proposed nonzero vector is tangent exactly when its dot product with the gradient at the point is zero.
5
Check the results.
Substitute the point into the tangent-plane equation, and verify the required dot products. These checks catch many sign and arithmetic errors.

Common Mistake: Differentiating a Quadratic Term Incorrectly

A frequent error is to differentiate \(x^2/4\) as \(x/4\). The coefficient \(1/4\) remains, but the power rule contributes a factor of \(2\):

$$ \frac{\partial}{\partial x}\left(\frac{x^2}{4}\right) = \frac{1}{4}(2x) = \frac{x}{2}, \qquad\text{not}\qquad \frac{x}{4}. $$

For the ellipsoid example, this incorrect derivative would produce the mistaken gradient \((x/4,2y/9,z/8)\), and at the given point its first component happens to be zero. That coincidence hides the error in the first component, but it does not make the differentiation rule correct. For example, at a point with \(x=2\), the correct first component is \(1\), while the mistaken one is \(1/2\). Always differentiate the full term before substituting the point.

Why These Checks Matter

The gradient packages local geometric information into a vector. Its direction determines the normal line, its perpendicular plane gives the tangent plane, and its zero dot product with a vector tests whether that vector lies in a tangent direction. These are related but distinct conclusions, so each should be checked in its own way.

The regular-point condition matters as much as the calculations. If \(\nabla F(\mathbf{p})=\mathbf{0}\), the point-normal equation would have a zero normal vector and would not define a plane. That does not prove there is no tangent geometry; it means this gradient method alone does not settle the question.

Check Your Understanding

For each question, use the gradient at the specified point and show the relevant substitution or dot product.

  1. For \(x^2+y^2+z^2=25\) at \((3,0,4)\), find the tangent plane and a normal line.
  2. For \(x^2/9+y^2/4+z^2=1\) at \((0,2,0)\), find two nonzero tangent directions.
  3. For \(xy+z^2=2\) at \((1,1,1)\), compute the gradient and write the tangent-plane equation.
  4. What condition on the gradient is required for the method in this tutorial to determine a tangent plane at a point?
  5. Why must a proposed tangent direction have zero dot product with the gradient?

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