Why Tangent Directions Have Zero First-Order Change
A level curve or level surface consists of points where a function has one fixed value. If you move along that curve or surface, the function value does not change. At a regular point, the gradient records the direction of greatest local increase and is perpendicular to the level set. Consequently, directions tangent to the level set are precisely the directions in which the directional derivative is zero.
This gives a practical way to find tangent directions without first parametrizing the whole curve or surface. The condition is local: it describes the instantaneous tangent direction at a point, not necessarily a straight path that remains on the level set.
Definitions and the Regular-Point Condition
Let \(F:\mathbb{R}^n\to\mathbb{R}\), and let \(c\) be a real number. The level set at height \(c\) is the set of all points \(\mathbf{x}\) satisfying \(F(\mathbf{x})=c\). A point \(\mathbf{a}\) on this set is called a regular point when \(\nabla F(\mathbf{a})\ne\mathbf{0}\).
A tangent vector to \(S\) at \(\mathbf{a}\) is a vector \(\mathbf{v}\) that occurs as the velocity of a differentiable curve \(\boldsymbol{\gamma}\) lying in \(S\), with \(\boldsymbol{\gamma}(0)=\mathbf{a}\) and \(\boldsymbol{\gamma}'(0)=\mathbf{v}\). A tangent direction is usually represented by a unit tangent vector. The zero vector is tangent in the vector-space sense, but it does not specify a direction.
For a differentiable function \(F\), the directional derivative in a unit direction \(\mathbf{u}\) is \(D_{\mathbf{u}}F(\mathbf{a})=\nabla F(\mathbf{a})\cdot\mathbf{u}\). Thus, when \(\mathbf{u}\) is a unit vector, its directional derivative measures the instantaneous rate of change of \(F\) per unit distance traveled in that direction.
The regularity assumption matters for describing the entire tangent space. If \(F\) is \(C^1\) on a neighborhood of \(\mathbf{a}\) and \(\nabla F(\mathbf{a})\ne\mathbf{0}\), the implicit function theorem says that the level set near \(\mathbf{a}\) is a \(C^1\) hypersurface. Its tangent space is the set of vectors perpendicular to the gradient. The term \(C^1\) means that the first partial derivatives are continuous; this assumption does not imply that the hypersurface is infinitely differentiable.
Why the Orthogonality Condition Holds
Suppose a differentiable curve \(\boldsymbol{\gamma}(t)\) lies in the level set and passes through \(\mathbf{a}\) at \(t=0\). Since every point of the curve has the same function value, \(F(\boldsymbol{\gamma}(t))=c\). Differentiating this constant expression using the multivariable chain rule gives the condition on its velocity.
For every parameter value near zero, \(F(\boldsymbol{\gamma}(t))=c\).
The chain rule gives \(\nabla F(\boldsymbol{\gamma}(0))\cdot\boldsymbol{\gamma}'(0)=0\), because the derivative of the constant \(c\) is zero.
Since \(\boldsymbol{\gamma}(0)=\mathbf{a}\) and \(\boldsymbol{\gamma}'(0)=\mathbf{v}\), it follows that \(\nabla F(\mathbf{a})\cdot\mathbf{v}=0\).
This proves that every tangent vector is perpendicular to the gradient. At a regular point of a \(C^1\) level set, the converse holds as well: every vector perpendicular to the gradient is tangent. To see why, choose a coordinate \(x_j\) for which \(F_{x_j}(\mathbf{a})\ne 0\). The implicit function theorem locally writes the level set as a graph \(x_j=g(\text{the other coordinates})\). Choose the other coordinate velocities to match those of any proposed vector \(\mathbf{v}\) with \(\nabla F(\mathbf{a})\cdot\mathbf{v}=0\), and follow this graph. Differentiating \(F(\boldsymbol{\gamma}(t))=c\) forces the remaining velocity to be \(v_j\), since \(F_{x_j}(\mathbf{a})\ne0\). Thus the graph has a curve with initial velocity \(\mathbf{v}\).
Worked Example: Tangent Directions to a Circle
Worked Example: Tangent Directions to a Circle
Find the unit tangent directions to the circle \(x^2+y^2=25\) at \((3,4)\).
Define \(F(x,y)=x^2+y^2\), so the circle is the level set \(F=25\). Its gradient is \(\nabla F(x,y)=(2x,2y)\), and hence \(\nabla F(3,4)=(6,8)\). If \(\mathbf{u}=(u_1,u_2)\) is a unit tangent direction, orthogonality requires $$ (6,8)\cdot(u_1,u_2)=6u_1+8u_2=0. $$ Dividing by \(2\) gives \(3u_1+4u_2=0\). Therefore \((u_1,u_2)\) is proportional to \((-4,3)\). The length of \((-4,3)\) is \(\sqrt{(-4)^2+3^2}=5\), so the two unit directions are $$ \mathbf{u}=\left(-\frac45,\frac35\right) \qquad\text{and}\qquad \mathbf{u}=\left(\frac45,-\frac35\right). $$ Check the first: its length is \(\sqrt{16/25+9/25}=1\), and its directional derivative is \(6(-4/5)+8(3/5)=(-24+24)/5=0\). The opposite direction also has zero dot product. Both are tangent directions, one in each orientation along the circle.
Worked Example: Tangent Directions to a Sphere
Worked Example: Tangent Directions to a Sphere
Find a unit tangent direction to the sphere \(x^2+y^2+z^2=9\) at \(\mathbf{a}=(1,2,2)\), and describe the full condition on tangent directions.
Set \(F(x,y,z)=x^2+y^2+z^2\). The point lies on the sphere because \(1^2+2^2+2^2=9\). The gradient there is $$ \nabla F(1,2,2)=(2,4,4). $$ A tangent vector \(\mathbf{v}=(v_1,v_2,v_3)\) must satisfy $$ (2,4,4)\cdot(v_1,v_2,v_3)=2v_1+4v_2+4v_3=0, $$ or \(v_1+2v_2+2v_3=0\). This equation describes a plane of tangent vectors. Choose \(\mathbf{v}=(2,-1,0)\); it satisfies \(2+2(-1)+2(0)=0\). Its length is \(\sqrt{2^2+(-1)^2}=\sqrt{5}\), so a unit tangent direction is \(\mathbf{u}=(2/\sqrt{5},-1/\sqrt{5},0)\). The directional derivative check is $$ (2,4,4)\cdot\mathbf{u} =\frac{4-4+0}{\sqrt{5}}=0. $$ For a direct curve check, let $$ \boldsymbol{\gamma}(s)=\mathbf{a}\cos(s/3)+3\mathbf{u}\sin(s/3). $$ Here \(\|\mathbf{a}\|=3\), \(\|\mathbf{u}\|=1\), and \(\mathbf{a}\cdot\mathbf{u}=(2-2)/\sqrt{5}=0\). Therefore \(\|\boldsymbol{\gamma}(s)\|^2=9\cos^2(s/3)+9\sin^2(s/3)=9\), so the curve remains on the sphere. Also, \(\boldsymbol{\gamma}(0)=\mathbf{a}\) and \(\boldsymbol{\gamma}'(0)=\mathbf{u}\), confirming that the direction is tangent.
Worked Example: An Implicit Surface
Worked Example: An Implicit Surface
For the surface \(z=x^2+y^2\), find a unit tangent direction at \((1,0,1)\), and verify it using a curve on the surface.
Write the surface as the level set \(F(x,y,z)=0\), where \(F(x,y,z)=z-x^2-y^2\). Its gradient is \(\nabla F(x,y,z)=(-2x,-2y,1)\), so \(\nabla F(1,0,1)=(-2,0,1)\). A tangent vector \(\mathbf{v}=(v_1,v_2,v_3)\) must satisfy $$ (-2,0,1)\cdot(v_1,v_2,v_3)=-2v_1+v_3=0. $$ Choose \(\mathbf{v}=(1,0,2)\); its length is \(\sqrt{1+0+4}=\sqrt{5}\). Thus \(\mathbf{u}=(1/\sqrt{5},0,2/\sqrt{5})\) is a unit tangent direction. The directional derivative is $$ \nabla F(1,0,1)\cdot\mathbf{u} =-\frac{2}{\sqrt{5}}+\frac{2}{\sqrt{5}}=0. $$ Now define \(\boldsymbol{\gamma}(s)=(1+s/\sqrt{5},0,(1+s/\sqrt{5})^2)\). Its points satisfy \(z=x^2+y^2\), and \(\boldsymbol{\gamma}(0)=(1,0,1)\). Differentiating gives \(\boldsymbol{\gamma}'(0)=(1/\sqrt{5},0,2/\sqrt{5})=\mathbf{u}\), which verifies the tangent direction directly.
Zero Directional Change Is an Instantaneous Statement
The condition \(D_{\mathbf{u}}F(\mathbf{a})=0\) means that the first-order rate of change at the point, in that direction, is zero. It does not mean that the function stays constant along the straight line in that direction. A tangent line or tangent direction approximates the level set locally; it need not lie on the level set.
For example, take \(F(x,y)=x^2+y^2\), the point \((1,0)\), and the unit direction \(\mathbf{u}=(0,1)\). The gradient is \((2,0)\), so the directional derivative is \((2,0)\cdot(0,1)=0\). But moving along the straight line gives \(F(1,t)=1+t^2\), which is not equal to \(1\) when \(t\ne0\). The derivative is zero at \(t=0\), while the function changes to second order as \(t\) moves away from zero.
The regular-point condition is also essential to the statement that every vector perpendicular to the gradient is tangent to the level set. At a point where the gradient is zero, the dot-product equation alone may not describe the actual geometry. The theorem above is intended for \(C^1\) functions at regular points.
How to Use the Condition
For a level curve in two variables, the tangent vectors form a line through the origin: they are all perpendicular to the nonzero gradient. There are two opposite unit tangent directions. For a level surface in three variables, the tangent vectors form a plane through the origin, so there are generally infinitely many unit tangent directions. In either case, the calculation has the same structure.
- Express the curve or surface as a level set \(F=c\).
- Compute \(\nabla F(\mathbf{a})\) and check that it is nonzero.
- Set \(\nabla F(\mathbf{a})\cdot\mathbf{v}=0\) and solve for tangent vectors.
- If a unit direction is requested, normalize a nonzero tangent vector by dividing by its length.
- Check the dot product and, when useful, verify the direction using a curve on the level set.
For a nonunit vector \(\mathbf{v}\), the orthogonality condition still determines whether it is tangent. Its dot product with the gradient is the rate along the parameterized straight motion \(\mathbf{a}+t\mathbf{v}\); the directional derivative per unit distance is instead computed with \(\mathbf{v}/\|\mathbf{v}\|\).
Check Your Understanding
Use the gradient-orthogonality condition and distinguish tangent information from claims about an entire path.
- For \(F(x,y)=x^2+y^2\) at \((0,5)\), find both unit tangent directions to the level curve through the point.
- For the level surface \(F(x,y,z)=x+y+z=6\), write the equation that tangent vectors satisfy at \((1,2,3)\).
- Why is a unit tangent direction at a regular point also a direction with zero directional derivative?
- What additional assumption on \(F\) ensures that the regular level set is locally a \(C^1\) hypersurface?
- If the directional derivative is zero, does the function necessarily remain constant along the corresponding straight line? Explain.