Writing u for f(P), the equation for the level curve (or surface) is u = u|P=P0. Writing Δu for f(P + ΔP) − f(P), a quantity that depends on both a point P and a vector ΔP, another equation for the level curve (or surface) is Δu|P=P0,ΔP=P−P0 = 0. That is, you take the expression for Δu, which says how much u changes between two points, put P0 in for the starting point P, and then put P − P0 in for the difference ΔP between the two points. Since the value of u shouldn't change on the level curve (or surface), this difference Δu should be zero. Notice that the meaning of P changes over the course of this substitution; originally it refers to the starting point, which we set to P0, but afterwards it refers to another point on the level curve (or surface), so we set the displacement ΔP between the two points to P − P0.
The tangent line (or plane) is given by a very similar equation, except that now we look at how the curve (or surface) is changing infinitesimally at P0 and extend this out to arbitrary distances. Thus, the equation Δu = 0 for the level curve (or surface) becomes du = 0 for the tangent line (or plane). However, we're still looking for the values of u in the same place, so the full equation is du|P=P0,dP=P−P0 = 0. If you follow the definition of differential from my earlier handout, then you'll see that this means precisely ∇f(P0) ⋅ (P − P0) = 0.
For example, if u = xy and P0 = (2, 3), then the level curve is xy = (2)(3), or simply xy = 6. (Replace x with 2 and y with 3 on the right-hand side.) Alternatively, Δu = (x + Δx)(y + Δy) − xy = y Δx + x Δy + Δx Δy, so the level curve is (3)(x − 2) + (2)(y − 3) + (x − 2)(y − 3) = 0. (Replace x with 2, y with 3, Δx with x − 2, and Δy with y − 3.) This also simplifies to xy = 6.
That was obviously more work than necessary for the level curve, but now apply the same technique to the differential to get the tangent line: du = y dx + x dy, so the tangent line is (3)(x − 2) + (2)(y − 3) = 0. (Replace x with 2, y with 3, dx with x − 2, and dy with y − 3.) This simplifies to 3x + 2y = 12, and now we learnt something that we didn't know before.
Because the normal line depends on the geometric notion of angle (to tell you what's perpendicular to what), this can't be done as slickly using only differentials. Now we really do want to think of the gradient vector. All the same, since this can be read off of the differential so easily, you can still start with du = y dx + x dy. First, replace only x with 2 and y with 3 to get 3 dx + 2 dy, then read off the gradient vector ⟨3, 2⟩. Since we started at the point (2, 3), the parametric equation is P = (2, 3) + t⟨3, 2⟩, or (x, y) = (3t + 2, 2t + 3) in more detail.
None of this (beyond the level curve (or surface) itself) works right if the gradient ∇f(P0) is zero or undefined. If the gradient is undefined, then of course we can't say anything using it; but if the gradient is zero, then these equations say that every point belongs to the tangent line (or plane) and only the point P0 belongs to the normal line. Of course, that would mean that they're not lines (or a plane and a line) at all! When the gradient is zero, the truth may be that there is no tangent or that there is a tangent but it really does consist of everything, or there may be an honest tangent line (or plane) after all; but in any case, these formulas won't help you know that!
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