Questions: Tangent Planes and Linear Approximation

Question 1

A function f(x,y) has partial derivatives f_x(0,0) = 2 and f_y(0,0) = 3 at the origin. A student concludes that the tangent plane z = f(0,0) + 2x + 3y must exist and provide a good local approximation. What assumption is this student making that may be wrong?

• A. The student forgot to evaluate the partial derivatives at the specific point — they need numerical values, not symbolic expressions
• B. The student is assuming differentiability, but the existence of both partial derivatives at a point does not guarantee differentiability — the tangent plane only gives a reliable approximation when the error vanishes faster than the distance to (0,0)
• C. The student should use the gradient vector ∇f rather than individual partial derivatives
• D. The tangent plane formula also requires second-order partial derivatives to be meaningful

Answer: (B)

This is the most important subtlety in multivariable differentiability. Partial derivatives measure slopes along the coordinate axes only. A function can have well-defined partial derivatives at a point while still having a crease or corner in other directions — the plane z = f(0,0) + 2x + 3y exists algebraically but may not be a good approximation to f near (0,0). Differentiability requires that the error |f(x,y) − L(x,y)| → 0 faster than the distance √(x² + y²) → 0, a condition that partial derivatives alone cannot guarantee.

Question 2

You know f(1, 3) = 5, f_x(1, 3) = 2, and f_y(1, 3) = −1. Using the linear approximation, what is the best estimate of f(1.01, 2.98)?

• A. 5 + 2(1.01) + (−1)(2.98) = 4.04 — substitute the full coordinates into the partial derivatives
• B. 5 + 2(0.01) + (−1)(−0.02) = 5.04 — substitute the changes Δx = 0.01 and Δy = −0.02
• C. 5 + 2(0.01) + (−1)(0.02) = 4.98 — treat both changes as positive displacements
• D. 5 × [1 + 2(0.01) − 1(0.02)] = 5.00 — the correction terms multiply the base value

Answer: (B)

The linear approximation is L(x,y) = f(a,b) + f_x(a,b)(x−a) + f_y(a,b)(y−b). The key is to substitute the *changes* Δx = x − a and Δy = y − b, not the raw coordinates. Here Δx = 1.01 − 1 = 0.01 and Δy = 2.98 − 3 = −0.02. So L = 5 + 2(0.01) + (−1)(−0.02) = 5 + 0.02 + 0.02 = 5.04. Option A is the classic error of plugging raw coordinates into the formula instead of the differences.

Question 3

If both partial derivatives f_x and f_y exist at a point (a,b), then f is differentiable at (a,b) and the tangent plane exists as a reliable local approximation.

• T. True
• F. False

Answer: False

This is explicitly listed as a common misconception. Partial derivatives measure the function's slope only along the x-axis and y-axis directions. A function can fail to be differentiable at a point where both partials exist if it has a crease, corner, or other irregularity in a diagonal direction. Differentiability requires the stronger condition that the linear approximation error goes to zero *faster than* the distance to (a,b) — partial existence alone is not enough.

Question 4

The tangent plane to z = f(x,y) at the point (a, b, f(a,b)) shares the same z-value and the same slopes in both the x and y directions as the surface f at that point.

• T. True
• F. False

Answer: True

This is precisely what makes the tangent plane the 'best' linear approximation. At x = a, y = b, the plane gives z = f(a,b) — matching the surface value. Its slope in the x-direction (partial derivative with respect to x) equals f_x(a,b), and its slope in the y-direction equals f_y(a,b). No other plane through (a, b, f(a,b)) matches both slopes simultaneously. This is the direct generalization of the tangent line in single-variable calculus, which matches both the function value and the slope at the point of tangency.

Question 5

A student says: 'I computed both partial derivatives at a point — that gives me the tangent plane.' Why might the student be wrong, and what additional condition is needed for the tangent plane to be a valid linear approximation?

Model answer: Partial derivatives only tell you the slope of the surface along two specific directions (the x-axis and y-axis). A surface could have well-defined partials but still be 'creased' or non-smooth in a diagonal direction, making the tangent plane formula valid algebraically but not a good local approximation. What's needed is differentiability: the error |f(x,y) − L(x,y)| must go to zero faster than the distance |(x,y) − (a,b)| as (x,y) → (a,b). Continuity of the partial derivatives at the point is a sufficient (though not necessary) condition for this.

The distinction matters in practice: for most smooth functions you encounter (polynomials, trigonometric functions, exponentials), continuity of partials is guaranteed and differentiability follows automatically. The pathological cases — piecewise-defined functions, functions with isolated discontinuities — are where the gap between partial existence and differentiability becomes real.