Introduction
A tangent plane provides a first-order estimate of a function near a point. But an estimate is not generally the exact function value. To use it responsibly, we need to ask two questions: how large is the error at a particular point, and what does differentiability tell us about the error as we move closer to the base point?
The central conclusion is precise: at a point where a function is differentiable, the error in its tangent-plane approximation, divided by the distance from that point, tends to zero as the distance tends to zero. This does not mean the error is always zero. It tells us how the error compares with distance sufficiently close to the base point.
Defining the Approximation and Its Error
Let \(f:\mathbb{R}^n\to\mathbb{R}\) be differentiable at a point \(\mathbf{a}\). Write an increment from the base point as \(\mathbf{h}\). The nearby input is \(\mathbf{a}+\mathbf{h}\), and its distance from the base point is the Euclidean norm \(\|\mathbf{h}\|\).
For \(f(x,y)\), this means that at \((a,b)\), with increments \(h=x-a\) and \(k=y-b\), the estimate is \(f(a,b)+f_x(a,b)h+f_y(a,b)k\). The increments are differences from the base point, not the coordinates \(x\) and \(y\) themselves.
When \(f\) is differentiable at \(\mathbf{a}\), its differentiability condition states that the difference between the function and this linear approximation is negligible relative to the size of the increment:
The numerator is precisely \(E(\mathbf{h})\). The limit therefore says that the signed error divided by distance tends to zero. Since absolute value is continuous, it also gives the corresponding statement for the absolute error.
Why the Error Is Smaller Than First Order
This estimate follows directly from the definition of differentiability, but it is useful to spell out the logic. The limit in the definition says that the error-to-distance ratio can be made smaller than any chosen positive tolerance by taking the point sufficiently close to the base point. Multiplying that ratio inequality by the positive distance gives a bound on the error itself.
The ratio \(\frac{|E(\mathbf{h})|}{\|\mathbf{h}\|}\) tends to zero as \(\mathbf{h}\to\mathbf{0}\), with \(\mathbf{h}\ne\mathbf{0}\).
For any \(\varepsilon>0\), the definition of the limit provides a \(\delta>0\) such that the ratio is less than \(\varepsilon\) whenever \(0<\|\mathbf{h}\|<\delta\).
Because \(\|\mathbf{h}\|>0\), the ratio inequality gives \(|E(\mathbf{h})|<\varepsilon\|\mathbf{h}\|\) for all such increments.
In particular, choosing \(\varepsilon=1\) shows that the absolute error is smaller than the distance at every sufficiently nearby nonzero point. The limit does not say that the error is zero, or that it is smaller than the distance at every nonzero point no matter how far from the base point. It does imply that the error is smaller than the distance at every sufficiently nearby nonzero point.
The key phrase is smaller than first order. Distance is the first-order scale \(\|\mathbf{h}\|\); differentiability says the error divided by that scale vanishes. It does not, by itself, specify an exact power of the distance or a fixed numerical error bound that works throughout a large region.
Worked Example: A Two-Variable Quadratic
Worked Example: A Two-Variable Quadratic
Estimate \(f(1.1,0.9)\) using the tangent plane to \(f(x,y)=x^2+xy+y^2\) at \((1,1)\), then calculate and check the error.
First evaluate the function at the base point:
Its partial derivatives are \(f_x=2x+y\) and \(f_y=x+2y\). At the base point, \(f_x(1,1)=3\) and \(f_y(1,1)=3\). The increments to \((1.1,0.9)\) are \(h=1.1-1=0.1\) and \(k=0.9-1=-0.1\). Thus the tangent-plane estimate is
Now compute the actual function value:
The signed error, actual value minus estimate, is \(3.01-3=0.01\), so the absolute error is \(0.01\). The distance from the base point is
The error-to-distance ratio is \(\frac{0.01}{\sqrt{2}/10}=\frac{\sqrt{2}}{20}\), approximately \(0.07071\) to five decimal places. As a check on the error calculation, expanding around the base point gives the exact remainder \(h^2+hk+k^2\). Substitution gives \((0.1)^2+(0.1)(-0.1)+(-0.1)^2=0.01-0.01+0.01=0.01\), matching the directly computed error.
Worked Example: A Three-Variable Quadratic
Worked Example: A Three-Variable Quadratic
For \(g(x,y,z)=x^2+2y^2+3z^2\), approximate \(g(1.1,-0.95,1.9)\) by linearizing at \((1,-1,2)\), and find the absolute error and its ratio to the distance.
At the base point, \(g(1,-1,2)=1+2+12=15\). The gradient is \(\nabla g=(2x,4y,6z)\), so \(\nabla g(1,-1,2)=(2,-4,12)\). The increments are \((h,k,\ell)=(0.1,0.05,-0.1)\). Therefore, the tangent-plane estimate is
Compute the actual value term by term:
Thus the signed error is \(13.845-13.8=0.045\), and the absolute error is \(0.045\). The distance is \(\sqrt{(0.1)^2+(0.05)^2+(-0.1)^2}=\sqrt{0.0225}=0.15\). The error-to-distance ratio is \(0.045/0.15=0.3\).
Check the remainder directly: subtracting the constant and linear terms from this quadratic leaves \(h^2+2k^2+3\ell^2\). At the chosen increments this is \(0.01+2(0.0025)+3(0.01)=0.01+0.005+0.03=0.045\), in agreement with the difference of the actual value and estimate.
Worked Example: An Error That Is Not Quadratic
Worked Example: An Error That Is Not Quadratic
Consider \(q(x,y)=(x^2+y^2)^{3/4}\) at the origin. Show that its tangent-plane estimate there is zero, verify differentiability at the origin, and calculate the error at \((0.04,0)\).
At the origin, \(q(0,0)=0\). For a nonzero increment \((h,k)\), the function value is \((h^2+k^2)^{3/4}\). Its size divided by the distance to the origin is
Consequently, \(q(h,k)=0+0h+0k+r(h,k)\), where \(r(h,k)/\sqrt{h^2+k^2}\to0\). This is the differentiability condition with zero linear part, so the tangent-plane estimate at the origin is \(L=0\).
At \((0.04,0)\), the actual function value is \((0.04^2)^{3/4}=(0.04)^{3/2}=0.04\sqrt{0.04}=0.04(0.2)=0.008\). The estimate is zero, so the signed and absolute errors are both \(0.008\). The distance is \(0.04\), and the ratio is \(0.008/0.04=0.2\). This example confirms that differentiability controls the error relative to distance without requiring the error to be a quadratic expression.
A Common Mistake: Using Coordinates Instead of Increments
The tangent-plane formula uses displacement from the base point. Substituting the coordinates themselves in place of those displacements changes the plane and can produce a meaningless error.
Worked Example: Correcting an Increment Error
Return to \(f(x,y)=x^2+xy+y^2\) at \((1,1)\), where \(f(1,1)=3\) and both partial derivatives equal \(3\). A common incorrect calculation at \((1.1,0.9)\) is
This is wrong because \(1.1\) and \(0.9\) are coordinates, not changes from \((1,1)\). The correct changes are \(1.1-1=0.1\) and \(0.9-1=-0.1\), so
Since the actual value is \(3.01\), the correct signed error is \(3.01-3=0.01\), not \(3.01-9=-5.99\). The check is that the tangent-plane estimate at the base point itself must equal the function value there: setting both increments to zero gives \(L=3\).
Interpreting an Error Estimate
At a specified point, subtracting the tangent-plane estimate from the actual value gives the exact signed error. The differentiability limit provides a different kind of information: it describes what happens to the ratio of error to distance as the point approaches the base point. The ratio need not be tiny at every point in a chosen region, and the limit alone does not supply a numerical bound for an entire region.
This distinction matters when deciding whether an approximation is adequate. A small absolute error may be useful, but the limit statement is relative to distance: for any chosen tolerance, sufficiently close points have error less than that tolerance times their distance. A numerical error bound over a particular region requires additional information beyond differentiability at one point.
Check Your Understanding
Use the definitions and reasoning from this tutorial to answer the following questions.
- For a differentiable function, what quantity tends to zero in the tangent-plane remainder condition?
- Why must the tangent-plane formula use increments from the base point rather than the input coordinates?
- For \(f(x,y)=x^2+y^2\) at \((1,0)\), calculate the tangent-plane estimate at \((1.1,0.2)\) and find the signed error.
- What does choosing \(\varepsilon=1\) in the error estimate imply for sufficiently nearby nonzero increments?
- Does differentiability at one point, by itself, give one fixed numerical error bound for every point in a large region? Explain.