The Tangent-Line Approximation

Close to the point, the tangent will do

Zoom in on a smooth curve at a point and it looks more and more like a straight line: its tangent. Near the point of contact the two are so close that the tangent, which is easy to work out, can stand in for the curve, which may not be.

Take $y=\sqrt{x}$y = √x at x = 4. The point is (4, 2) and the gradient is $\frac{1}{2\sqrt{4}}=\frac{1}{4}$1/(2√4) = 1/4, so the tangent is $y=\frac{x}{4}+1$y = x/4 + 1. At x = 4.1 the tangent gives 2.025, and $\sqrt{4.1}=2.024846$√4.1 = 2.024846 to six places.

The tangent as a formula

The tangent at x = a passes through (a, f(a)) with gradient f'(a), so it is y − f(a) = f'(a)(x − a). Rearranged, it is the linear approximation L(x) = f(a) + f'(a)(x − a), also called the linearization of f at a.

Read it as a walk along the tangent: start at the height f(a), then add the gradient times the step x − a. Near a, $f\left(x\right)\approx L\left(x\right)$f(x) ≈ L(x).

Estimating $\sqrt{4.1}$√4.1

Choose a = 4, close to 4.1 and with a known root. f(4) = 2 and $f^{\prime }\left(4\right)=\frac{1}{4}=0.25$f'(4) = 1/4 = 0.25, and the step is 0.1. So L(4.1) = 2 + 0.25 × 0.1 = 2.025.

The true value is $\sqrt{4.1}=2.024846$√4.1 = 2.024846, so the error, L minus the true value, is 2.025 − 2.024846 = 0.000154. The estimate is too big, by less than two ten-thousandths.

Going the other way works the same: L(3.9) = 2 − 0.025 = 1.975, and $\sqrt{3.9}=1.974842$√3.9 = 1.974842, again too big, by 0.000158.

Further from the point, a bigger error

The estimate gets worse as the step grows. At x = 4.4 the tangent gives 2.1 against $\sqrt{4.4}=2.097618$√4.4 = 2.097618, an error of 0.002382. At x = 5 it gives 2.25 against $\sqrt{5}=2.236068$√5 = 2.236068, an error of 0.013932.

A step ten times as long, 1 instead of 0.1, gave an error about ninety times as large. Near the point the error grows roughly with the square of the step, which is why a tangent estimate is very good close in and poor further out.

Too big or too small

The curve $y=\sqrt{x}$y = √x bends down: its gradient $\frac{1}{2\sqrt{x}}$1/(2√x) shrinks as x grows. A curve that bends down lies below its tangent on both sides of the point of contact, so the tangent gives values that are too big. That is why both $\sqrt{4.1}$√4.1 and $\sqrt{3.9}$√3.9 came out over.

The curve $y=\frac{x^2}{4}$y = x²/4 bends up. At x = 2 its point is (2, 1) and its gradient is $\frac{x}{2}=1$x/2 = 1, so the tangent is y = x − 1. At x = 2.2 the tangent gives 1.2, and the curve is $\frac{2.2^2}{4}=1.21$2.2²/4 = 1.21; at x = 1.8 the tangent gives 0.8 and the curve is 0.81. A curve that bends up lies above its tangent, so both estimates are too small, here by 0.01 each.

The curve $y=\sqrt{x}$y = √x and its tangent $y=\frac{x}{4}+1$y = x/4 + 1 at (4, 2). The curve bends down and lies below the tangent on both sides, so every value read off the tangent near x = 4 is a little too big.

The curve $y=\frac{x^2}{4}$y = x²/4 and its tangent y = x − 1 at (2, 1). The curve bends up and lies above the tangent on both sides, so every value read off the tangent is a little too small.

The second derivative decides

Which way a curve bends is the sign of its second derivative. If f''(a) < 0 the curve bends down near a, the tangent lies above it, and L overestimates. If f''(a) > 0 the curve bends up, the tangent lies below it, and L underestimates.

For $\sqrt{x}$√x, $f^{\prime \prime }\left(x\right)=-\frac{1}{4x\sqrt{x}}$f''(x) = −1/(4x√x), which at x = 4 is $-\frac{1}{32}$−1/32, negative: an overestimate. For $\frac{x^2}{4}$x²/4, $f^{\prime \prime }=\frac{1}{2}$f'' = 1/2, positive: an underestimate.

Two more. For f(x) = sin x at a = 0, the tangent is L(x) = x, so $\sin 0.1\approx 0.1$sin 0.1 ≈ 0.1; the true value is 0.099833, and the estimate is over by 0.000167, as f'' = −sin x, negative just right of 0, says. For $f\left(x\right)=x^{10}$f(x) = x¹⁰ at a = 1, f'(1) = 10, so $1.02^{10}\approx 1+10\times 0.02=1.2$1.02¹⁰ ≈ 1 + 10 × 0.02 = 1.2; the true value is 1.218994, and the estimate is under by 0.018994, as $f^{\prime \prime }=90x^8$f'' = 90x⁸, positive, says.

And a cube root: $\sqrt[3]{8.1}$∛8.1 from a = 8, where f(8) = 2 and $f^{\prime }\left(8\right)=\frac{1}{12}$f'(8) = 1/12, gives $L=2+\frac{0.1}{12}=2.008333$L = 2 + 0.1/12 = 2.008333 against the true 2.008299, over by 0.000034. The cube root bends down for x > 0, so the over was expected.

The curve $y=x^3-3x$y = x³ − 3x and its tangent at $x_0$x₀. At $x_0=1$x₀ = 1 the tangent is the level line y = −2, the second derivative 6x is 6, and the curve lies above the tangent near the point, so the tangent underestimates there. Drag $x_0$x₀ to −1: the second derivative is −6, the curve lies below, and the tangent overestimates. At $x_0=0$x₀ = 0 the second derivative is 0 and the curve crosses its tangent.

The usual mistakes

Adding the step without the gradient. From f(2) = 5 and f'(2) = 3, $f\left(2.1\right)\approx 5+3\times 0.1=5.3$f(2.1) ≈ 5 + 3 × 0.1 = 5.3, not 5.1.

Adding a whole unit of rise: f(a) + f'(a). With f(2) = 5 and f'(2) = 3 that gives 8, as if x had moved a whole unit instead of 0.1.

Starting from a instead of f(a). For $\sqrt{9.6}$√9.6 from a = 9, the tangent starts at the height $\sqrt{9}=3$√9 = 3, not at 9: $3+0.6\times \frac{1}{6}=3.1$3 + 0.6 × 1/6 = 3.1.

Taking a far from the point wanted. The estimate is good only near a; the error grows fast with the step.

A concrete base

In the application below, a square base must cover 101 square meters, so its side is $\sqrt{101}$√101. With only $\sqrt{100}=10$√100 = 10 to hand, the tangent to $\sqrt{x}$√x at 100 gives the side, and the bend of $\sqrt{x}$√x says the estimate is a little too long.