Welcome! Today, we'll explore a powerful idea that sits at the heart of calculus: differentiable functions are locally linear. This means that if we zoom in close enough on any smooth curve, it starts to look like a straight line. We will learn how to exploit this fact to approximate complex functions and to estimate errors in scientific measurements.
How does your calculator compute something like $\sqrt{4.01}$ instantly? It doesn't have a giant table of square roots memorized. It uses a clever, fast approximation. The simplest and most important of these approximations is the one we'll learn today, and it all starts with a visual idea.
Consider the graph of a smooth function. From a distance, it's clearly a curve. But as we zoom in closer and closer to any single point, its curvature becomes less and less apparent until it's virtually indistinguishable from a straight line—its tangent line.
In the graph below, use the zoom button to zoom in on the point $(0.5, f(0.5))$.
Let's look at the function $f(x) = x^2$ near the point $x=1$. The tangent line is given by the equation $y - f(1) = f'(1)(x-1)$, which equals $y - 1 = 2(x-1)$, or $y = 2x - 1$. Let's see how the values of the function and the tangent line compare for points very close to 1.
The tangent line gives us a fast, easy way to get a very good approximation of the function's value, as long as we stay close to the point of tangency.
In the graph below, zoom in to both points to see how close the functional value is to the tangent line value.
Which of the following functions would be best approximated by its tangent line at $x=0$? Why?
a) $f(x) = e^x$
b) $g(x) = |x|$
We can formalize this idea by starting with the equation of the tangent line. The point-slope form for the line tangent to $f(x)$ at $x=a$ is:
Solving for $y$, we get the equation of the tangent line. We call this function the linearization of $f(x)$ at $a$, denoted $L(x)$.
The linearization of $f$ at $a$ is the tangent line function:
$$L(x) = f(a) + f'(a)(x-a)$$For values of $x$ near $a$, we can say that $f(x) \approx L(x)$. This is called the linear approximation or tangent line approximation of $f$ at $a$.
In the graph below, use the sliders for $a$ and $x_1$ to see how the error between the function and the tangent line gets smaller as $x_1$ gets closer to $a$.
Let's use our new tool to approximate $\sqrt[3]{8.06}$. Can you compute this in your head? Probably not. But we can approximate it easily using a tangent line.
Step 1: Find the equation of the tangent line, $L(x)$.
We need to approximate $f(x) = \sqrt[3]{x}$ near a "nice" point. The obvious choice is $a=8$, since we know $\sqrt[3]{8}$ perfectly.
First, we need the ingredients: $f(a)$ and $f'(a)$.
Step 2: Evaluate the tangent line at the desired point.
We want to approximate $\sqrt[3]{8.06}$, so we evaluate our tangent line at $x=8.06$. $$L(8.06) = 2 + \frac{1}{12}(8.06 - 8) = 2 + \frac{1}{12}(0.06) = 2 + \frac{1}{200} = 2 + 0.005 = 2.005$$ So, our approximation is $\sqrt[3]{8.06} \approx 2.005$.
Step 3: Compare with a calculator and find the error.
A calculator gives $\sqrt[3]{8.06} \approx 2.004987...$ The error in our approximation is $|2.004987... - 2.005| \approx 0.000013$. That's incredibly accurate for a calculation we could do by hand!
Find the linearization of $f(x) = (x+3)^2$ at $a=1$. Then use it to approximate $(3.99)^2$.
One of the most famous and useful linearizations in all of science and engineering is the approximation for $\sin(\theta)$ when $\theta$ is small and, critically, measured in radians.
Step 1: Find the linearization of $f(\theta) = \sin(\theta)$ at $a=0$.
This gives us the famous small-angle approximation: for $\theta$ near 0 (in radians), $\sin(\theta) \approx \theta$.
Step 2: Let's use it to estimate $\sin(0.05 \text{ rad})$.
Using our approximation, the answer is simply $\sin(0.05) \approx 0.05$.
Step 3: Let's check the result.
A calculator gives $\sin(0.05) \approx 0.049979...$. Our estimate is excellent!
Step 4: Application
A ground station tracks a satellite with an altitude of 200 km. The distance to the satellite is $D = h / \sin(\theta)$, where $\theta$ is the angle above the horizon. If the measured angle is $\theta=0.04$ radians, estimate the distance $D$ to the satellite.
Using the small-angle approximation, $\sin(0.04) \approx 0.04$. $$D = \frac{200 \text{ km}}{\sin(0.04)} \approx \frac{200 \text{ km}}{0.04} = \frac{20000}{4} = 5000 \text{ km}$$ This approximation allows for a quick mental calculation that is highly accurate.
A surveyor measures a small angle to be $2^{\circ}$. Estimate the value of $\sin(2^{\circ})$ without a calculator by first converting the angle to radians and then using the small-angle approximation. (Recall that $180^{\circ} = \pi$ radians.)
Differentials offer a slightly different notation for the same core idea of linear approximation. They focus on the change in the function's value.
Let's start with a change in $x$, which we call $\Delta x$. To bring calculus into this, we define a new, independent variable called $dx$, the differential of x. We are free to choose the value of $dx$, so for convenience, we set it to be equal to the actual change in $x$. So, by definition, $dx = \Delta x$.
Now, we consider the consequences. The actual change in the function $y$ along the curve is $\Delta y = f(x+\Delta x) - f(x)$. The change along the tangent line is what we call the differential of y, $dy$.
The key takeaway is that while we set $dx = \Delta x$ exactly, the relationship for $y$ is an approximation: the actual change $\Delta y$ is well-approximated by the differential $dy$. That is, $\Delta y \approx dy$.
The interactive graph below provides a visual representation of this key idea. Use the slider for $dx$ to see how the actual change, $\Delta y$, and the estimated change, $dy$, become nearly identical as the change in $x$ gets smaller.
For the function $y = f(x) = x^2 + 3x$, let's compare $\Delta y$ and $dy$ as $x$ changes from $2$ to $2.1$. Here, $x=2$ and $\Delta x = dx = 0.1$.
Step A: Find the Exact Change ($\Delta y$).
This is the actual change in the function's value.
Step B: Find the Estimated Change ($dy$).
First, we find the general differential. $f'(x) = 2x+3$, so $dy = (2x+3)dx$.
Now, we plug in our specific values, $x=2$ and $dx=0.1$:
$$dy = (2(2)+3)(0.1) = (7)(0.1) = 0.7$$
Step C: Compare.
The actual change was $\Delta y = 0.71$. The estimated change using differentials was $dy = 0.7$. They are very close, confirming that $dy$ is a good approximation for $\Delta y$.
For the function $y = 1/x$, find the differential $dy$. Use it to estimate the change in $y$ as $x$ changes from $5$ to $5.2$.
In science and engineering, we often measure one quantity and use it to calculate another. But measurements are never perfect! They always have some small error. Differentials are the perfect tool for estimating how a small measurement error (like $dr$) propagates into the final calculated value (like $dV$).
The radius of a spherical ball bearing is measured to be $10$ cm, with a possible measurement error of $\pm 0.05$ cm. Use differentials to estimate the maximum possible error in the calculated volume.
Step 1: State the Goal.
We want to find the error in the Volume, which we'll call $dV$, that results from a known error in the radius, $dr = 0.05$ cm.
Step 2: Find the Governing Formula.
The volume of a sphere is given by $V = \frac{4}{3}\pi r^3$.
Step 3: Find the Differential Relationship.
We differentiate with respect to $r$ to find the relationship between the differentials.
$$\frac{dV}{dr} = 4\pi r^2 \implies dV = 4\pi r^2 dr$$
This equation tells us how the volume error ($dV$) is related to the radius error ($dr$).
Step 4: Plug In and Calculate the Propagated Error.
We substitute our measured values: $r=10$ and the possible error $dr=0.05$.
$$dV = 4\pi (10)^2 (0.05) = 4\pi (100) (0.05) = 400\pi (0.05) = 20\pi \text{ cm}^3$$
Step 5: Interpret the Result.
A tiny error of just $0.05$ cm in the radius measurement can lead to a potential error of $20\pi \approx 62.8$ cm³ in the calculated volume! This is the propagated error.
To better understand the significance, we can calculate the relative error:
$$\frac{\text{error in volume}}{\text{total volume}} = \frac{dV}{V} = \frac{20\pi}{\frac{4}{3}\pi (10)^3} = \frac{20\pi}{\frac{4000\pi}{3}} = \frac{20 \cdot 3}{4000} = \frac{60}{4000} = 0.015$$
The percentage error is just $0.015 \times 100 = 1.5\%$.
The side of a square is measured to be $15$ cm with a possible error of $\pm 0.1$ cm. Use differentials to estimate the maximum error in the calculated area.
Use a linear approximation to estimate the value of $\ln(1.05)$.
The period $T$ of a simple pendulum of length $L$ is given by $T = 2\pi\sqrt{L/g}$, where $g$ is a constant. Use differentials to estimate the percentage error in $T$ if the length $L$ is measured with a percentage error of 2%.
A company's cost function is $C(x) = 1000 + 5x + 0.01x^2$ dollars, where $x$ is the number of units produced. If the current production level is 500 units, use differentials to estimate how much the cost will increase if 10 more units are produced.
Today we explored two sides of the same powerful coin. Whether we call it a linear approximation or use the language of differentials, the core idea is that we can use the tangent line to a function to approximate its behavior near the point of tangency.
After this lecture, you should be able to: