I know this may seem like a really broad question, but I will narrow it down. I really want to know the purpose of some of the things my teacher is emphasizing in my calc class.

For example why it so important to know: \begin{align} \frac{d}{dx}\sin\left(x\right)&=\cos\left(x\right),\\ \frac{d}{dx}\left(\sec\left(x\right)\right)&=\sec\left(x\right)\tan\left(x\right), \text{ or }\\ \lim _{x\to 0}\left(\frac{\sin\left(x\right)}{x}\right)&=1? \end{align}

I am looking for a real world benefit or application to knowing these things. Yes, its cool to be able to prove things like $\lim\limits _{x\to -\infty \:}\left(\sqrt{4\cdot \:x^2-5\cdot \:x}+2\cdot \:x\right) = \frac{5}{4}$, but how does all these toplics like: End Behavior,Limits, difference quotients, derivatives come into play in real world applications

So, far we learned the difference quotient to find the average rate of change. We learned the end behavior, so we can learn what occurs if a parameter approaches infinity. We learned limits so we can work around computing values at points where the function is not defined. Ex: Dividing by 0. Now we are learning derivatives which is the limit of a difference quotient as $h\to 0$.

But we haven't had any questions yet, that this is used in a practical problem. The questions we are asked in class are purely proofs and computations.

Ex: Find $\frac{d}{dx}\left(\frac{d}{dx}\left(\frac{1}{\sqrt{2\pi \:}}e^{-\frac{x^2}{2}}\right)\right)$

Ex:$$ f(x) = \left\{ \begin{array}{ll} 4x^2+1 & \quad x > 2 \\ 17 & \quad x = 2 \\ 16x-15 & \quad x < 2 \end{array} \right. $$

I asked my teacher this question and he mentioned the purpose of calculus is to find the global and local extrema, finding roots,calculating instantaneous rates of change, but didn't really go into to many real world applications. So far my Calc class has been prove blablabla because you have the math skills to do so. In classes like algebra 2 we didn't just learn algebra, but we learned many real world practical applications for it. My proffessor mentioned Calculus is used a lot in the real world to find area under curves and rates of change. Can you give me some examples on real world applications where I would need to find the area under the curve? Or find the instantaneous rate of change?

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – user642796
    Oct 20, 2015 at 17:28

10 Answers 10


The higher you go in math, the less the class will focus on real life applications. The math professors leave that to the science and engineering professors to teach. This might be fore some of the following reasons:

  • Calculus and higher mathematics topics are very dense, and there just isn't time to teach both concepts, methods, and applications.

  • Calculus and higher mathematics are not "everyday math" for most people who use it. By this, I mean you won't use calculus to figure out how much change you should have received from the cashier, or calculating the amount of wallpaper you need to buy. Calculus is applied quite often, but in highly specialized settings. If calculus teachers also taught applications, such applications would be useless to most students. By leaving these topics to the science professors, these topics are taught just to students who will use them. In other words, a physicist, an engineer, and a chemist all use calculus, but they do it in different ways.

  • Mathematicians sometimes pride themselves in how disconnected from application they can be. I've even heard a mathematician say "If there's an application to this, I don't want to know it." (though he did not say that about calculus).

    So now you know why you haven't learned applications, here are some:

  • Calculus is the study of change. Situations in science or engineering where nothing is changing are pretty boring, so we use calculus to study questions that do change. For example: Newton's second law of motion is $$F=ma$$ where $F$ is force, $m$ is mass, and $a$ is acceleration. Though it is not immediately apparent how this equation can be "solved," solutions to this equation describe the motion of an object when the force $F$ is applied it. Acceleration is the change in velocity per change in time: $$a=\frac{dv}{dt}.$$ Velocity, in turn, is the change in position per change in time, so $$a=\frac{d^2x}{dt^2}.$$ Thus, we are left with $$F=m\frac{d^2x}{dt^2}.$$ Often, force itself is a function of $x$, and possible time $t$ , so we have to solve: $$F(x,t)=m\frac{d^2x}{dt^2}.$$ So position in terms of time, $x(t)$, is a function whose second derivative times $m$ is equal to $F(x,t)$. This is called a differential equation, and is somewhat more difficult to solve than an algebraic equation. But by solving such, we can predict how an object will move in time. I have heard differential equations called the "language of physics," since almost every physical situation can be described in terms of a differential equation, and solving these differential equation is key to understanding how the physical situation evolves in time and space.

  • If we already know how an object moves through time, that is we know $x(t)$, we can find $v(t)$ and $a(t)$ by taking derivatives.

  • A very important problem in all of science, engineering, and even economics and finance is that of optimization. Figuring out what choices to make to get the best results is very applicable. A company can decide what price to sell their product at to make the most profit. An urban power can decide the best way to set up utilities. A chemist can calculate the optimal amount of reagent to use to complete an experiment. Optimization almost always means calculus because differentiable functions may reach minima or maxima when their derivatives are zero. So the challenge is to take the derivative, set it equal to zero, and solve for the value of $x$ that makes this true.

  • Calculus also involves an operation which is, in some sense, the opposite of differentiation called integration. You will probably discuss this in your class. Integration is useful when we must find a "total effect" of a constantly changing cause. For instance, a dam can only hold back a certain force of water behind it. The force from the water depends on water pressure. But the water pressure towards the bottom of the dam is more than at the top. So how can we find the total force? Integration is the key.

  • It can almost be said that multivaribale calculus was developed with the express purpose of studying fluid mechanics and electricity and magnetism.

  • Calculus is also heavily used in biology. This isn't really my area, so I don't know all the specifics, but I do know there are differential equations used to predict how a population will grow or shrink in time. This can be used to find the optimal amount of animals to hunt without risking extinction.

  • 1
    $\begingroup$ "But the water pressure towards the bottom of the dam is less than the top", its the other way around. The water pressure towards the bottom of the dam is more than the top. I know it has nothing to do with the context of the problem you were trying to solve but as a physicist, it hurts. $\endgroup$
    – iCantC
    Oct 19, 2019 at 15:05
  • $\begingroup$ @iCantC thanks for pointing out the typo. $\endgroup$
    – Alex S
    Oct 19, 2019 at 15:14

Somewhat perversely, I choose to respond to precisely your explicit examples. Mostly because the "surprising thing" is that all of the material you are learning comes from the "Real World".

  • $\frac{\mathrm{d}}{\mathrm{d}x}\sin(x) = \cos(x)$ -- position and velocity for object in simple harmonic motion. Protons in the LHC, electrons in a radio antenna, masses dangling from springs, pendula of small displacement, infrared oscillations of molecules -- these are all (approximately) harmonic motions.
  • $\frac{\mathrm{d}}{\mathrm{d}x}\sec(x) = \sec(x)\tan(x)$ -- rate of change of cost to shingle a roof versus roof angle, or rate of Mercator latitudinal compression versus angle above equator (although the fascinating half of this problem (distortion in Mercator projection) is the secant integral, both discussed at the link.)
  • $\lim_{x \rightarrow 0} \frac{\sin(x)}{x} = 1$ -- The sinc function is the Fourier transform of a rectangle; the limit you give informs us that rectangular windowing faithfully reproduces the DC component of a signal.
  • $\frac{\mathrm{d}}{\mathrm{d}x} \frac{\mathrm{d}}{\mathrm{d}x} \frac{1}{\sqrt{2 \pi \,}} \mathrm{e}^{- \frac{x^2}{2}}$ -- this is the second derivative of the normal distribution (a.k.a., Gaussian distribution). Second derivatives of distributions are commonly used to improve fitting parameters in spectroscopy.
  • $f(x) = \begin{cases} 4x^2 + 1, &x>2; \\ 17, &x=2; \text{ and} \\ 16x-15, &x < 2 \end{cases}$ -- at low velocities, the drag experienced by nearly spherical particles is linear in their velocity (see Stokes' Law), but at higher velocities, the drag is quadratic in velocity (see Newtonian drag) (with much complexity, because: fluid dynamics over a wide range of Reynolds numbers). While it is a stretch to say that your example is directly applicable, practice with these simplified model problems is necessary before tackling real examples.

We use integrals (areas under curves) to compute things which accumulate and whose rates we know. For instance, absorbed dose is an integral of the emission rate of a radiation source. Light scattering in clouds and fog is modelled by integrals versus penetration depth. Also, to go from densities to quantities -- mass density to weight, stress-strain to fracture specific energy, pressure to force, surface tension to surface energy, et al...


Just some examples for the topics you mentioned:

  1. Motion: The standard kinematics equations give $v = \frac{dx}{dt}$ and $a = \frac{dv}{dt} = \frac{d}{dt}(\frac{dx}{dt})$
  2. Thermodynamics: Energy, pressure, temperature, etc vary with each other, so you'll have to deal with equations like $dE = TdS - PdV + \mu dN$
  3. Stoichiometry: Amounts of chemical reactants change over time as the reaction takes place, changing the ratios.
  4. Chemistry: The speed of a chemical reaction changes over time.
  5. Electricity: For example, electrical induction is described by the equation $\nabla \times E = - \frac{\partial B}{\partial t}$

The primary reason to learn calculus is that the laws of nature are written in her language. For example, to answer why you should know that the derivative of $\sin x$ is $\cos x$, suppose you want to model the motion of a spring. The force experienced by the spring is proportional to its displacement from equilibrium, with the force pointing in the opposite direction. Letting $x = x(t)$ denote the displacement of the spring from equilibrium as a function of time, using Newton's law that $F = ma$, and recognizing that $F = -kx$ (where $k$ is some positive constant) and $a = \frac{d^2x}{dt^2}$, you reduce the problem to finding a function $x(t)$ satisfying $x''(t) = -\omega^2 x$, where $\omega = \sqrt{k/m}$ is chosen in hindsight once we see the solution. What functions have second derivative that are equal to their negatives? Why, $\sin t$ and $\cos t$ fit the bill. Extending this idea a bit, we see that $x(t) = c_1 \sin (\omega t) + c_2 \cos (\omega t)$ satisfies the differential equation $x''(t) = - \omega^2 x(t)$ (plug in and check that this is true!). With some more tools to work with (specifically linear algebra), you can show that this is the most general possible solution. If you specify the initial position and initial velocity of the spring, you can determine the constants $c_1$ and $c_2$ explicitly.

The upshot is that, using the tools of calculus, you can completely understand the motion of a simple spring. If that was all, it would not be so impressive. But with further study and continued refinement of your calculus theory and problem-solving skills, you can understand all manner of more complicated physical systems whose underlying laws are governed by differential equations -- equations that specifically invoke derivatives to formulate them. Mind you, there are still lots of interesting unsolved problems involving complicated physical systems, but calculus is almost always necessary to even understand the formulation of such problems.


Two important ideas live in Calculus. One is that of instantaneous rate of change. The instantaneous rate of change in position is velocity. How do you compute this? This was the critical question Newton wrestled with in trying to understand the dynamics of moving bodies.

The other is this. If you have an equation entailing rates of changes (imagine: bacteria grow on a plate proportional to their population), how do you solve this and obtain a function describing the phenomenon.

Calculus is one of the most important problems solving tools ever created human beings. Suspend your disbelief, leap in, and you will be rewarded with important insights.


In addition to learning how the world works (i.e. the language of physics is Calculus), you are also learning how to think and solve problems. You are learning how to see patters in the problems, how to combine ideas from different areas of maths to solve problems, to learn patters and prototypes for solving problems. You are learning how to take a practical problem and express it in symbols which can then be understood, simplified, and solved. You will never solve "book" problems in real-life, but by taking calculus and other higher maths courses, you are exercising your mind in ways that will be beneficial to your future life as a student and beyond.


The answers above all concentrate on the theoretical applications of calculus. That is, the use of calculus in developing and understanding theories in various fields. But what about the application side?

For the most part, you will never directly use calculus in applications. This is not because calculus doesn't apply. It is because someone else has already applied calculus to the problem, found the solution, and left algebraic formulas for everyone else to use for their particulars. So once calculus has been used, it rarely needs to be used again to do the same thing. I have only rarely found uses for calculus in my engineering job. But rarely is not the same as never. In those cases where I have made use of it, I used it to produce formulas which I then used in programs. Everyone who runs those programs is making use of the results of calculus, but they themselves never need to do calculus, because I have already done it.

The same is true generally. Almost all technical products were developed and designed through extensive use of calculus. Without calculus, they would have been impossible. But you can use them without knowing calculus (provided you are dull enough not to wonder how everything works...).


In probability and statistics, all continuous (smooth) distributions must be handled in terms of integrals. As one example of hundreds, from an elementary statistics course: the normal (bell-shaped) distribution which describes most biological and mechanical measurement processes, and more importantly an approximation to the sampling distribution of all means (averages), is defined by a probability density function (pdf). To answer any probability questions for this or other continuous curves, one finds the area under the curve (cumulative density function, cdf) -- the area under the curve being equal to the probability of getting an element from that region.

Thus: Effectively all probability questions for continuous measurements, and thus all estimation and prediction procedures, are predicated on integrals (areas under the curve).

Example 1: See the following article on "probability density functions" and note that the definitions are all in terms of integrals:

Example 2: See this list of continuous probability functions; click on any of them; notice in the right margin the first two things communicated are graphs for PDF and CDF -- the relationship between all of these is an integral.


Calculus is the backbone of physics. Here's a good example, the 1D wave equation for electric field:

$$\frac{\partial^2 E}{\partial x^2} = \frac{1}{c^2}\frac{\partial^2E}{\partial t^2} $$

This is a partial differential equation for the electric field $E$. On the left you see two position derivatives and on the right are two time derivatives. Relating them is a constant factor $\frac{1}{c^2}$, where $c$ is the speed of light. Coincidentally, the magnetic field $B$ satisfies a similar equation.


I just remembered this problem which appeared some months ago, from an Scottish official exam. People complained because it was 'too difficult'. Actually you just need a little calculus to do it and it a good example of the applications of derivatives for finding minima:

(source: gaussianos.com)

EDIT: It seems that I didn't make my point clear. Finding the path that minimizes time IS a very important application of calculus. I suppose that you can use it in real life problems such as designing a GPS to calculate the best route or something like that. What I do know it that most physics can be formulated in that way. The simplest example is optics.

Fermat principle tells you that the light travels in a way that minimizes time. If water and land were two media in which light travels with different speed, that would be precisely the path of a light ray.

Apart from that, classical mechanics problems can be solved by minimizing a Lagrangian function. In relativity, the path of an object is the one that minimizes proper time (time in the object's reference frame). Quantum mechanics can also be seen as minimizing certain action, using Feynman path integrals.

  • 2
    $\begingroup$ I don't think this really counts as a real-world application of calculus. $\endgroup$
    – user66698
    Oct 20, 2015 at 4:35
  • $\begingroup$ Not as phrased here it is not. However, the principles involved are regularly invoked in real world problems. This is an example of the "Least Action Principle" which is the foundation of Legendre's formulation of classical mechanics, and Schodinger's formulation of quantum mechanics. $\endgroup$ Oct 20, 2015 at 16:41

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