# Calculus and Physics

#### PrudensOptimus

Perhaps someone who has great knowledge in Physics and Calculus can answer these:

(1). How will mastering Calculus bring beneficial understandings, ones that cannot be seeing easily through other mathematics, to the learning to physics?

(2). What is the true meaning of derivatives, other than just performing some calculations based on some specific rules, at least that is what it appears to be to me, right now.

(3). Physics B, which is a highschool course for students who had trigonometry, will Calculus be a special part here or in Physics C? Assuming the student has very good algebra/geometry/trigonometry background and is enrolled in calculus while taking Physics B.

(4). What are somethings in Physics that only with Calculus knowledge can be proven or solved?

(5). Some tips for learning Calculus and Physics B, XOR C. Assuming the student has great algebra/geometry/trig background and has enrolled in Physics PAP in the freshman year.

-- Thanks
-Tom

Related Other Physics Topics News on Phys.org

#### ahrkron

Staff Emeritus
Gold Member
Originally posted by PrudensOptimus
(1). How will mastering Calculus bring beneficial understandings, ones that cannot be seeing easily through other mathematics, to the learning to physics?
Calculus is extremely useful for many things (not only for physics).

The reason is that, in many cases, you can measure the rate at which something accumulates, and you want to describe how the accumulated quantity behaves.

When such rates are constant, it is possible to solve things without calculus, but they rarely are constant. Calculus gives you a tool to relate rates with quantities.

Examples:
1. you know that the more people there are, the faster the population increases. This simple description corresponds exactly to a differential equation, whose solution can tell you what population you can expect to have in 10 years.
2. Say you drive your car or 2hrs at 50 miles/hour. How far did you go? 100 miles. That is easy and does not need calculus. But, how often do you keep your speed at exactly 50 mi/h for 2 hours? A varying speed can be used to obtain the total displacement using calculus.
3. Say you have a rocket engine. You can measure how much gas it uses per second. This can be translated in thrust, which gives you the rate of change of speed, which in turn is the rate of change of distance. If you need to determine how much gas you need to cover a specific distance, you definitely need calculus.
4. You probably know F=ma already. Most probably you learned how to use it on problems where a, m and F are constant, which is never the case in real problems.

(2). What is the true meaning of derivatives, other than just performing some calculations based on some specific rules, at least that is what it appears to be to me, right now.
Derivative = rate of change.

Your car's speed is a derivative (of the position of your car with respect to time),

Your bank's interest rate is a derivative (of the ammount you have in your account, again wrt time),

The acceleration of a rocket is a derivative (it tells you how much the speed changes per second)

A pan's handle usually does not get as hot as the pan itself. It is made of a different material. How do you select such material? you want one whose rate of temperature increase with respect to heat received is not as big (for experts: ok, it works a bit differently, but its quite late here ).

(4). What are somethings in Physics that only with Calculus knowledge can be proven or solved?
Pretty much everything once you step out of the introductory book's (frictionless, perfect materials, constant everything, simple relations) examples.

As for course content, you probably would be better off asking people at your school.

#### PrudensOptimus

OK, suppose you have a car travelling at

V(t) = t^2 m/s.

Its derivative is obviously 2t.

How does V(t) relate to 2t?? I couldn't understand this concept, so please someone explain.

#### ahrkron

Staff Emeritus
Gold Member
Originally posted by PrudensOptimus
OK, suppose you have a car travelling at

V(t) = t^2 m/s.
This mieans that the speed at t=0 is 0,
at t=2 is 4 m/s,
at t=100 sec, the speed is 10,000 m/s.

Its derivative is obviously 2t.
i.e., the acceleration of the car is
A(t) = 2t m/s2:
At t=1, the acceleration is 2 m/s2; i.e., it has an acceleration that, if held constant, would increase the speed by 2 m/s every second.

At t=5, the acceleration is such that it would add 10 m/s every second.

If the acceleration was constant, say a = 3 m/s^2, then the speed would be increasing by 3 m/s every second. It could be

V(t) = 3t
and the acceleration would be
A(t) = 3
(which is precisely such constant acceleration, and also the derivative of the constantly-increasing speed).

In your problem, the speed grows more that a staright line, signaling that the acceleration is constantly increasing (instead of constant).

#### HallsofIvy

Homework Helper
Here is the problem that actually led to the development of calculus:
Imagine that you are in a rocket ship well above the plane of the ecliptic (the planets) in the solar system. You take a "snapshot" that includes both the sun and the earth.

From that snap shot you could measure the distance from the sun to the earth AT THAT INSTANT. Of course, since that is a "still" shot you couldn't determine either the speed or the acceleration of the earth from a single snapshot (at a single instant). Both speed and acceleration require movement which means they require measurement at different times.

Yet people of Newton's era (and well before) knew that the force of gravity must depend on distance in order to give Kepler's laws. They also knew that "Force equals mass time acceleration".

There is the puzzle: If you can measure distance AT A SPECIFIC INSTANT, then you can calculate the force AT THAT INSTANT and so you can calculate the acceleration AT THAT INSTANT. But what in the world does "acceleration at an instant" mean?

It was necessary to develop calculus to give meaning to the basic concepts of "speed" and "acceleration" AT A SPECIFIC INSTANT that were necessary for "force equals mass times acceleration" to make sense!

#### jeff

Originally posted by HallsofIvy
Here is the problem that actually led to the development of calculus:
Imagine that you are in a rocket ship well above the plane of the ecliptic (the planets) in the solar system. You take a "snapshot" that includes both the sun and the earth.

From that snap shot you could measure the distance from the sun to the earth AT THAT INSTANT. Of course, since that is a "still" shot you couldn't determine either the speed or the acceleration of the earth from a single snapshot (at a single instant). Both speed and acceleration require movement which means they require measurement at different times.

Yet people of Newton's era (and well before) knew that the force of gravity must depend on distance in order to give Kepler's laws. They also knew that "Force equals mass time acceleration".

There is the puzzle: If you can measure distance AT A SPECIFIC INSTANT, then you can calculate the force AT THAT INSTANT and so you can calculate the acceleration AT THAT INSTANT. But what in the world does "acceleration at an instant" mean?

It was necessary to develop calculus to give meaning to the basic concepts of "speed" and "acceleration" AT A SPECIFIC INSTANT that were necessary for "force equals mass times acceleration" to make sense!
Where did you get this from?

#### NEOclassic

The secret is in logarithms

Among others, Napier was looking for a function whose derivative was equal to the function. Logarithms of the usual kinds depended upon the "base"; e.g., the popular logarithm is called "log to the base 10; the log of 2 = 0.30103.. and that means that 10^(0.30103..) = 2. Logarithms to bases other than 10 (such as the base 2 in boolian math)are possible but because of the decimal numbering system, base 10 is popular. Napier discovered that there was an irrational (and also called transcendental) base called "e" that has numerical value equal to 2.7182818.. The Logarithm to the base "e" is labled "ln" to distinguish it from "log to the base 10". For the function y = e^x, the derivative dy = e^x dx. The real value of this simple interdependence resulted in the postulation of the "integration factor" that simplifies cluimsy integrations. Cheers, Jim
PS: "e" can be evaluated with an infinite series.

#### MartinW

The derivative gives the slope of a curve at the location where it is taken. It's real calculation, for a function f(x), is:

df(x)/dx = lim(h->0)[f(x+h) - f(x)]/h

For an easy example, using f(x) = x^2:

df(x)/dx = lim(h->0)[(x+h)^2 - x^2]/h
= lim(h->0)[x^2 + h^2 + 2xh - x^2]/h
= lim(h->0)[h^2 + 2xh]/h
= lim(h->0)h(h + 2x)/h
= lim(h->0)[h + 2x]
= 2x

The formula for the derivative gives the slope of the secant through the function at x and the function at x+h. As h->0, the secant approaches the tangent.

Martin

#### PrudensOptimus

So derivatives is simply another "mathematical trick"?

How do nx^(x-1) estimate the rate of change for a term, say x^2?

#### chroot

Staff Emeritus
Gold Member
I'm not really sure what qualifies a piece of mathematical machinery to be called a "trick." :)

- Warren

Calculus is going to be what describes things that are changing. Or any dynamic system. You know, or maybe you don't if you in high school, but that's ok, how the number pi appears in many equations that don't seem to have anything to do with circumfrences or diameters; and how pi also appears in the definition of some physical constants. When you see pi, it is a clue that the result you're looking at came from something that is periodic in nature like a sine wave for instance; or it means that somewhere in arriving at this result you used some kind of circular or spherical symmetry.

Derivatives pop up in the same way. Almost everything you are going to want to study in physics is going to evolve over time. When this happens there will always be derivatives sprinkled everywhere. I haven't been in a physics class in college yet where I haven't encountered derivatives, and I only have 2 physics classes left. They are as common as pi :)

#### frankR

Once you take Calculus it will make sense. The concepts are of biggest importance. If you don't know the concepts, the mathematical techniques are essential useless. It sounds like you know the power rule, but have no idea what it means or what it's used for. Once you learn Calculus you'll find it to be elegant and useful.

Check this out: http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html

#### PrudensOptimus

I have been practicing all the questions in the book relating to Power Rules, Product Rules, Quotient Rules, and Chain rules. And I have get to know them all and know how to do most of them.

However, concept is the most important, if no concept, then no new creation. So I want to understand why must the change in things relate to "derivatives." How do they measure and why is it that we can find the change using the same rules?(Prod, Quo, Chain,...)

- Tom

#### Hurkyl

Staff Emeritus
Gold Member
So I want to understand why must the change in things relate to "derivatives."
For a particular example, compare the formula for the average velocity of an object over the interval of time [a, a + h]:

vav = (x(a + h) - x(a)) / h

with the formula for the instantaneous velocity of an object at time a:

v(a) = x'(a) = limh->0 (x(a + h) - x(a)) / h

How do they measure and why is it that we can find the change using the same rules?(Prod, Quo, Chain,...)
Look at the proofs in your calculus text, and/or try proving them directly from the limit definition of the derivative... continuing the analogy given above between average rate of change and instantaneous rate of change might help give insight as well.

So I want to understand why must the change in things relate to "derivatives."
the rate of change is the derivative!!!

To understand the rules of derivatives, you must go through their derivation. Then they will make sense as to why you can take these "shortcuts" instead of doing all the work on the problem. Essentially, when you use a rule for finding a derivative, you are skipping the derivation of that rule in that problem, which you would otherwise have to do to get the answer. Since mathematicians are lazy by nature, they realised there is no need to go through these long proofs when an exploitation of a pattern will save them a lot of time and work.

#### anomaly

(1). How will mastering Calculus bring beneficial understandings, ones that cannot be seeing easily through other mathematics, to the learning to physics?
Hello, I am new to this forum. As evidenced by this being my first post. But I saw this question and was compelled to reply because I have been asked this on many occassions while tutoring students at my Alma Mater in both Physics and Mathematics.

So, how will mastering calculus help your career in physics? I think that calculus really opened up a new level to my problem solving abilities, including the many different ways to approach a problem and think my way through it to come to the solution. It wasn't apparent to me what I was learning nor how important it was to my educational goals until I was enrolled in caluulus III. Things began to 'click' for me in other areas of study, especially physics. The mastery of the mathematics portion helped me to understand not only how a solution was determined in physics, but why -- especially the proofs.

However, concept is the most important, if no concept, then no new creation. So I want to understand why must the change in things relate to "derivatives." How do they measure and why is it that we can find the change using the same rules?(Prod, Quo, Chain,...)
Study ahrkron's posts carefully. The reason that the derivative works can maybe best be understood first with geometry. If you graph y=x, a simple diagonal line, then mark out two values of x, the area under the curve of the graph will be the integral of the function. You might know that the integral is inverse of the derivative,
I mean:
&int;x dx = x2/2 ; d(x2/2)/dx = x
Likewise, the derivative of a function is the slope of that function at the point of evaluation. Let's take y=x:
dy/dx at 0 = 1
or y=x2/2
dy/dx at 0 = 0, the slope at 0.
As for a deeper answer as to why it works, I really can;t give you one. Try number theory. My number theory is terrible.
But as for what class you should take, I'd advise taking calculus with physics, esp. if you;ve already had trig. Calculus was invented by people thinking about physics, ya know. They compliment each other, even if the physics class you're taking is introductory.

#### PrudensOptimus

It is harder for me to understand somethings without proofs. In trig, even though it was rather easy to understand, but I still looked for the proof of the Heron's Formula, identities, logs rules...

Derivative is really easy to solve, but the notion part of it is very important, so I want to know how did derivative came up.

#### Mr. Robin Parsons

Originally posted by frankR
(SNIP) The funamental principle in Calculus is the limit. I'm surprised no one has explained what a limit is.(SNoP)
That is one I remember well, limits, but I cannot seem to remember the equation(s) that was used, anyone?

#### HallsofIvy

Homework Helper
If a person drives a car at a constant 40 mph, then his speed at any instant is 40 mph! That's pretty clear!

If you graph "distance traveled" against "time", then the slope of the line is the speed: that's because you calculate speed the same way you calculate slope: how much the distance has changed divided by how much the time has changed.

Suppose you start at 0 mph (standing still) and then accelerate up to 40 mph in, say 20 seconds. What is your speed at t= 5 seconds?

One could argue that "speed at a specific instant" doesn't make sense because you have to have SOME change in time before you can divide "change in distance" by "change in time" (That's one of Zeno's paradoxes).

If you were to graph distance against time in this case, "slope" wouldn't make sense because the graph is not a straight line. Geometrically, however, we could draw a tangent line to the curve and argue that it's slope gives the same result- the "instantaneous rate of change" is the slope of the tangent line: that's the geometric definition of the derivative.

#### Integral

Staff Emeritus
Gold Member
Let's see if I can do this off the top of my head

Let xn be a sequence of numbers. Some number l is said to be the limit of xn if

|xn-l| < &epsilon; where &epsilon; is a arbitray Real number, This must hold for all n>N;

In english, once you get past the Nth element in xn the rest of the members of the sequence will remain "close" to l, where close means that we can find elements of xn closer to l then the arbitrary choosen small real number &epsilon;

#### relinquished™

Hello ^_^ I'm pretty new here in the Physics Forums, but I really got interested on the topic about derivatives.

It seems like you're discussing the basis of Derivatives (rate of change of the value of a function) which has lead you to the concept of the limit.

The definition of the limit given by the Integral is correct, but I believe there is an easier way of saying it without having to resort to symbols ^_^

For me to understand the limit, I simply define it as a number the value of a function can never attain, but can get closer and closer to that number but not be that number ^_^

Now, according to my understanding of Leithold's book "The Calculus 7",The tangent line of the graph of the function is the line having the slope

m(x) = lim [f(x + ∆x) - f(x)] / ∆x as ∆x approcahes 0. If you further study the tangent line, you will discover that it is the derivative (meaning that it is the analytic view of the tangent line) That is why the limit is the basis of the derivative (to the fullest of my understanding ^_^`)

#### arivero

Gold Member
relinquished, you re right in the classical interpretation, but perhaps the historical way is not the more sensible way. Some mathematicians stress Barrow's rule, the fact of integration and derivation being inverse operators one of other. In this point of view, if you are able to build two such operators, you will already have a lot of the known properties of calculus. An ex-cambridge, Majid, did recently some interesting work in this line.

Another very physical basis could be a kind of "pseudorenormalisation". You know that lim $$[f(x+\Delta x)]/\Delta x$$ is infinite, so you look for another infinite to substract it :-)

#### zeronem

I think a graph interpretation is alot more explainable of the reason for the derivative.

Lets say you have a curve, There are two points plotted on this curve. A straight line runs through the two points. This labels that straight line, "the secant to the curve". Now we must approximate the secant's slope, by the two points on the curve that the secant runs through.

You know that the slope of a straight line connecting two points is this.

(y_2 - y_1)/(x_2 - x_1). Now find the slope of the secant of that curve.

Lets say dy = (y_2 - y_1), and dx = (x_2 - x_1). then slope of secant is dy/dx.

Apply the limit to this secant, lim_dx->0. Well, what happens when dx moves towards zero, the slope of the secant changes of course. Remarkably, it changes to the tangent of the curve. This tangent symbolizes the rate of change of this curve. The curve has tangents at all points to the curve, which shows the change in slope of the curve.

y = f(x)

y + dy = f(x+dx)

x = x_1

x + dx = x_2

y = y_1

y + dy = y_2

in Cartesian coordinates, the two points on the particular curve our found on

(x , f(x)), (x + dx , f(x + dx))

remembering that f(x) = y and f(x + dx) = y + dy.

Anyways, the slope of the tangent remarkably resembles the rate of change in the curve. If you look at the graph it will really come to you. When I studied particularly the derivative, this is how I understood the true meaning of the Derivative in Calculus.

So when you apply calculus to physics, it gets really interesting. You know what the rate of change in velocity is? Acceleration. So when you have a equation describing the velocity through the independent time varible, you can differentiate that equation with respect to time, and find the change in velocity which is acceleration.

From the formalism above, just change x = t, and y = V.

"V" being for Velocity, and "t" for time.

I would have used the special math language, so it would look alot better, but apparently I was too lazy to do so.

Please, if anyone finds an error in my post let me know about it.

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