The one-hour calculus course

 

(Calculus = the mathematics of derivatives and integrals)

 

1. Derivative = gradient function

 

The gradient shows how steeply inclined a graph is. For a straight line, the gradient has one specific value. For y = x its 1, for y = 2x it is 2, for y = ½x it is ½; for a horizontal line the gradient is 0.

 

Fig. calc1: Graphs of y = 0, y = ½x, y = x, y = 2x,

 

For a nonlinear curve, the gradient is changing. Below is the graph of y = x2 :

Fig. calc2 : y = x2 with tangents at x = 0, x =1 and x =2

 

The gradient in any point shows how the curve is inclined right there; it can be found by making a very large graph millimeter graph paper and drawing a tangent at the point in question; that is a line showing roughly how the curve is going very near the point. The gradient of this tangent is the gradient of the parabola at this point. For the y = x2 curve above, one can find that the gradient is 0 when x = 0, it is 2 when x = 1 and it is 4 when x = 2.

Fig. calc3: The points (0,0) , (1,2) , (2,4) making the graph of y = 2x

 

 We notice that the gradients make the graph of a new function, y = 2x. This - another function which shows the gradient in the first function - is the "derivative".

 

·        A function takes a value gives a new (or the same) value; y = x2 changes 1 to 1, 2 to 4, 3 to 9,....

·        The derivative takes a function and gives another function: it changes y = x2 to y = 2x

 

2. Some derivative rules

 

We found that y = 2x is the derivative of y = x2. Instead of finding the gradient function = derivative graphically it can be proved that certain rules give the derivative of any function we have. (This will be left to later mathematics lessons).

When we take the derivative of a function it can be symbolised with "D" or "(d/dx)". So for example:

 

Dx2 = 2x or (d/dx)x2 = 2x

 

Rule A:         Dxn = nx(n-1)

 

·        Dx2 = 2x is an example with n = 2 where Dx2 = 2x(2-1) = 2x1 = 2x

·        Dx3 = 3x(3-1) = 3x2 

·        Dx4 = 4x(4-1) = 4x3 

Note that

·        Dx = Dx1 = 1x(1-1) = x0 = 1

 

Rule B: Constants are not involved in the derivation

 

·        D5x2 = 5Dx2 = 5*2x = 10x

·        D7x4 = 7Dx4 = 7*4x3 = 28x3 

 

 

Rule C: If we have several terms, we take the derivative of one at a time

 

·        D(x3 + x2) = Dx3 + Dx2 = 3x2 + 2x

·        D(2x5 - 7x) = 2Dx5 - 7Dx = 2*5x4 - 7*x0 = 10x4 - 7

 

3. Derivatives in physics

 

The  derivative tells what the gradient of something else is. This is common in physics; for example the velocity is the gradient of the displacement when both are graphed with the time on the horisontal axis. From Mechanics we know that

 

s = ut +½at2 and v = u + at

 

Here a is a constant (it is uniformly accelerated motion) and u is a constant (we can reach higher final velocities by accelerating for a longer time, but that does not change what the initial velocity was). We can take the derivative of the s-function with t instead of x as the variable:

 

·        D(ut + ½at2) = Dut + D(½at2) = uDt + ½aDt2 = u*t0 + ½a*2t = u + at as expexted

 

4. Integral = area function or antiderivative

 

Some physical quantities are the gradient of something else, like the velocity is the gradient of the displacement. Others are the area under a graph of the other, for example the displacement is the area under the velocity graph. Since from a velocity graph we get back the same quantity (displacement) as we used when taking the derivative we may accept that the integral is the "opposite" of the derivative.

 

Fig. calc 4: Graphs of y = 2x with areas from x = 0 to x = 0, 1 and 2 shadowed

 

If we make a function to show how much area we have under the graph we can use the triangles above.

·        from x = 0 to x = 0 no area is yet found

·        from x = 0 to x = 1 we have the area (1*2)/2 = 1

·        from x = 0 to x = 2 we have the area (2*4)/2 = 4

 

If we now make a graph of the accumulated area under the y = 2x graph we get:

Fig. calc5: Graph of (0,0) , (1,1), (2,4) or y = x2

 

The function we get in this way is, as expected, y = x2. We can write this as

 

ò2xdx = x2

 

This looks a bit confusing, but just note that the integral symbol consists of two parts, a  ò and a dx which tells that x is the variable (may be useful if we have several letters involved). Whatever is in between the ò and the dx is the function to integrate.

 

Summary:

 

·        derivative: D(functiontoderive) or (d/dx)functiontoderive

·        integral: ò(functiontointegrate)dx 

 

Like the derivative, the integral takes one function and returns another function.

 

5. Some integration rules

 

Rule A: òxndx= x(n+1)/(n+1)

 

·        òxdx = òx1dx = x(1+1)/(1+1) = ½x2  

 

Rule B: Constants are not involved in integration

 

·        ò2xdx = 2òx1dx = 2*½x2 = x2 as expected

·        ò5x4dx = 5òx4dx = 5*x(4+1)/(4+1) = x5

 

 

 

Rule C: If several terms are integrated, we can take one at a time

 

·        ò(3x2 - 7x3)dx = ò3x2dx - ò7x3dx = 3òx2dx - 7òx3dx =

 3*x(2+1)/(2+1) - 7*x(3+1)/(3+1) = x3 - 7x4/4

 

 

6. Integrals in physics

 

We expect the integral of the velocity as a function of time (v = u +at) to be the displacement as a function of time (s = ut + ½at2)

 

·        ò(u + at)dt = òudt + òatdt = ò(u*1)dt + aòtdt = uòt0dt + aòt1dt

=u*t(0+1)/(0+1) + a*t(1+1)/(1+1) = ut + ½at2 as expected

 

We may also use integrals in other cases where an area under a graph is used. The work W = Fs when the force F is constant, but if it is not, then W = the area under the graph of F as a function of time. For example, the work done in stretching a spring against the force  F = kx requires the work given by the area of the triangle under the F-graph; streching it from s = 0 to s = x gives W = kx*x/2 = ½kx2, which is then the elastic potential energy stored in the stretched spring. This could also have been found with an integral:

 

W = òkxdx = kòxdx = kòx1dx = k*x(1+1)/(1+1) = kx2/2 = ½kx2 = Ep,elas 

 

 

In Mechanics, we had for objects far out in space that the force of gravity is

 

F = Gm1m2/r2

 

where r is the variable, the distance from the center of the planet, and the others are constants. We also had that

 

Ep = - Gm1m2/r

 

We can sort of understand this if we think of Ep as the amount of work done when an object is falling or rising in the graviational field; so it should have something to do with the integral of the force function. Here Gm1m2 is a constant, call it k. Now

 

·        ò(k/r2)dr = kòr-2dr = k*r(-2+1)/(-2+1) = -kr-1 = -k/x

 

which may explain the minus sign and the change from r2 to r. Note that we use r = infinity as the zero level since that is where the force function is zero; at r = 0 we would have F = infinite. (When we integrated y = 2x we could start from x = 0 as a "zero level" since y = 2x = 0 when x = 0).

 

 

 

 

 

7. Some other rules

 

Integration constants

 

If we take the derivative of a function y = k where k is a constant, then

 

Dk = D(k*1) = D(k*x0) = kDx0 = k*0*x0-1 = 0 independent of x

 

Therefore any constant will disappear in derivation, for example

 

Dx2 = 2x, D(x2 + 1) = 2x, D(x2 + 2) = 2x, D(x2 -127) = 2x etc.

 

On the other hand, if we integrate ò2xdx = x2, the result could as well have been x2 +1 or x2 +2 or x2 - 127 or anything similar. We could then write x2 + C where C is an unknown (positive or negative) integration constant. In physics we can mostly set C = 0, for example as in: ò(u + at)dt = ... =  ut + ½at2 but we should really have ... = s0 + ut + ½at2 where s0 is the displacement at the time t = 0. But this can usually be set to zero by defining s = 0 when t = 0; that is we start measuring the time when an object passes a chosen starting line.

 

The integral of y = 1/x

 

We would get òx-1dx = x(-1+1)/(-1+1) which involves division by zero. It can be shown in mathematics that òx-1dx = ln x (the natural logarithm of x). This is useful in thermodynamics where we find the work done in an  isothermal process as the area under a graph in a diagram with pressure P as a function of volume V:

 

PV = nRT gives P = nRT/V so the work is found using W = ò(nRT/V)dV = nRTòV-1dV = nRTlnV

 

when n, R and T are constants.

 

Trigonometric functions

 

Most useful are :

 

Dsin x = cos x                 Dcos x = - sin x

òcos x dx = sin x               òsin x dx = - cos x

 

We can also differentiate (= take the derivative of) or integrate a function more than once. If we have a displacement function (with the time as a variable) and differentiate it once we get a velocity function. If we differentiate that again, we get an acceleration function. If we try trigonometric functions, then Dsinx = cosx, and then Dcosx = -sin x. We got the original function back, only with an extra minus sign! Now think of a mass oscillating on a spring. Then F = ma and F = kx or -ks (with directions) for the force on it. Then

 

 ma = -ks gives a = -(k/m)s

 

that is, the acceleration function is the displacement function, give or take a negative constant. This may help explain why sine-functions are important for oscillating things, like the water molecules in an ocean wave. More about this in the Waves topic.