CBSE Class 11 Physics Calculus for Physics Notes Set 01

Functions

Constants

Constants are fixed real numbers whose value does not change. Examples: 3, e, a, −1, etc.

Variable

A variable is something that is likely to vary, something that is subject to variation; or, a quantity that can assume any of a set of values.
Types of variables:
(i) Independent variable: the variable which is being manipulated or changed.
(ii) Dependent variable: the variable whose value is the result of the change in the independent variable.
Example: in y = x², y is the dependent variable and x is the independent variable.

Function

A function is a rule of relationship between two variables in which one is assumed to be dependent and the other independent.
The temperature at which water boils depends on the elevation above sea level (the boiling point drops as you ascend). Here elevation above sea level is the independent variable and temperature is the dependent variable.
The interest paid on a cash investment depends on the length of time the investment is held. Here time is the independent variable and interest is the dependent variable.
In each case, the value of one variable quantity (dependent variable), which we might call y, depends on the value of another variable quantity (independent variable), which we might call x. Since the value of y is completely determined by the value of x, we say that y is a function of x and write it mathematically as y = f(x).

All possible values of the independent variable (x) are called the domain of the function.
All possible values of the dependent variable (y) are called the range of the function.
Think of a function f as a kind of machine that produces an output value f(x) in its range whenever we feed it an input value x from its domain (figure).
When we study circles, we usually call the area A and the radius r. Since area depends on radius, we say that A is a function of r, A = f(r). The equation A = πr² is a rule that tells how to calculate a unique (single) output value of A for each possible input value of the radius r.
A = f(r) = πr² (here the rule which describes the function may be described as 'square and multiply by π').
If r = 1, A = π;  if r = 2, A = 4π;  if r = 3, A = 9π.
The set of all possible input values for the radius is called the domain of the function. The set of all output values of the area is the range of the function.
We usually denote functions in one of two ways:
1. By giving a formula such as y = x² that uses a dependent variable y to denote the value of the function.
2. By giving a formula such as f(x) = x² that defines a function symbol f to name the function. Strictly speaking, we should call the function f and not f(x).
In y = sin x, the function is y, since x is the independent variable.

Example 1. The volume V of a ball (solid sphere) of radius r is given by the function \( V(r) = \frac{4}{3}\pi r^3 \). The volume of a ball of radius 3 m is?
Answer:
\( V(3) = \frac{4}{3}\pi (3)^3 = 36\pi \) m³

Example 2. Suppose that the function F is defined for all real numbers r by the formula F(r) = 2(r − 1) + 3. Evaluate F at the input values 0, 2, x + 2 and F(2).
Answer:
In each case we substitute the given input value for r into the formula for F:
F(0) = 2(0 − 1) + 3 = −2 + 3 = 1
F(2) = 2(2 − 1) + 3 = 2 + 3 = 5
F(x + 2) = 2(x + 2 − 1) + 3 = 2x + 5
F(F(2)) = F(5) = 2(5 − 1) + 3 = 11

Example 3. A function f(x) is defined as f(x) = x² + 3. Find f(0), f(1), f(x²), f(x + 1) and f(f(1)).
Answer:
f(0) = 0² + 3 = 3
f(1) = 1² + 3 = 4
f(x²) = (x²)² + 3 = x⁴ + 3
f(x + 1) = (x + 1)² + 3 = x² + 2x + 4
f(f(1)) = f(4) = 4² + 3 = 19

Differentiation

Finite Difference

The finite difference between two values of a physical quantity is represented by the Δ notation. For example, the difference in two values of y is written as Δy, as given in the table below.

Infinitely Small Difference

An infinitely small difference means a very, very small difference, and this difference is represented by the 'd' notation instead of 'Δ'. For example, an infinitely small difference in the values of y is written as 'dy'.
If y₂ = 100 and y₁ = 99.9999999999999…, then dy = 0.0000000000000…00001

Definition of Differentiation

Another name of differentiation is derivative. Suppose y is a function of x, or y = f(x). Differentiation of y with respect to x is denoted by the symbol f′(x),
where \( f'(x) = \frac{dy}{dx} \); dx is a very small change in x and dy is the corresponding very small change in y.

Notation

There are many ways to denote the derivative of a function y = f(x). The most common notations are:

Average Rate of Change

Given an arbitrary function y = f(x), we calculate the average rate of change of y with respect to x over the interval (x, x + Δx) by dividing the change in the value of y, i.e. Δy = f(x + Δx) − f(x), by the length of the interval Δx over which the change occurred.
The average rate of change of y with respect to x over the interval [x, x + Δx] \( = \frac{\Delta y}{\Delta x} = \frac{f(x + \Delta x) - f(x)}{\Delta x} \)

Geometrically, \( \frac{\Delta y}{\Delta x} = \frac{QR}{PR} = \tan\theta \) = slope of the line PQ.
In triangle QPR, \( \tan\theta = \frac{\Delta y}{\Delta x} \).
Therefore we can say that the average rate of change of y with respect to x is equal to the slope of the line joining P and Q.

The Derivative of a Function

We know that the average rate of change of y with respect to x is \( \frac{\Delta y}{\Delta x} = \frac{f(x + \Delta x) - f(x)}{\Delta x} \).
If the limit of this ratio exists as Δx → 0, then it is called the derivative of the given function f(x) and is denoted as
\( f'(x) = \frac{dy}{dx} = \lim_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{\Delta x} \)

Geometrical Meaning of Differentiation

The geometrical meaning of differentiation is very useful in the analysis of graphs in physics. To understand the geometrical meaning of derivatives we should have knowledge of the secant and the tangent to a curve.
Secant: A secant to a curve is a straight line which intersects the curve at any two points.

Tangent: A tangent is a straight line which touches the curve at a particular point. A tangent is the limiting case of a secant which intersects the curve at two overlapping points.
In the figure of the previous section (Figure 1), if the value of Δx is gradually reduced, then the point Q will move nearer to the point P. If the process is continuously repeated (Figure 2), the value of Δx will be infinitely small and the secant PQ to the given curve will become a tangent at point P.

Therefore \( \lim_{\Delta x \to 0}\left(\frac{\Delta y}{\Delta x}\right) = \frac{dy}{dx} = \tan\theta \)
We can say that the differentiation of y with respect to x, i.e. \( \left(\frac{dy}{dx}\right) \), is equal to the slope of the tangent at point P(x, y), or \( \tan\theta = \frac{dy}{dx} \).
(From Figure 1, the average rate of change of y from x to x + Δx is identical with the slope of the secant PQ.)

Rule 1: Derivative of a Constant

The first rule of differentiation is that the derivative of every constant function is zero. If c is a constant, then \( \frac{d}{dx}c = 0 \).

Example 4. Find the derivatives of the constants 8, −1/2 and √3.
Answer:
\( \frac{d}{dx}(8) = 0 \),   \( \frac{d}{dx}\left(-\frac{1}{2}\right) = 0 \),   \( \frac{d}{dx}\left(\sqrt{3}\right) = 0 \)

Rule 2: Power Rule

If n is a real number, then \( \frac{d}{dx}x^n = nx^{n-1} \).
To apply the power rule, we subtract 1 from the original exponent (n) and multiply the result by n.

Example 6. Find (i) \( \frac{d}{dx}\left(\frac{1}{x}\right) \) (ii) \( \frac{d}{dx}\left(\frac{4}{x^3}\right) \)
Answer:
(i) \( \frac{d}{dx}\left(\frac{1}{x}\right) = \frac{d}{dx}(x^{-1}) = (-1)x^{-2} = -\frac{1}{x^2} \)
(ii) \( \frac{d}{dx}\left(\frac{4}{x^3}\right) = 4\frac{d}{dx}(x^{-3}) = 4(-3)x^{-4} = -\frac{12}{x^4} \)

Example 7. Find (a) \( \frac{d}{dx}(x^{1/2}) \) (b) \( \frac{d}{dx}(x^{1/5}) \)
Answer:
(a) \( \frac{d}{dx}(x^{1/2}) = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}} \)
The function is defined for x ≥ 0; the derivative is defined only for x > 0.
(b) \( \frac{d}{dx}(x^{1/5}) = \frac{1}{5}x^{-4/5} \)
The function is defined for all real x; the derivative is not defined at x = 0.

Rule 3: The Constant Multiple Rule

If u is a differentiable function of x, and c is a constant, then \( \frac{d}{dx}(cu) = c\frac{du}{dx} \).
In particular, if n is a positive integer, then \( \frac{d}{dx}(cx^n) = cnx^{n-1} \).

Example 8. Explain the meaning of the derivative formula \( \frac{d}{dx}(3x^2) = 3(2x) = 6x \).
Answer:
It says that if we rescale the graph of y = x² by multiplying each y-coordinate by 3, then we multiply the slope at each point by 3.

Example 9. A useful special case: find the derivative of the negative of a differentiable function u.
Answer:
The derivative of the negative of a differentiable function is the negative of the function's derivative. Rule 3 with c = −1 gives
\( \frac{d}{dx}(-u) = \frac{d}{dx}(-1 \cdot u) = -1 \cdot \frac{d}{dx}(u) = -\frac{du}{dx} \)

Rule 4: The Sum Rule

The derivative of the sum of two differentiable functions is the sum of their derivatives.
If u and v are differentiable functions of x, then their sum u + v is differentiable at every point where u and v are both differentiable, and
\( \frac{d}{dx}(u + v) = \frac{du}{dx} + \frac{dv}{dx} \)
Similarly, \( \frac{d}{dx}(u - v) = \frac{d}{dx}[u + (-1)v] = \frac{du}{dx} + (-1)\frac{dv}{dx} = \frac{du}{dx} - \frac{dv}{dx} \)
The sum rule also extends to sums of more than two functions, as long as there are only finitely many functions in the sum. If u₁, u₂, …, u_n are differentiable at x, then so is u₁ + u₂ + … + u_n, and
\( \frac{d}{dx}(u_1 + u_2 + \ldots + u_n) = \frac{du_1}{dx} + \frac{du_2}{dx} + \ldots + \frac{du_n}{dx} \)

Example 10. Find dy/dx for (a) y = x⁴ + 12x (b) \( y = x^3 + \frac{4}{3}x^2 - 5x + 1 \)
Answer:
(a) \( \frac{dy}{dx} = \frac{d}{dx}(x^4) + \frac{d}{dx}(12x) = 4x^3 + 12 \)
(b) \( \frac{dy}{dx} = \frac{d}{dx}x^3 + \frac{d}{dx}\left(\frac{4}{3}x^2\right) - \frac{d}{dx}(5x) + \frac{d}{dx}(1) = 3x^2 + \frac{4}{3}\cdot 2x - 5 + 0 = 3x^2 + \frac{8}{3}x - 5 \)
Notice that we can differentiate any polynomial term by term, the way we differentiated the polynomials in the above example.

Rule 5: The Product Rule

If u and v are differentiable at x, then so is their product uv, and \( \frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx} \).
The derivative of the product uv is u times the derivative of v plus v times the derivative of u. In prime notation, (uv)′ = uv′ + vu′.
While the derivative of the sum of two functions is the sum of their derivatives, the derivative of the product of two functions is not the product of their derivatives. For instance,
\( \frac{d}{dx}(x \cdot x) = \frac{d}{dx}(x^2) = 2x \), while \( \frac{d}{dx}(x) \cdot \frac{d}{dx}(x) = 1 \cdot 1 = 1 \), which is wrong.

Example 11. Find the derivative of y = (x² + 1)(x³ + 3).
Answer:
Using the product rule with u = x² + 1 and v = x³ + 3, we find
\( \frac{d}{dx}[(x^2 + 1)(x^3 + 3)] = (x^2 + 1)(3x^2) + (x^3 + 3)(2x) \)
= 3x⁴ + 3x² + 2x⁴ + 6x = 5x⁴ + 3x² + 6x
This example can be done as well (perhaps better) by multiplying out the original expression for y and differentiating the resulting polynomial. We now check:
y = (x² + 1)(x³ + 3) = x⁵ + x³ + 3x² + 3
\( \frac{dy}{dx} = 5x^4 + 3x^2 + 6x \)
This is in agreement with our first calculation. There are times, however, when the product rule must be used, as in the following example where we have only numerical values to work with.

Example 12. Let y = uv be the product of the functions u and v. Find y′(2) if u(2) = 3, u′(2) = −4, v(2) = 1 and v′(2) = 2.
Answer:
From the product rule, in the form y′ = (uv)′ = uv′ + vu′, we have
y′(2) = u(2)v′(2) + v(2)u′(2) = (3)(2) + (1)(−4) = 6 − 4 = 2

Rule 6: The Quotient Rule

If u and v are differentiable at x, and v(x) ≠ 0, then the quotient u/v is differentiable at x, and \( \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2} \)
Just as the derivative of the product of two differentiable functions is not the product of their derivatives, the derivative of the quotient of two functions is not the quotient of their derivatives.

Example 13. Find the derivative of \( y = \frac{t^2 - 1}{t^2 + 1} \).
Answer:
We apply the quotient rule with u = t² − 1 and v = t² + 1:
\( \frac{dy}{dt} = \frac{(t^2 + 1)\cdot 2t - (t^2 - 1)\cdot 2t}{(t^2 + 1)^2} \)   \( \left[\text{As } \frac{d}{dt}\left(\frac{u}{v}\right) = \frac{v(du/dt) - u(dv/dt)}{v^2}\right] \)
\( = \frac{2t^3 + 2t - 2t^3 + 2t}{(t^2 + 1)^2} = \frac{4t}{(t^2 + 1)^2} \)

Rule 7: Derivative of the Sine Function

\( \frac{d}{dx}(\sin x) = \cos x \)

Example 14. Differentiate (a) y = x² − sin x (b) y = x² sin x (c) y = (sin x)/x
Answer:
(a) \( \frac{dy}{dx} = 2x - \frac{d}{dx}(\sin x) = 2x - \cos x \)   (Difference rule)
(b) \( \frac{dy}{dx} = x^2\frac{d}{dx}(\sin x) + 2x\sin x = x^2\cos x + 2x\sin x \)   (Product rule)
(c) \( \frac{dy}{dx} = \frac{x\cdot\frac{d}{dx}(\sin x) - \sin x\cdot 1}{x^2} = \frac{x\cos x - \sin x}{x^2} \)   (Quotient rule)

Rule 8: Derivative of the Cosine Function

\( \frac{d}{dx}(\cos x) = -\sin x \)

Example 15. Differentiate (a) y = 5x + cos x (b) y = sin x cos x
Answer:
(a) \( \frac{dy}{dx} = \frac{d}{dx}(5x) + \frac{d}{dx}(\cos x) = 5 - \sin x \)   (Sum rule)
(b) \( \frac{dy}{dx} = \sin x\frac{d}{dx}(\cos x) + \cos x\frac{d}{dx}(\sin x) \)   (Product rule)
= sin x(−sin x) + cos x(cos x) = cos² x − sin² x = cos 2x

Rule 9: Derivatives of Other Trigonometric Functions

Because sin x and cos x are differentiable functions of x, the related functions \( \tan x = \frac{\sin x}{\cos x} \), \( \sec x = \frac{1}{\cos x} \), \( \cot x = \frac{\cos x}{\sin x} \) and \( \text{cosec}\, x = \frac{1}{\sin x} \) are differentiable at every value of x at which they are defined. Their derivatives, calculated from the quotient rule, are given by the following formulas:
\( \frac{d}{dx}(\tan x) = \sec^2 x \);   \( \frac{d}{dx}(\sec x) = \sec x\tan x \)
\( \frac{d}{dx}(\cot x) = -\text{cosec}^2 x \);   \( \frac{d}{dx}(\text{cosec}\, x) = -\text{cosec}\, x\cot x \)

Example 16. Find dy/dx if y = tan x.
Answer:
\( \frac{d}{dx}(\tan x) = \frac{d}{dx}\left(\frac{\sin x}{\cos x}\right) = \frac{\cos x\frac{d}{dx}(\sin x) - \sin x\frac{d}{dx}(\cos x)}{\cos^2 x} \)
\( = \frac{\cos x\cos x - \sin x(-\sin x)}{\cos^2 x} = \frac{\cos^2 x + \sin^2 x}{\cos^2 x} = \frac{1}{\cos^2 x} = \sec^2 x \)

Example 17. Find (a) \( \frac{d}{dx}(3x + \cot x) \) (b) \( \frac{d}{dx}\left(\frac{2}{\sin x}\right) \)
Answer:
(a) \( \frac{d}{dx}(3x + \cot x) = 3 + \frac{d}{dx}(\cot x) = 3 - \text{cosec}^2 x \)
(b) \( \frac{d}{dx}\left(\frac{2}{\sin x}\right) = \frac{d}{dx}(2\,\text{cosec}\, x) = 2\frac{d}{dx}(\text{cosec}\, x) = 2(-\text{cosec}\, x\cot x) = -2\,\text{cosec}\, x\cot x \)

Rule 10: Derivatives of Logarithm and Exponential Functions

\( \frac{d}{dx}(\log_e x) = \frac{1}{x} \),   \( \frac{d}{dx}(e^x) = e^x \)

Example 18. Differentiate \( y = e^x \cdot \log_e x \).
Answer:
\( \frac{dy}{dx} = \frac{d}{dx}(e^x)\cdot\log_e x + \frac{d}{dx}[\log_e x]\cdot e^x \)
\( \Rightarrow \frac{dy}{dx} = e^x\log_e x + \frac{e^x}{x} \)

Rule 11: Chain Rule or 'Outside-Inside' Rule

\( \frac{dy}{dx} = \frac{dy}{du}\cdot\frac{du}{dx} \)
It sometimes helps to think about the chain rule in the following way. If y = f(g(x)), then \( \frac{dy}{dx} = f'[g(x)]\cdot g'(x) \).
In words: to find dy/dx, differentiate the 'outside' function f and leave the 'inside' g(x) alone; then multiply by the derivative of the inside.
We now know how to differentiate sin x and x² − 4, but how do we differentiate a composite like sin(x² − 4)? The answer is the chain rule, which says that the derivative of the composite of two differentiable functions is the product of their derivatives evaluated at appropriate points. The chain rule is probably the most widely used differentiation rule in mathematics. We begin with examples.

Example 19. The function y = 6x − 10 = 2(3x − 5) is the composite of the functions y = 2u and u = 3x − 5. How are the derivatives of these three functions related?
Answer:
We have \( \frac{dy}{dx} = 6 \), \( \frac{dy}{du} = 2 \), \( \frac{du}{dx} = 3 \)
Since 6 = 2 × 3, \( \frac{dy}{dx} = \frac{dy}{du}\cdot\frac{du}{dx} \)
Is it an accident that this relation holds? If we think of the derivative as a rate of change, our intuition allows us to see that this relationship is reasonable. For y = f(u) and u = g(x), if y changes twice as fast as u and u changes three times as fast as x, then we expect y to change six times as fast as x.

Example 20. Let us try this again on another function: y = 9x⁴ + 6x² + 1 = (3x² + 1)² is the composite of y = u² and u = 3x² + 1. Compare the derivatives.
Answer:
\( \frac{dy}{du}\cdot\frac{du}{dx} = 2u\cdot 6x = 2(3x^2 + 1)\cdot 6x = 36x^3 + 12x \)
and \( \frac{dy}{dx} = \frac{d}{dx}(9x^4 + 6x^2 + 1) = 36x^3 + 12x \)
Once again, \( \frac{dy}{du}\cdot\frac{du}{dx} = \frac{dy}{dx} \).
The derivative of the composite function f(g(x)) at x is the derivative of f at g(x) times the derivative of g at x.

Example 21. Find the derivative of \( y = \sqrt{x^2 + 1} \).
Answer:
Here y = f(g(x)), where \( f(u) = \sqrt{u} \) and u = g(x) = x² + 1. Since the derivatives of f and g are \( f'(u) = \frac{1}{2\sqrt{u}} \) and g′(x) = 2x, the chain rule gives
\( \frac{dy}{dx} = \frac{d}{dx}f(g(x)) = f'(g(x))\cdot g'(x) = \frac{1}{2\sqrt{g(x)}}\cdot g'(x) = \frac{1}{2\sqrt{x^2 + 1}}\cdot(2x) = \frac{x}{\sqrt{x^2 + 1}} \)

Example 22. Find the derivative of sin(x² + x).
Answer:
\( \frac{d}{dx}\sin(x^2 + x) = \cos(x^2 + x)\cdot(2x + 1) \)
Here sin is the outside function and x² + x is the inside. cos(x² + x) is the derivative of the outside with the inside left alone, and (2x + 1) is the derivative of the inside.

Example 23. Find (a) \( \frac{d}{dx}(1 - x^2)^{1/4} \) (b) \( \frac{d}{dx}\sin 2x \) (c) \( \frac{d}{dt}(A\sin(\omega t + \phi)) \)
Answer:
(a) \( \frac{d}{dx}(1 - x^2)^{1/4} = \frac{1}{4}(1 - x^2)^{-3/4}(-2x) \)   (u = 1 − x² and n = 1/4)
\( = \frac{-x}{2(1 - x^2)^{3/4}} \)   (function defined on [−1, 1]; derivative defined only on (−1, 1))
(b) \( \frac{d}{dx}\sin 2x = \cos 2x\,\frac{d}{dx}(2x) = \cos 2x \cdot 2 = 2\cos 2x \)
(c) \( \frac{d}{dt}(A\sin(\omega t + \phi)) = A\cos(\omega t + \phi)\frac{d}{dt}(\omega t + \phi) = A\cos(\omega t + \phi)\cdot\omega = A\omega\cos(\omega t + \phi) \)

Rule 12: Power Chain Rule

If u is a differentiable function of x, then \( \frac{d}{dx}u^n = nu^{n-1}\frac{du}{dx} \)

Example 24. Find \( \frac{d}{dx}\left(\frac{1}{3x - 2}\right) \).
Answer:
\( \frac{d}{dx}\left(\frac{1}{3x - 2}\right) = \frac{d}{dx}(3x - 2)^{-1} = -1(3x - 2)^{-2}\frac{d}{dx}(3x - 2) \)
\( = -1(3x - 2)^{-2}(3) = -\frac{3}{(3x - 2)^2} \)
Here we could also have found the derivative with the quotient rule.

Example 25. Find (a) \( \frac{d}{dx}(Ax + B)^n \) (b) \( \frac{d}{dx}\sin(Ax + B) \) (c) \( \frac{d}{dx}\log(Ax + B) \) (d) \( \frac{d}{dx}\tan(Ax + B) \) (e) \( \frac{d}{dx}e^{(Ax + B)} \)
Answer:
Here u = Ax + B, \( \frac{du}{dx} = A \)
(a) \( \therefore \frac{d}{dx}(Ax + B)^n = n(Ax + B)^{n-1}\cdot A \)
(b) \( \frac{d}{dx}\sin(Ax + B) = \cos(Ax + B)\cdot A \)
(c) \( \frac{d}{dx}\log(Ax + B) = \frac{1}{Ax + B}\cdot A \)
(d) \( \frac{d}{dx}\tan(Ax + B) = \sec^2(Ax + B)\cdot A \)
(e) \( \frac{d}{dx}e^{(Ax + B)} = e^{(Ax + B)}\cdot A \)
Note: These results are important.

Double Differentiation

If f is a differentiable function, then its derivative f′ is also a function, so f′ may have a derivative of its own, denoted by (f′)′ = f″. This new function f″ is called the second derivative of f because it is the derivative of the derivative of f. Using Leibniz notation, we write the second derivative of y = f(x) as
\( \frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{d^2y}{dx^2} \)
Another notation is \( f''(x) = D^2 f(x) \).

Example 26. If f(x) = x cos x, find f″(x).
Answer:
Using the product rule, we have \( f'(x) = x\frac{d}{dx}(\cos x) + \cos x\frac{d}{dx}(x) = -x\sin x + \cos x \)
To find f″(x) we differentiate f′(x):
\( f''(x) = \frac{d}{dx}(-x\sin x + \cos x) = -x\frac{d}{dx}(\sin x) + \sin x\frac{d}{dx}(-x) + \frac{d}{dx}(\cos x) \)
= −x cos x − sin x − sin x = −x cos x − 2 sin x

Applications of Differentiation

Differentiation as a Rate of Change

\( \frac{dy}{dx} \) is the rate of change of 'y' with respect to 'x'. For example:
(i) \( v = \frac{dx}{dt} \): velocity 'v' is the rate of change of displacement 'x' with respect to time 't'.
(ii) \( a = \frac{dv}{dt} \): acceleration 'a' is the rate of change of velocity 'v' with respect to time 't'.
(iii) \( F = \frac{dp}{dt} \): force 'F' is the rate of change of momentum 'p' with respect to time 't'.
(iv) \( \tau = \frac{dL}{dt} \): torque 'τ' is the rate of change of angular momentum 'L' with respect to time 't'.
(v) \( P = \frac{dW}{dt} \): power 'P' is the rate of change of work 'W' with respect to time 't'.

Example 27. The area A of a circle is related to its diameter by the equation \( A = \frac{\pi}{4}D^2 \). How fast is the area changing with respect to the diameter when the diameter is 10 m?
Answer:
The (instantaneous) rate of change of the area with respect to the diameter is \( \frac{dA}{dD} = \frac{\pi}{4}\cdot 2D = \frac{\pi D}{2} \)
When D = 10 m, the area is changing at the rate (π/2)(10) = 5π m²/m. This means that a small change ΔD m in the diameter would result in a change of about 5π ΔD m² in the area of the circle.

Example 28. (Physical example) Boyle's law states that when a sample of gas is compressed at a constant temperature, the product of the pressure and the volume remains constant: PV = C. Find the rate of change of volume with respect to pressure.
Answer:
V = C/P = CP⁻¹, so \( \frac{dV}{dP} = -\frac{C}{P^2} \)

Example 29. (a) Find the average rate of change of the area of a circle with respect to its radius r as r is changed from (i) 2 to 3 (ii) 2 to 2.5 (iii) 2 to 2.1
(b) Find the instantaneous rate of change when r = 2.
(c) Show that the rate of change of the area of a circle with respect to its radius (at any r) is equal to the circumference of the circle. Try to explain geometrically why this is true by drawing a circle whose radius is increased by an amount Δr. How can you approximate the resulting change in area ΔA if Δr is small?
Answer:
(a) Average rate \( = \frac{\pi r_2^2 - \pi r_1^2}{r_2 - r_1} \): (i) \( \frac{\pi(9 - 4)}{1} = 5\pi \) (ii) \( \frac{\pi(6.25 - 4)}{0.5} = 4.5\pi \) (iii) \( \frac{\pi(4.41 - 4)}{0.1} = 4.1\pi \)
(b) \( \frac{dA}{dr} = 2\pi r = 4\pi \) at r = 2
(c) dA/dr = 2πr, which is the circumference. When the radius increases by Δr, the added area is a thin ring of length ≈ 2πr and width Δr, so ΔA ≈ 2πr Δr.

Maxima and Minima

Suppose a quantity y depends on another quantity x in a manner shown in the figure. It becomes maximum at x₁ and minimum at x₂. At these points the tangent to the curve is parallel to the x-axis and hence its slope is tan θ = 0. Thus, at a maximum or a minimum, slope \( \Rightarrow \frac{dy}{dx} = 0 \)

Maxima: Just before the maximum the slope is positive, at the maximum it is zero and just after the maximum it is negative. Thus, \( \frac{dy}{dx} \) decreases at a maximum and hence the rate of change of \( \frac{dy}{dx} \) is negative at a maximum, i.e. \( \frac{d}{dx}\left(\frac{dy}{dx}\right) < 0 \) at a maximum. The quantity \( \frac{d}{dx}\left(\frac{dy}{dx}\right) \) is the rate of change of the slope. It is written as \( \frac{d^2y}{dx^2} \).
Conditions for maxima are: (a) \( \frac{dy}{dx} = 0 \) (b) \( \frac{d^2y}{dx^2} < 0 \)

Minima: Similarly, at a minimum the slope changes from negative to positive. Hence, with the increase of x, the slope is increasing, which means that the rate of change of slope with respect to x is positive. Hence \( \frac{d}{dx}\left(\frac{dy}{dx}\right) > 0 \).
Conditions for minima are: (a) \( \frac{dy}{dx} = 0 \) (b) \( \frac{d^2y}{dx^2} > 0 \)
Quite often it is known from the physical situation whether the quantity is a maximum or a minimum. The test on \( \frac{d^2y}{dx^2} \) may then be omitted.

Example 30. Find the maximum or minimum values of the functions: (A) y = 25x² + 5 − 10x (B) y = 9 − (x − 3)²
Answer:
(A) For a maximum or minimum value, we put \( \frac{dy}{dx} = 0 \): \( \frac{dy}{dx} = 50x - 10 = 0 \;\therefore\; x = \frac{1}{5} \)
Further, \( \frac{d^2y}{dx^2} = 50 \), i.e. \( \frac{d^2y}{dx^2} \) has a positive value at x = 1/5. Therefore y has a minimum value at x = 1/5. Substituting x = 1/5 in the given equation,
\( y_{min} = 25\left(\frac{1}{5}\right)^2 + 5 - 10\left(\frac{1}{5}\right) = 4 \)
(B) y = 9 − (x − 3)² = 9 − x² − 9 + 6x, or y = 6x − x²
\( \therefore \frac{dy}{dx} = 6 - 2x \). For a maximum or minimum value of y, put \( \frac{dy}{dx} = 0 \): 6 − 2x = 0, x = 3
To check whether y is maximum or minimum at x = 3, we check the sign of \( \frac{d^2y}{dx^2} \): \( \frac{d^2y}{dx^2} = -2 \), which is negative at x = 3. Hence y is maximum.
This maximum value of y is y_max = 9 − (3 − 3)² = 9

Integration

Definitions

A function F(x) is an antiderivative of a function f(x) if F′(x) = f(x) for all x in the domain of f. The set of all antiderivatives of f is the indefinite integral of f with respect to x, denoted by \( \int f(x)\,dx \).
The symbol ∫ is an integral sign. The function f is the integrand of the integral and x is the variable of integration.
For example, if f(x) = x³, then f′(x) = 3x². So the integral of 3x² is x³. Similarly, if f(x) = x³ + 4, then f′(x) = 3x²; therefore the general integral of 3x² is x³ + c, where c is a constant.
Once we have one antiderivative F of a function f, the other antiderivatives of f differ from F by a constant. We indicate this in integral notation in the following way:
\( \int f(x)\,dx = F(x) + C \)   …(i)
The constant C is the constant of integration or arbitrary constant. Equation (i) is read, "The indefinite integral of f with respect to x is F(x) + C." When we find F(x) + C, we say that we have integrated f and evaluated the integral.

Example 31. Evaluate \( \int 2x\,dx \).
Answer:
\( \int 2x\,dx = x^2 + C \), where x² is an antiderivative of 2x and C is the arbitrary constant.
The formula x² + C generates all the antiderivatives of the function 2x. The functions x² + 1, x² − π and x² + √2 are all antiderivatives of the function 2x, as you can check by differentiation.
Many of the indefinite integrals needed in scientific work are found by reversing derivative formulas.

Integral Formulas

Example 32. Evaluate, using the above formulas: (a) ∫dx (b) ∫x⁵ dx (c) \( \int \frac{1}{\sqrt{x}}dx \) (d) ∫sin 2x dx (e) \( \int \cos\frac{x}{2}dx \)
Answer:
(a) \( \int dx = x + C \)
(b) \( \int x^5\,dx = \frac{x^6}{6} + C \)   (Formula 1 with n = 5)
(c) \( \int \frac{1}{\sqrt{x}}dx = \int x^{-1/2}dx = 2x^{1/2} + C = 2\sqrt{x} + C \)   (Formula 1 with n = −1/2)
(d) \( \int \sin 2x\,dx = \frac{-\cos 2x}{2} + C \)   (Formula 2 with k = 2)
(e) \( \int \cos\frac{x}{2}dx = \int \cos\frac{1}{2}x\,dx = \frac{\sin(1/2)x}{1/2} + C = 2\sin\frac{x}{2} + C \)   (Formula 3 with k = 1/2)

Example 33. Which of the following is right: ∫x cos x dx = x sin x + cos x + C, or ∫x cos x dx = x sin x + C?
Answer:
Right: ∫x cos x dx = x sin x + cos x + C
Reason: the derivative of the right-hand side is the integrand.
Check: \( \frac{d}{dx}(x\sin x + \cos x + C) = x\cos x + \sin x - \sin x + 0 = x\cos x \)
Wrong: ∫x cos x dx = x sin x + C
Reason: the derivative of the right-hand side is not the integrand.
Check: \( \frac{d}{dx}(x\sin x + C) = x\cos x + \sin x + 0 \ne x\cos x \)

Rule 1: Constant Multiple Rule

A function is an antiderivative of a constant multiple k of a function f if and only if it is k times an antiderivative of f.
\( \int kf(x)\,dx = k\int f(x)\,dx \)

Example 34. Evaluate \( \int 5x^n\,dx \).
Answer:
\( \int 5x^n\,dx = 5\int x^n\,dx = \frac{5x^{n+1}}{n+1} + C \) (n ≠ −1)

Rule 2: Sum and Difference Rule

A function is an antiderivative of a sum or difference f ± g if and only if it is the sum or difference of an antiderivative of f and an antiderivative of g.
\( \int [f(x) \pm g(x)]\,dx = \int f(x)\,dx \pm \int g(x)\,dx \)

Example 35. (Term-by-term integration) Evaluate \( \int (x^2 - 2x + 5)\,dx \).
Answer:
If we recognize that \( \frac{x^3}{3} - x^2 + 5x \) is an antiderivative of x² − 2x + 5, we can evaluate the integral as \( \int (x^2 - 2x + 5)\,dx = \frac{x^3}{3} - x^2 + 5x + C \)
If we do not recognize the antiderivative right away, we can generate it term by term with the sum and difference rule:
\( \int (x^2 - 2x + 5)\,dx = \int x^2\,dx - \int 2x\,dx + \int 5\,dx = \frac{x^3}{3} + C_1 - x^2 + C_2 + 5x + C_3 \)
This formula is more complicated than it needs to be. If we combine C₁, C₂ and C₃ into a single constant C = C₁ + C₂ + C₃, the formula simplifies to \( \frac{x^3}{3} - x^2 + 5x + C \) and still gives all the antiderivatives there are.
For this reason we recommend that you go right to the final form even if you elect to integrate term by term. Find the simplest antiderivative you can for each part and add the constant at the end.

Example 36. We can sometimes use trigonometric identities to transform integrals we do not know how to evaluate into integrals we do know how to evaluate. The integral formulas for sin² x and cos² x arise frequently in applications. Evaluate (a) ∫sin² x dx (b) ∫cos² x dx
Answer:
(a) Using \( \sin^2 x = \frac{1 - \cos 2x}{2} \): \( \int \sin^2 x\,dx = \int \frac{1 - \cos 2x}{2}dx = \frac{1}{2}\int (1 - \cos 2x)\,dx = \frac{1}{2}\int dx - \frac{1}{2}\int \cos 2x\,dx \)
\( = \frac{x}{2} + \left(-\frac{1}{2}\right)\frac{\sin 2x}{2} + C = \frac{x}{2} - \frac{\sin 2x}{4} + C \)
(b) Using \( \cos^2 x = \frac{1 + \cos 2x}{2} \): \( \int \cos^2 x\,dx = \int \frac{1 + \cos 2x}{2}dx = \frac{x}{2} + \frac{\sin 2x}{4} + C \)
As in part (a), but with a sign change.

Some Standard Indefinite Integrals

(An arbitrary constant C should be added to each of these integrals.)
(a) \( \int (ax + b)^n\,dx = \frac{(ax + b)^{n+1}}{a(n + 1)} + C \) (provided n ≠ −1)
(b) \( \int \frac{1}{x}dx = \ln x + C \)
(c) \( \int \frac{dx}{a + bx} = \frac{1}{b}\ln(a + bx) + C \)
(d) \( \int e^{ax}dx = \frac{1}{a}e^{ax} + C \)
(e) \( \int \sin(ax + b)\,dx = -\frac{1}{a}\cos(ax + b) + C \)
(f) \( \int \cos(ax + b)\,dx = \frac{1}{a}\sin(ax + b) + C \)

Example 37. Evaluate: (a) ∫(3x + 2)³ dx (b) \( \int \frac{2\,dx}{x} \) (c) \( \int \frac{dx}{5 + 2x} \) (d) \( \int \frac{dx}{3 - 5x} \) (e) ∫e^3x dx (f) ∫e^−x/2 dx (g) ∫sin(3x + 5) dx (h) ∫cos(2x − 5) dx
Answer:
(a) \( \int (3x + 2)^3\,dx = \frac{(3x + 2)^4}{4 \times 3} + C = \frac{(3x + 2)^4}{12} + C \)
(b) \( \int \frac{2\,dx}{x} = 2\ln x + C \)
(c) \( \int \frac{dx}{5 + 2x} = \frac{1}{2}\ln(5 + 2x) + C \)
(d) \( \int \frac{dx}{3 - 5x} = -\frac{1}{5}\ln(3 - 5x) + C \)
(e) \( \int e^{3x}dx = \frac{1}{3}e^{3x} + C \)
(f) \( \int e^{-x/2}dx = -2e^{-x/2} + C \)
(g) \( \int \sin(3x + 5)\,dx = -\frac{1}{3}\cos(3x + 5) + C \)
(h) \( \int \cos(2x - 5)\,dx = \frac{1}{2}\sin(2x - 5) + C \)

Definite Integration (Integration with Limits)

In \( \int_a^b f(x)\,dx \), the integral sign is ∫, a is the lower limit of integration, b is the upper limit of integration, the function f(x) is the integrand and x is the variable of integration. The whole expression is read as the integral of f from a to b.
If g(x) is an antiderivative of f(x), then \( \int_a^b f(x)\,dx = [g(x)]_a^b = g(b) - g(a) \)

Example 38. Evaluate (i) \( \int_{-1}^{4} 3\,dx \) (ii) \( \int_0^{\pi/2} \sin x\,dx \)
Answer:
(i) \( \int_{-1}^{4} 3\,dx = 3\int_{-1}^{4} dx = 3[x]_{-1}^{4} = 3[4 - (-1)] = (3)(5) = 15 \)
(ii) \( \int_0^{\pi/2} \sin x\,dx = [-\cos x]_0^{\pi/2} = -\cos\frac{\pi}{2} + \cos 0 = -0 + 1 = 1 \)

Example 39. Evaluate (1) \( \int_0^a x^2\,dx \) (2) \( \int_3^5 x\,dx \) (3) \( \int_0^b x^{3/2}\,dx \)
Answer:
(1) \( \int_0^a x^2\,dx = \left[\frac{x^3}{3}\right]_0^a = \frac{a^3}{3} \)
(2) \( \int_3^5 x\,dx = \left[\frac{x^2}{2}\right]_3^5 = \frac{5^2 - 3^2}{2} = 8 \)
(3) \( \int_0^b x^{3/2}\,dx = \left[\frac{x^{5/2}}{5/2}\right]_0^b = \frac{2}{5}b^{5/2} \)

Applications of Definite Integral

Calculation of Area Under a Curve

From the graph shown in the figure, divide the whole area into infinitely small strips of width dx. Take a strip at position x of width dx.
Small area of this strip, dA = f(x) dx
So, the total area between the curve and the x-axis = sum of the areas of all strips = \( \int_a^b f(x)\,dx \)
Let f(x) ≥ 0 be continuous on [a, b]. The area of the region between the graph of f and the x-axis is \( A = \int_a^b f(x)\,dx \)

Example 40. (Using an area to evaluate a definite integral) Evaluate \( \int_a^b x\,dx \), 0 < a < b.
Answer:

We sketch the region under the curve y = x, a ≤ x ≤ b (figure) and see that it is a trapezium with height (b − a) and parallel sides a and b. The value of the integral is the area of this trapezium:
Thus \( \int_a^b x\,dx = (b - a)\cdot\frac{a + b}{2} = \frac{b^2}{2} - \frac{a^2}{2} \)
For example, \( \int_1^{\sqrt{5}} x\,dx = \frac{(\sqrt{5})^2}{2} - \frac{(1)^2}{2} = 2 \), and so on.
Notice that x²/2 is an antiderivative of x, further evidence of a connection between antiderivatives and summation.

To Find Impulse

\( F = \frac{dp}{dt} \), so impulse \( = \int F\,dt \). Impulse is the area under the F–t curve.

Example 41. If F = kt, then find the impulse from t = 0 to t = 3 s.
Answer:

The impulse is the area under the F–t curve:
\( I = \int_0^3 kt\,dt = k\left[\frac{t^2}{2}\right]_0^3 \)
\( \Rightarrow I = \frac{9k}{2} \)

To Calculate Work Done by a Force

\( W = \int \vec{F}\cdot d\vec{x} \)
So the area under the F–x curve gives the value of the work done.