NEET Physics Solutions For Class 11 Chapter 10 Mathematical Tools

NEET Physics Class 11 Chapter 10 Mathematical Tools

Mathematics is the language of physics. It becomes easier to describe, understand, and apply the physical principles if one has a good knowledge of mathematics.

Mathematical Tools

To solve the problems of physics Newton made significant contributions to Mathematics by inventing differentiation and integration.

Appropriate Choice Of Tool Is Very Important

What are mathematical tools?

This chapter includes all the necessary mathematics knowledge that we require to possess to study physics efficiently Importance of mathematical tools. This is the most important chapter of physics as it will be repeatedly used in all the upcoming chapters.

Importance of Mathematical Tools in Physics: This chapter is the foundation of physics that needs to be studied. Here we will learn about the mathematics that will be involved in Physics.

NEET Physics Class 11 Chapter 10 Function

The function is a rule of relationship between two variables in which one is assumed to be dependent and the other independent variable, for example,

For example: The temperatures at which water boils depend on the elevation above sea level (the boiling point drops as you ascend). Here elevation above sea level is the independent and temperature is the dependent variable

For example:  The interest paid on a cash investment depends on the length of time the investment is held. Here time is the independent and interest is the dependent variable.

• In each of the above examples, 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 represent it mathematically as y = f(x).

Here f represents the function, x is the independent variable & y is the dependent variable.

• All possible values of independent variables (x) are called the domain of a function.
• All possible values of a dependent variable (y) are called a range of functions.
• 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.

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 of relationship that describes the function may be square and multiplied by π).

If r = 1 A = π ; if r = 2A = 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 the two ways:

1. A formula such as y = x² 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),

y = sin x. Here the function is sine, and x is the independent variable.

Question 1. The volume V of a ball (solid sphere) of radius r is given by the function V(r) = (4/3)π(r³). What is the volume of a ball with a radius of 3m?

Answer:

V= 4/3π(3)³ = 36π m³.

Function Of A Function:

Suppose we are given 2 functions, f (x) = x² and g (x) = x + 1

If we are required to find f(g(x))

i.e., value of f(x) at x = g(x)

f (g(x)) = (x + 1)² [put g (x) in place of x in f(x)]

g (f (x)) = x² + 1 [put f (x) in place of x in g (x)]

Question 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:

Given formula F(r) = 2(r – 1) + 3.

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(1) + 3 = 2 + 3 = 5

F(x + 2) = 2(x +2 – 1) + 3 = 2x + 5 ;

F(F(2)) = F(5) = 2(5 – 1) + 3 = 2(4) + 3 = 8 + 3 = 11.

Question 3. A function ƒ(x) is defined as ƒ(x) = x² + 3, Find ƒ(0), ƒ(1), ƒ(x²), ƒ(x+1) and ƒ(ƒ(1)).

Answer:

f(x) = x² + 3

We have to find f(0), f(1), f(x²), f(x + 2) and f(f(1))

ƒ(0) = 0² + 3 = 3 ;

ƒ(1)= 1² + 3 = 4 ;

ƒ(x²) = (x²)²+3 = x⁴+3

ƒ(x+1) = (x + 1)² + 3 = x² + 2x +1 + 3 = x² + 2x + 4;

ƒ(ƒ(1)) = ƒ(4)= 4²+3 = 16 + 3 = 19

Question 4. If function F is defined for all real numbers x by the formula F(x) = x². Evaluate F at the input values 0, 2, x + 2, and F

Answer:

F(x) = x²

We have to find f(0), f(2), f(x + 2) and f(f(2)).

F(0) = 0 ;

F(2)= 2² = 4 ;

F(x+2) = (x+2)² = x² + 8x + 4;

F(F(2)) = F(4) = 4² = 16

NEET Physics Class 11 Chapter 10 Mathematical Tools – Geometry

Formulae For Determination Of Area:

1. Area of a square = (side)²
2. Area of rectangle = length × breadth
3. Area of a triangle = \(\frac{1}{2}\) (base × height)
4. Area of trapezoid = \(\frac{1}{2}\)(distance between parallel side) × (sum of parallel side)
5. Area enclosed by a circle = π r²(r = radius)
6. Surface area of a sphere = 4πr²(r = radius)
7. Area of a parallelogram = base × height
8. Area of curved surface of cylinder = 2πr l(r = radius and l = length)
9. Area of ellipse = π ab (a and b are semi-major and semi-minor axes respectively)
10. Surface area of a cube = 6(side)²
11. Total surface area of cone = \(\pi r^2+\pi r \ell\) where \(\pi r \ell=\pi r \sqrt{r^2+h^2}\)lateral area

Formulae For Determination Of Volume:

1. Volume of rectangular slab = length × breadth × height = abt
2. Volume of a cube = (side)³
3. Volume of a sphere = \(\frac{4}{3} \pi r^3 \ell\) (r = radius)
4. Volume of a cylinder = \(\pi r^2 \lambda \ell\) (r = radius and l is length)
5. Volume of a cone = \(\frac{1}{3} \pi r^2 h\) (r = radius and h is height) 3

Note:

⇒ \(\pi=\frac{22}{7}\)= 3.14, π² = 9.8776 ≈10

and \(\frac{1}{\pi}\)

= 0.3182 ≈0.3

NEET Physics Class 11 Chapter 10 Mathematical Tools – Differentiation

Finite Difference

The finite difference between two values of a physical quantity is represented by Δ notation. For example:

The difference in two values of y is written as Δy as given in the table below.

Infinitely Small Difference

The infinitely small difference means a very small difference. And this difference is represented by ‘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= 99.99999999……..

then dy = 0.000000……………….00001

NEET Physics Class 11 Chapter 10 Definition Of Differentiation

Another name for differentiation is derivative. Suppose y is a function of x or y = f(x) Differentiation of y concerning x is denoted by the symbol f’(x)

where F(x) = \(\frac{d y}{d x}\)

dx is a very small change in x and dy corresponds very small change in y.

Notation: There are many ways to denote the derivative of a function y = f(x). Besides f ’(x), the most common notations are these:

NEET Physics Class 11 Chapter 10 Mathematical Tools – Slope Of A Line

It is the tan of the angle made by a line with the positive direction of the x-axis, measured in an anticlockwise direction.

Slope = tan θ ( In 1^st quadrant tan θ is +ve and 2nd quadrant tan θ is –ve )

In Figure – 1 slope is positive In Figure – 2 slope is negative θ < 90° (1st quadrant) θ > 90° (2nd quadrant)

NEET Physics Class 11 Chapter 10 Mathematical Tools – Average Rates Of Change

Given an arbitrary function y = f(x) we calculate the average rate of change of y concerning 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 interval Δx over which the change occurred.

The average rate of change of y concerning x over the interval

[x, x + Δx] = \(\frac{\Delta y}{\Delta x}=\frac{f(x+\Delta x)-f(x)}{\Delta x}\)

Geometrically, \(\frac{\Delta y}{\Delta x}=\frac{Q R}{P R}\)= tan θ = Slope of the line PQ

therefore we can say that the average rate of change of y concerning x is equal to the slope of the line joining P and Q.

In triangle QPR tanθ = \(\frac{\Delta y}{\Delta x}\)

The Derivative Of A Function

We know that, average rate of change of y w.r.t. 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

\(f^{\prime}(x)=\frac{d y}{d x}=\Delta x \rightarrow 0 \frac{\lim _x(x+\Delta x)-f(x)}{\Delta x}\)

 

NEET Physics Class 11 Chapter 10 Mathematical Tools – 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 know the secant and tangent to a curve

Secant And 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. Tangent is a limiting case of secant which intersects the curve at two overlapping points.

1. In the figure-1 shown, if the value of Δx is gradually reduced then the point Q will move nearer to the point P.
2. If the process is continuously repeated (Figure – 2) value of Δx will be infinitely small and the secant PQ to the given curve will become a tangent at point P.

Therefore \(_{\Delta x \rightarrow 0}\left(\frac{\Delta y}{\Delta x}\right)=\frac{d y}{d x}=\tan \theta\)= tan θ

we can say that differentiation of y with respect to x, i.e. \(\left(\frac{d y}{d x}\right)\) is equal to slope of the tangent at point P (x, y) or tanθ = \(\frac{d y}{d x}\)

(the average rate of change of y from x to x + Δx is identical to the slope of the second PQ.)

NEET Physics Class 11 Chapter 10 Integration

In mathematics, for each mathematical operation, there has been defined an inverse operation. For example- The inverse operation of addition is subtraction, the inverse operation of multiplication is division and the inverse operation of a square is a square root. Similarly, there is an inverse operation for differentiation which is known as integration

Antiderivatives Or Indefinite Integrals 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 concerning x, denoted by

The symbol ∫is an integral sign. The function f is the integrand of the integral and x is the variable of integration.

For example f(x) = x³ then f′(x) = 3x²

So the integral of 3x² is x³

Similarly if f(x) = x³ + 4 then f′(x) = 3x²

So the integral of 3x² is x³ + 4

there for general integral of 3x² is x³ + c where c is a constant

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:

∫f(x)dx F(x) C. = + ∫………….(1)

The constant C is the constant of integration or arbitrary constant, Equation (1) is read, “The indefinite integral of f concerning x is F(x) + C.” When we find F(x)+ C, we say that we have integrated f and evaluated the integral.

Example. Evaluate ∫2x dx.

Answer:

The formula x² + C generates all the antiderivatives of the function 2x. The function x² + 1, x² – π, and \(x^2+\sqrt{2}\) 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.

NEET Physics Class 11 Chapter 10 Integral Formulas

Indefinite Integral

⇒ \(\int x^n d x=\frac{x^{n+1}}{n+1}+C, n \neq-1, n \text { rational }\)

⇒ \(\int d x=\int 1 d x=x+C \text { (special case) }\)

⇒ \(\int \sin (A x+B) d x=\frac{-\cos (A x+B)}{A}+C\)

⇒ \(\int \cos k x d x=\frac{\sin k x}{k}+C\)

Reversed Derivation Formula

⇒ \(\frac{d}{d x}\left(\frac{x^{n+1}}{n+1}\right)=x^n\)

⇒ \(\frac{d}{d x}(x)=1\)

⇒ \(\frac{d}{d x}\left(-\frac{\cos k x}{k}\right)=\sin k x\)

⇒ \(\frac{d}{d x}\left(\frac{\sin k x}{k}\right)=\cos k x\)

Question 1. Examples based on the above formulas:

1. \(\int x^5 d x=\frac{x^6}{6}+C\) Formula 1 with n = 5
2. \(\int \frac{1}{\sqrt{x}} d x=\int x^{-1 / 2} d x=2 x^{1 / 2}+C=2 \sqrt{x}+C\) Formula 1 with n = –1/2
3. \(\int \sin 2 x d x=\frac{-\cos 2 x}{2}+C\) Formula 2 with k = 2
4. \(\int \cos \frac{x}{2} d x=\int \cos \frac{1}{2} x d x=\frac{\sin (1 / 2) x}{1 / 2}+C=2 \sin \frac{x}{2}+C\) Formula 3 with k = 1/2

Question 2. Right: ∫x cosx dx = x sin x + cos x + C

Reason: The derivative of the right-hand side is the integrand:

Check: \(\frac{\mathrm{d}}{\mathrm{dx}}\) (x sin x + cos x + C) = x cos x + sin x – sin x + 0 = x cos x

Wrong: ∫x cosx dx = x sin x + C

Reason: The derivative of the right-hand side is not the integrand:

Check: \(\frac{\mathrm{d}}{\mathrm{dx}}\)(x sin x + C) = x cos x + sin x + 0 x cos x.