Calculus of Single & Multiple Variables

# Calculus of Single & Multiple Variables Notes | Study Mathematical Models - Physics

## Document Description: Calculus of Single & Multiple Variables for Physics 2022 is part of Mathematical Models preparation. The notes and questions for Calculus of Single & Multiple Variables have been prepared according to the Physics exam syllabus. Information about Calculus of Single & Multiple Variables covers topics like Limits, Continuity, Differentiability, Partial Differentiation, Jacobian, Taylor’s Series and Maclaurine Series Expansion and Calculus of Single & Multiple Variables Example, for Physics 2022 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for Calculus of Single & Multiple Variables.

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Limits

The number A is said to be the limit of f (x) at x = a if for any arbitrary chosen positive

numbers, however small but not zero, there exists a corresponding number ε greater than zero such that
For all values of x for which 0

Right hand and Left hand Limit

If x approaches a from larger values of x than a, then
"Put a + h for x in f(x) and make h approaches zero'’. In short, we have

If x approaches a from smaller values of x than a, then In this case

If both right hand and left hand limits of f(x), as x → a, exist and are equal in value, their common value, evidently, will be the limit of f(x), as x → a .

Theorem of Limits

If f (x) and g (x) are two functions then
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L’Hospital’s Rule

If a function f (x) takes the form then we say that f (x) is indeterminate at x = a. If ∅(x) and ψ(x) are functions of x such that ∅(a) = 0 and ψ(a) = 0, then

Some Important Standard Limits
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Example 1: Find the limit

Example 2: Find the value of

Continuity

f (x) is said to be continuous at x = a if R = L = V i.e.,

In case a function is not defined at x = a , i.e. f(a) does not exist or in case

theR.H.L ≠ L.H.L , then we say that the function is discontinuous at x = a. Its graph will show a break at x = a.

Example:
(i) is discontinuous at x = a as f (a) does not exist.
(ii)  is discontinuous at x = 0 because

(iii) f (x) = 1 - cos e1x is discontinuous at x = 0 as f (0) = 1 - cos e1/0 is undefined. It is oscillating.
Example: Let the function f (x) be defined by

f(x) will be continuous at those points only where rational and irrational values coincide i.e. ex = e1 - x ⇒ ex = So /(x) is continuous at x = l/2 only.

Differentiability

f (x) is said to be differentiable at x = a if R' = L' i.e,

Example: f(x) = e-x2 is differentiable but g(x) =  is not differentiable.

Right hand Limit
Left hand Limit
Thus R' ≠ L' means g(x)is not differentiable.

Tangents and Normal

Let y = f (x) be a given curve and P(x, y)and Q(x + δx, y + δy) be two neighbouring points on it. Equation of the line PQ is

This line will be tangent to the given curve at P if Q → P which in tem means that δx → 0 and we know that lim
Therefore the equation of the tangent is Y - y =
Normal at (x, y)
The normal at (x, v) being perpendicular to tangent will have its slope as and hence its equation is Y - y =
Geometrical meaning of dy/dx
dv/dx represents the slope of the tangent to the given curve y = f(x) at any point (x, y)
where ψ is the angle which the tangent to the curve makes with +ve

direction of x-axis.
In case we are to find the tangent at any point (x1, y1) then i.e. the value of  at (x1, y1) will represent the slope of the tangent and hence its equation in this case will be

Normal

Condition for tangent to be parallel or perpendicular to x-axis

If tangent is parallel to x-axis or normal is perpendicular to x-axis then
If tangent is perpendicular to x-axis or normal is parallel to x-axis then

Maxima and Minima

For the function y = f (x) at the maximum as well as minimum point the tangent is parallel to x-axis so that its slope is zero.
Calculate  and solve for x. Suppose one root of is at x = a .
If = negative for x = a, then maximum at x = a .
If positive for x = a, then minimum at x = a .
If at x = a , then find
If0 at x = a, neither maximum nor minimum at x = a.
If
If i.e., positive at x = a, then y is minimum at x = a and if i.e -ve at

x = a , then y is maximum at x = a and so on.

Partial Differentiation

If a derivative of function of several independent variables be found with respect to any one of them, keeping the others as constants it is said to be partial derivative. The operation of finding the partial derivative of function of more than one independent variable is called partial differentiation.

Euler Theorem of Homogeneous Function

If u is homogeneous function of x and y of degree n then
For n variable function f = f(x1,x2,x3.........xn) of degree n then

Total derivative u = f(x,y) x = ∅(t) and y = ψ(t) then
Differentiation of implicit functions f(x, y) = c be an implicit relation between x and y which defines as a differential function of x then

Change in variable u = f(x,y), x = ∅(s,t) and y = ∅(s,t)

Maxima and Minima (of function of two independent variables)

The necessary condition that f (x, y) should have maximum or minimum at x = a, y = b is that
Sufficient condition for maxima and minima

Case 1: if f(x, y) will have maximum or minimum at x = a, y = b if rt >s2 further f(x, y) is maximum or minimum according as r in negative or positive.
Case 2: if f(x, y) will have maximum or minimum at x = a, y = b if rt < s2 further /(x, y) is saddle point.
Case 3: if rt = s2 this case is doubtful case and further advanced investigation is needed to determined whether f(x,y) a maximum or minimum at or not.

Example 3: For what values of x and y , does the integral dt attain its maximum?

l(x,y) or f(x,y) =
fx = -(6-x-x2) = x2+x-6 = (x + 3)(x-2), fy =6 - y - y2 = -(y + 3)(y - 2)
f= 0 ⇒ x = 2,-3, f= 0 ⇒ y = 2,-3

so, stationary points are (2, 2),(2,-3),(-3,2)&(-3,-3)

fxx = 2x +1, fyy = -2y -1, fxy = 0

fxxfyy - (fxy)2  = - (2x +1) (2y +1)

At x = -3 and y = 2 and fxxfyy - (fxy)2 > 0

So maximum value of f (x, y) is obtained at x = -3 and y = 2

Example 4: If z = z(x, y) then prove that

Example 5: Find the first and second partial derivatives of z = x3 + y3 - 3axy and prove that

We have z = x3 + y3 - 3axy

Also

We observe that

Example 6: If u = x2
Show that  and

We have

Similarly,
and

Example 7: If z = f (x + ct) + ∅(x - ct), prove that

We have

and
Again
and
From (i) and (ii), it follows that
This is an important partial differential equation, known as wave equation.

Example 8: If u = log (x3 + y3 +z3 - 3xyz) , show that

we have

Now

Example 9: If u = f (r) and x = r cos θ, y = r sin θ, prove that

we have
Similarly,

now to find etc,we write r = (x+ v2)1/2

Similarly,
Substituting the values of etc. in (i), we get

Example 10: Show that  where

∴ z is a homogeneous function of degree 2 in x and y .
By Euler’s theorem, we get

Hence (i) becomes

Example 11: If z is a homogeneous function of degree n in x and y , show that

By Euler's theorem, (i)
Differentiating (i) partially w.r.t. x, we get
(ii)
Again differentiating (i) partially w.r.t. y , we get
(iii)

Multiplying (ii) by x and (iii) by y and adding, we get

Example 12: Given u = sinx = et and v = t2, find du/dt as a function of t. verify your dt result by direct substitution.

We have

Example 13: If x increases at the rate of 2 cm /sec at the instant when x = 3 cm . and at what rate must y be changing in order that the function 2xy - 3x2y shall be neither increasing nor decreasing?

Let u = 2xy-3x2y, so that

when x = 3 and y = 1, and u is neither increasing nor decreasing ,i.e.,
∴ (i) becomes 0 = (2-6x3)2 + (2x3-3x9)
or cm/sec. Thus y is decresing at the rate of 32/21 cm/sec.

Example 14: If μ = F(x-y,y -z,z- x) prove that

Put x-y = r,y-z = s and z-x = t, so that u = f (r,s,t) .

(i)
Similarly, (ii)
(iii)
Adding (i), (ii) and (iii), we get the required result.

Jacobian

If u and v are two function of two independent variable x and y then the determinant

is called the Jacobian of u and v with respect to x and y which is written as or

If u, v and w are functions of independent variable x, y and z then the determinant

is called the Jacobian of u , v and w with respect to of x, y z which is written as

Properties of Jacobian

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2. Chain rule for Jacobian if u,v arc function of r,s are function of x,y then

3. If u1, u2, u3 instead of being given explicitly in terms x1, x2, x3 be connected with them equations such as

then
((-1)3 is for three variable system)
4. If u1, u2, u3 be functions of x1, x2, x3 then the necessary and sufficient condition for existence of a functional relationship of the form f1 (u1u2, u3) = 0 is
Example 15: In a polar coordinates x = r cos θ y = r cos θ then find

Example 16:

Example 17: u = x2 - y2,v = 2xy x = rcosθ,y = rsinθ find

Use the formula

Example 18: u = xyz,v = x2 + y2 + z2,w = x + y + z find

We are using property three
f1 = u - xyz, f= v - x2 - v- z2, f3 = w-x-y-z

Taylor’s Series and Maclaurine Series Expansion

If a function f(x) has continuous derivatives up to (n + 1)th order, then this function can be expanded in the following way:

where Rn called the remainder after (n + 1) terms, is given by

When this expansion converges over a certain range of x that is, then the expansion is called Taylor Series of f(x) expanded about a.
If a = 0, the series is called Maclaurin Series:

Maclaurine’s Development

Changing a to 0 and h to x one will get

Where fn (ξ) is identified as reminder term ξ = θx and 0 < θ < 1

Example 19: In the Taylor series expansion of exp x + sin(x) about x = π then what is coefficient of (x - π)2.

Taylor series about x = π

f(x) = exp x + sin(x)

f(π) = exp π + sin(π) = exp π

f' (x) = exp x + cos x ⇒ f'(π) = exp π + cos π = exp π -1
f'' (x) = expx - sinx ⇒ f''(π) = expπ  sinπ ⇒ f'' (π) = expπ
The coefficient of (x - π)2 is

Example 20: Expand Maclaurin's series expand of tan x up to x3

f(x) = tanx ⇒ f(0) = 0
f'(x) = sec2 x ⇒ f'(0) = 1

f''(x) = 2 tan x sec2 x ⇒ f''(0) = 0

f''' (x) = 2sec2 x +6tan2 xsec2 x ⇒ f''' (0) = 2

Some Important Expansions
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Example 21: Expand exp(sin x) by maclaurin series upto the term containing x4

Taylor 's theorem for function in two variable .

Puth h = x - a ,k = y - b
f(x, y) = f(a, b) + [(x - a)fx (a, b) + (y - b)fy (a, 6)]

Example 22: Expand ex log(1 + y) upto powers of x and y upto term of two degree.

f (x, y) = ex log(1 + y) ⇒ f (0,0) = 0
fx(x, y) = elog(1 + y) ⇒ fx (0,0) = 0,
fxx (x, y) = elog(1+y) ⇒ fxx  (0,0) = 0,

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