Partial Derivative

Geometrical Interpretation of CMVT: A geometrical interpretation may be given for the t=c y theorem as follows. t=b Let a plane curve be represented Parametrically by the equations x=g(t) and y=f(t), where a
t=a 0

The slope of the curve for a given t is (1)

x

(2)The constant is the slope of the straight line joining the points on the curve corres ondin to t=a and t=b respectively. The theorem says that two slopes (1) and (2) are equal for at least one value of t between a and b.

(2)The constant is the slope of the straight line joining the points on the curve corres ondin to t=a and t=b respectively. The theorem says that two slopes (1) and (2) are equal for at least one value of t between a and b.

Q1. in CMVT, if that c is independent of x.

, show

Solution. Since both f(x) and g(x) are continuous on [a, b] and f’(x) and g’(x) exist in (a, b) and ’ ≠ According to CMVT, we get

=>a+b = 2c =>c= (a+b)/2 Therefore c is independent of x.

Partial Differentiation: Consider u= f(x,y). Here x and y are independent variables and u is dependent variable. In practical many situation arises where a variable depends on more than one independent variable.

The differentiation u w.r.t. x treating y as a constant is called the partial derivative of u w.r.t. x and denoted as Anal ticall

if exists.

Similarly,

if exists.

Homogenous Function: Definition: A function f(x,y) is said to be homogenous of degree n if it can be express in the form or

For example, Let z = x2+xy+y2 = x2(1 + y/x + y2/x2) =x2.Ф(y/x) Or, z = y2(x2/y2 + x/y + 1) = y2.Ф(x/y) In either cases z is homogenous of degree 2.

For more than 2 variables, it can be expressed as, f(x,y,z)= xn.Ф1(y/x, z/x) n.Ф (x/y, z/y) Or y 2 n. Ф (x/z, y/z) Or z 3

EULER’S THEOREM: If f(x,y) be homogenous function of x and y of degree n, then

Proof: Since f(x,y) is homogenous function of degree n. Let f(x,y) = = xn.Ф(t), where t = y/x. So,

And

Therefore,

Implicit Function: Definition: An implicit function x and y is an equation of the form f(x,y)=0 which is not necessarily be solved for one of the variables in terms of others explicitly that is say x in terms of y.

For example, X3+y3-3axy=0 is an implicit function of x and y. In contract, when y is directly expressed in terms of x, say y=f(x), then y is called as explicit function.

Partial Differentiation of an implicit function: Let f(x,y,z)=0 represents an equation of an implicit function of three variables x, y and z. Assuming z=f(x,y) then differentiating partially f(x,y,z)=0 with respect to x and y respectively, we get. and Provided

(b) Determination of

:

Assuming x=f(y,z) then by differentiating f(x,y,z) partially with respect to y and z respectively , we get . an Provided

(c) Determination of Assuming y=f(x,z) then by differentiating partially f(x,y,z)=0 w.r.t. x and z respectively, we get. and Provided

Example 1. Find

Solution: Let, f(x, y,z)=

Differentiating (1) partially w.r.t. x, we get

Differentiating (1) partially w.r.t. z, we get

Differentiating (1) partially w.r.t. y, we get

………………………(1)

Total Derivative: Let u = f(x,y) be a function of two independent variables x and y, which has continuous partial derivatives w.r.t. x and y. Let be differentiable function of t, then the composite function is differentiable function of t and

Here , is called total derivative of u. Equation (1) is called chain rule for two independent variables. In general form of this rule for functions of several independent variables like can be written as,

Corollary 1. If u=f(x,y) is a function of two independent variables x and y, y is a function of x, y=g(x), then chain rule (1) can be written as follows,

Corollary 2. If u=f(x,y) is a function of two independent variables x and y, and if x=Ф1(t1, t2), y=Ф2(t1, t2) then the composite function u=f[Ф1(t1, t2), Ф2(t1, t2)] is a function of t1 and t2.

If x, y and f are differentiable function, then we get,

Generalizing, chain rule (4) for

We get,

………………………………

Example.1 Find as total derivative for u= 3x+xy-y2 where, x=cos3t, y=sin3t. Solution: Here

u= 3x+xy-y2

Also, x= cos3t y= s n

= (3+y).(-3sin3t) + (x-2y).(3cos3t) = (3+sin3t)(-3sin3t) + (cos3t-2sin3t)(3cos3t) = -9sin3t-3sin23t + 3cos23t - 6sin3t.cos3t = -9sin3t-3sin23t+3(1-sin23t) – 3(2sin3t.cos3t) = -9sin3t – 6sin23t + 3 - 3sin6t = - s n + cos – s n . Example 2. If z=f(x,y), prove that if x=eu+e -v, y=e-u- ev then

Solution: Here we can write the chain rule as

z

x

u

y

v Fig.1.2

Now (1)-(2), we get,

Hence proved.

Laplace Operator or Laplacian: The Laplace operator is a second order differential operator in the n-dimensional Euclidean space, defined as the divergence of the gradient .Thus if f is a twice-differentiable real-valued function,

f Equivalently, the Laplacian of is the sum of all the unmixed second partial derivatives in the Cartesian coordinates x i … … … … … (2)

Notes: (1) Let x 1, x 2,… … x n be a system of Cartesian coordinates on a n-dimensional Euclidean space, and let i1, i2,… … in be the corresponding basis of unit vectors. The diver ence of a continuousl differentiable vector field f = g1 i1 + g2 i2 + … …+ gn in is defined to be the scalar-valued function:

(2) The gradient (or gradient vector field) of a scalar function is denoted where (the nabla symbol) denotes the vector differential operator, del. The notation grad(f) is f is also used for the gradient. The gradient of defined to be the vector field whose components are t e part a er vat ves o f. T at s: …. …. … … … (4) Here the gradient is written as a row vector, but it is often taken to be a column vector.

(3) Vector differential operator symbol.

called nabla

(4) Laplace operator or Laplacian s mbol.

called tebla

Applications of Partial Differentiation: 1. Jacobians and their Applications. 2. Taylor’s Theorem for functions of two variables. . . 4. Maxima and Minima. 5. Saddle points. 6. Lagrange’s method of undetermined multipliers.

1. Jacobians and their Applications: Defination1: (2-D case): If u = f(x, y) and v = g(x, y) be differentiable functions of two independent variables x and y, then the jaconian of u and v w.r.t. x and y is denoted by and is defined as follows,

Defination2:(3-D Case): If u = f(x, y, z), v = (x, y, z) and w = (x, y, z) be differentiable functions of three independent variables x, y and z, then the Jacobian of u, v, w w.r.t. x, y, z is denoted by

and is defined as follows,

Thus in general (n-D Case) if u1 = f 1(x1, x2, … …, xn) u2 = f 2(x1, x2, … …, xn), … … … … … … … … … …, un = f n(x1, x2, … …, xn), be differentiable functions of n-independent variables x1, x2, … …, xn then

Example1: Find

when x = r.cosθ, y = r.sinθ.

Solution: Here x = r.cosθ and y = r.sinθ. Now we have,

= r.cos2θ+r.sin2θ =r(cos2θ+sin2θ) =r.

Example2: If x= r.sinθ.cosφ, y = r.sinθ.sinφ, z = r.cosθ, find . Solution: Here given that, x= r.sinθ.cosφ, y = r.sinθ.sinφ, z = r.cosθ. Now we have,

= r2sinθ.

Properties of Jacobians: 1). If J is the Jacobian of u, v w.r.t. x, y and J’ is the Jacobian of x, y w.r.t. u, v, then JJ’ = 1 that is,

Proof: Let, u = f(x, y) and v = g(x, y) ………………..(1) Solving (1) for x and y, we get x = φ(u, v) and y = ψ(u, v) ……………….(2) Differentiating (1) partially w.r.t. u and v gives,

N

,

Using (3) to (6), we get Hence proved.

In general, the property can be written as

If u1 = f 1(x1, x2, … … xn), u2 = f 2(x1, x2, … … xn), … … … … n= n , ,…… n . Note: J’ is known as inverse transformation Jacobian of J and vice versa.

2). Chain rule of Jacobian (Jacobians of composite functions): If u, v are functions of r, s and r, s are functions of x, y then

roo : and y.

ere, u an

v are compos e unc ons o x

Differentiating u and v partially w.r.t. x and y, we get

Similarly,

Now,

using (1) to (4) Hence proved.

In general, the property can be written as

If u1, u2, … …, un are functions of r1, r2, … …, rn are unc ons o x1, x2, … …, xn. 3). If functions u, v, w of three independent variables x, y, z are functionally related or dependent iff

4). If the functions u1, u2, …, un of the variables x1, x2, …, xn be defined by the relations u1 = f 1(x1), u2 = f 2(x1,x2), u3 = f 3(x1, x2, x3) and so on un = f n(x1, x2, … …, xn), then

Proof: Given, u1 = f 1(x1). u2 = f 2(x1,x2). and so on.

Now,

=

Example: If x = u(1-v), y = uv, evaluate and . Also Solution: Here, x = u(1-v) and y = uv.

Now, x=u(1-v), y=uv solving them we get, u = x+y and v=y/(x+y)

Hence verified.

2. Taylor’s Theorem for functions of two variables. Taylor’s series is useful to approximate a function in the neighborhood of a point. It is widely used in many disciplines like engineering, science, etc. to establish continuous models. It is also the mainsta of many numerical techniques. Statement: If f(x, y) and all its partial derivatives upto the nth order are finite and continuous for all points (x, y), where a ≤ x ≤ a+h, b ≤ y ≤ b+k, then

Alternate forms of Taylor’s series: 1.) In suffix notation, we get

2.) Putting x = a and y = b in (1), we have

3.) Putting a+h = x and b+k = y in (2), we get

Which expands f(x,y) in powers of (x-a) and (y-b) and is known as Taylor’s series or Taylor’s expansion or Taylor’s series expansion of f(x,y) about the point (a,b). Example: Use Taylor’s series to expand sinx.cosy in powers of (x-π/3) and (y-π/4). Solution: Here, x-a = x - π/3. and y-b =y - π/4 => a = π/3

=> b = π/4

Let, f(x, y) = sinx.cosy => f(π/3, π/4) = √3/2√2. f x(x, y) = cosx.cosy => f x(π/3, π/4) = 1/2√2. f y(x, y) = -sinx.siny => f y(π/3, π/4) = -√3/2√2.

f xx (x, y) = -sinx.cosy => f xx(π/3, π/4) = -√3/2√2. f xy (x, y) = -cosx.siny => f xy (π/3, π/4) = - 1/2√2. f yy (x, y) = -sinx.cosy => f yy(π/3, π/4) = -√3/2√2. ... … …

………

……………

………

Now, the Taylor’s series in powers of (x - π/3) and (y - π 4 is iven b

3.Errors and Approximations: Let u be some physical quantity of interest depends on two measurable quantities x and y by the relation u = f(x, y). It is obvious that any change ∆x in x and ∆y in y will propagate the overall error ∆u in u and is given by ∆u = ∆f = f(x+∆x, y+∆y) – f(x,y). Now , expanding f(x+∆x, y+∆y) using Taylor’s series, we have

For small values approximately

of

∆x

and

∆y,

we

have

f(x,y) if the small errors in x and y occurs. It can easily be generalized for functions more than two variables approximately as

Where f is the function of x, y, z, t, …. … As we know that the error occurs in ∆x and ∆y may be positive or negative, therefore, the maximum error in f if given by

And in general,

If f is a function of x, y, z, t, … … …

Note: (i) df in (1) and (2) above known as approximate error in f while dx, dy, … … are known as actual (Absolute) errors in x, y, …., respectively. (ii) dx/x, dy/y, … …, are known as relative errors ro ortinal error in x …. …. res ectivel . (iii) (dx/x).100 and (dy/y).100 are known as percentage errors in x and y respectively respectively..

Exam ampl ple1 e1:: Us Usin ing g di difffer eren enttia ialls fi find nd the approximate value of √8(65)1/3. Solution: Let, f(x, y) = x1/2y1/3. Choosing x = 9, y = 64 with decrease 1 in x and increase 1 in y. ,

Here ∆x = -1 and ∆y = 1, putting these values we have,

Thus, change in x and y, decrease the value of the function by . Therefore, Example2: The two resistors x and y are connected in ar aral alle lell so th that at th the e tot otal al res esis isttan ance ce R is iv iven en b xy/(x+y). If x and y are measured to be 200Ω and 300Ω respectively, with the increase of 1.5Ω in x and decrease of 4Ω in y, find the change in R. Solution: According to the given question, we get x=200Ω

y=300Ω

Here, Similarly, Now , Here, x = 200Ω, y = 300Ω, ∆x = 1.5Ω, ∆y = -4Ω putting these values we get,

Therefore the decrease in R is 0.1Ω.

4.Maxima and minima of a function of two variables: Definition: Consider a function f(x, y) of two variables x and y and a point (a, b) and the neighborhood points (a+h, b+k). The value f(a, b) is (i) maximum if f(a, b) > f(a+h, b+k) and (ii) minimum if f(a, b) < f(a+h, b+k). Extremum point: A point (a, b) which either a maximum or a minimum is called an extremum point.

Theorem1: The necessary conditions for f(a, b) to be an extremum value of “f” are that f x(a, b) = 0 and f y(a, b) = 0 The point (a, b) is called a stationary point and the Theorem2: The sufficient conditions for the stationary value f(a, b) to be an extremum value as follows: Let f xx(a, b) = A, f xy(a, b) = B, f yy(a, b) = C , then

1. If AC-B2 > 0 and A (or C) < 0, then f(a, b) is a maximum value. 2. If AC-B2 > 0 and A (or C) > 0, then f(a, b) is a minimum value. 3. If AC-B2 < 0, then f(a, b) is neither a maximum nor a minimum value. But it a saddle point. 4. If AC-B2= 0 then no conclusion can be drown about maximum or minimum and further investigation is needed. Note: Every extremum value is a stationary value but converse is not true.

Stationary point or Critical point: A point (a, b) is said to be stationary point of “f” if (i) at (a, b) both f x and f y are zero, or (ii) it is a boundary point of the domain. Working rules for finding extremum values of a 1. Find

.

2. Find points (x, y) for which This gives the stationary points. 3. At each stationary point, evaluate values of

.

Saddle point or minimax point: In mathematics, a saddle point is a point in the domain of a function of two variables which is a stationary point but not a local extremum. At such a point, in general, the surface resembles a saddle that curves up in one direction, and curves down in a different direction . of contour lines, a saddle point can be recognized, in general, by a contour that appears to intersect itself. For example, two hills separated by a high pass will show up a saddle point, at the top of the pass, like a figure-eight contour line.

Example1: Examine the function f(x, y) =

2x4+y2-x2-2y for maxima and minima. Solution: Here, f(x, y) = 2x4+y2-x2-2y. 3 f = 8x -2x, and f x y = 2y-2.

For stationar

oints we have f =0 f = 0.

Therefore, 8x3-2x = 0 and 2y-2 = 0. Which implies points (0, 1), (1/2, 1), (-1/2, 1) 2-2, B = f = 0, C = f = 2. Now A = f = 24x xx xy yy

Case1: At the point (0, 1), AC-B2 = (-2)(2)= -4<0. Therefore (0. 1) is a saddle point.

Case2: At the point (1/2, 1), AC – B2 = (4)(2)-0=8>0 and A=4>0. Therefore (1/2, 1) is a minimum point and the minimum value of the function is -9/8. Case3: At the point (-1/2, 1), AC-B2 = (4)(2)-0= 8>0 and A=4>0. Therefore (-1/2, 1) is a minimum point and the minimum value of the function is -9/8.

6.Lagrange’s Method of Undetermined Multipliers: This method is more useful for finding the stationary points of the function subject to the given constraint with the introduction of unknown . , The main drawback of this method is that it gives no information regarding the nature of the stationary points. This fact is to be determined from the physical or geometrical consideration of the problem and which in some cases may be difficult to comprehend.

Let, u=f(x, y, z), be a function of three independent variables x, y, z which are connected by the relation g(x, y, z) = 0. [ g(x, y, z)=0 is called constraint.]. The Lagrange’s function is given by , ,

=

, ,

, ,

……………………

Where λ is the arbitrary constant. Then dF = 0 implies that, f x +λgx=0, f y +λgy=0, f z +λgz=0 ……………………(2) This procedure is only for one constraint.

If there are 3 constraints, then the Lagrange’s function is given by, F(x, y, z) = f(x, y, z)+λ1g1(x, y, z)+λ2 g2(x, y, z)+λ3g3(x, y, z) ………………………………………………...(3) Where, g1(x, y, z)=0, g2(x, y, z)=0,and gn(x, y, z) =0 are constraints. Then dF=0 implies that f x+λ1g1x +λ2g2x + λ3g3x = 0 ………………………………..(4) f y+λ1g1y +λ2g2y + λ3g3y = 0 ……………………………….(5) f z+λ1g1z +λ2g2z + λ3g3z = 0 …………………………………(6) Then solve the equations (4), (5) and (6) to find out the values of λ1, λ2 and λ3. Put the values of λ1, λ2 and λ3 in the equations (4), (5) and (6) to get the values of x, y and z. Finally put the values of x, y and z in the given function.

Working rule for the Lagrange’s Method of undetermined multipliers: Step1: Construct the Lagrange’s function[F(x, y, z)]. Step2: Find the differential of Lagrange’s function (dF). Step3: Consider, dF = 0 for stationary values. Here we w ge equa ons , an . Step4: Solve the equations (4), (5) and (6) to get the values of λ. Step5: Put the values of λ in equations (4), (5) and (6) to get the values of x, y and z. Step6: Finally put the values of x, y and z in the given function.

Example1: Find the minimum value of x2+y2+z2, given that ax+by+cz = p. Using Lagrange’s method. Solution: Let u = x2+y2+z2 and g(x,y,z) = ax+by+cz-p=0. ………………………..(i) The Lagrange’s function is given by F(x, y, z) = (x2+y2+z2)+λ(ax+by+cz-p) For stationary values, dF=0 gives, 2x+λa=0 ……………………………………………….(ii) 2y+λb=0 ………………………………………………(iii) 2z+λc=0 ……………………………………………….(iv)

Now , (ii).x+(iii).y+(iv).z implies that 2(x2+y2+z2)+λ(ax+by+cz)=0 => 2u+λp=0, using (i) =>λ=- 2u/p Putting this value in (ii), (iii) and (iv), we get x = au/p, y = bu/p, z = cu/p. Substituting these values of x, y, and z in the given function, we have u= u2(a2+b2+c2)/p2 => u = p2/(a2+b2+c2), neglecting the trivial case u = 0.

As we know that the perpendicular distance is always minimum. Therefore, the length of the perpendicular from origin o on the plane ax+by+cz=p is given by . Hence u(min) = p2/(a2+b2+c2).

O(0,0,0)

(x, y, z)

ax+by+cz=p

Example2: Find the maximum value of x2y3z4 given that 2x+3y+4z=a. using Lagrange’s method. Solution: The Lagrange’s function is given by F(x, y, z) = (x2y3z4)+λ(2x+3y+4z-a). For sationary values, dF = 0 gives z+ . …………………………………………….. 3x2y2z4+3λ = 0. …………………………………………….(ii) 4x2y3z3+4λ = 0. ……………………………………………..(iii) Now (i).x+(ii).y+(iii).z, implies that λ(2x+3y+4z) = -9x2y3z4 => λ = -9x2y3z4 /a, using 2x+3y+4z=a.

Partial Derivative

Recommend Documents