MATH 265 (MATHEMATICAL METHODS) · 2018. 8. 8. · @x = 2x + y @u @y = x + 2y 2 Find the partial...

MATH 265

(MATHEMATICAL METHODS)

Rev. Dr. W. Obeng Denteh

November 20, 2015

Rev. Dr. W. Obeng Denteh (KNUST) MATH 265(MATHEMATICAL METHODS) November 20, 2015 1 / 20

Outline

1 Partial Differentiation Of Function Several Variables

2 Differentiation of Implicit Functions

3 Theorem and Application

4 Jacobians

5 Differentiation of a Vector Function of several variables

6 The Tangent Vector

7 Curvilinear Coordinates

8 Multiple Surface and Volume Integral

9 Gradient, Divergence and Curls.

10 The theorems of Green, Gauss and Stokes.

11 Applications to physical and Geometrical problems.


Partial Differentiation Of Function Several

Variables

The Volume V of a Cylinder of radius r and height h is given by:

V = πr 2h

i.e V depends on two quantities r and h.The volume V increases if r is kept constant as h is increased.Considering the derivative of V with respect to h when r is keptconstant is written as

∂V

∂h= πr 2

For vice versa,the derivative of V with respect to r when r is keptconstant is written as

∂V

∂r= 2πrh


Therefore,∂V

∂h

and∂V

∂r

are called the Partial Derivatives of h and r respectively. V is thedependent variable whereas r and h are the independent variables.V is therefore a function of (x , y) which can be written asV = (x , y) The following can be defined from V = (x , y)∂v∂x

∂v∂y

∂2v∂x2

∂2v∂x2

∂2v∂y .∂x

∂2v∂x .∂y

And also, ∂2v∂y .∂x

∂2v∂x .∂y


Examples1 Given u = x2 + xy + y 2, find du

dxand du

dy.

Solution

∂u

∂x= 2x + y

∂u

∂y= x + 2y

2 Find the partial derivatives of the function, z = x3 + y 3 − 2x2ywith respect to the variables x and y.Solution

∂z

∂x= 3x2 − 4xy

∂z

∂y= 3y 2 − 2x2


Exercise 1

Find ∂z∂x, ∂z∂y

of the following functions:

1 z = (2x − y)(x + 3y)

2 z = 2x−yx+y

3 f (x , y , z) = xsin(y + 3z).

4 z = tan(3x + 4y)

5 z = sin(3x+2y)xy

6 Given V = ln(x2 + y 2), prove that ∂2v∂x2

+ ∂2v∂y2 = 0

7 find the first and second partial derivatives of the function:z = x3 + y 3 − 2x2y

8 If V = f (x2 + y 2), Show that x ∂V∂y− y ∂V

∂x= 0


Differenciation Of Implicit Functions

If y = x2 − 4x + 2, y is completely defined in terms of x , and y iscalled an explicit function of x .When the relationship between x and y is more involved, it may notbe possible to separate y completely on the left-hand side, eg.xy + siny = 2. In such a case as this, y is called an implicit functionof x because a relationship of the form y = f (x) is implied in thegiven equation.


Example

If x2 + y 2 − 2− 6y + 5 = 0, find dydx

and d2ydx2

at x = 3, y = 2.Solution.

2x + 2ydy

dx− 2− 6

dy

dx= 0

(2y − 6)dy

dx= 2− 2x

dy

dx=

2− 2x

2y − 6=

1− x

y − 3

∴ at(3, 2),dy

dx=

1− (3)

(2)− 3= 2

Thend2y

dx2=

d

dx

{1− x

y − 3

}=

(y − 3)(−1)− (1− x)dydx

(y − 3)2


Example Cont’.

=(3− y)− (1− x)dy

dx

(y − 3)2

at(3, 2)d2y

dx2=

(3− 2)− (1− 3)2

(y − 3)2=

1− (−4)

1= 5

∴ At (3, 2), dydx

= 2 and d2ydx2

= 5


Exercise 21 Find df

dxand df

dyof the following:

(i) f (x , y) = logy x(ii) f (x , y) =

∫ yx g(t)dt

(iii) f (x , y) = 2x2 − 3y − 4(iv) f (x , y) = sin2(x − 3x)

2 Find fy , fx and fz of the following:

(i) f (x , y , z) = (x2 + y2 + z2)−12

(ii) f (x , y , z) = e(x2+y2+z2)

3 Find the value of dzdx

at the point (1,-1,-3) if the equationxz + ylnx − z2 + 4 = 0 defines x as a function of the twoindependent variables y and z and the partial derivatives exists.


Kay_Dee

Comment on Text

df/dx = (1/ x lny) df/dx = - (ln |x| ) / y(ln |y|)^2

Theorem and Application

Clairaut’s TheoremThis theorem states that: Given f (x , y) such that its partialderivatives fx , fy , fxy and fyx are defined throughout an open regioncontaining a point (a, b) and are all continuous at (a, b) then;

fxy (a, b) = fyx(a, b)


Example

Verify Clairaut Theorem for f (x , y) = yln(4x + y 2)Solution

fx =4y

4x + y 2

fxy = (4x + y 2)4− 4y(2y)

=16x + 4y 2 − 8y 2

(4x + y 2)2

fyx = 2y 2(−1)(4x + y 2)−24 +4

4x + y 2


Application/Interpretation of Partial Differential

Equation

Geometrical Interpretation Of Partial Derivatives

Let z = f (x , y) be a surface.z = f (x , y0) is a curve (where y0 is a constant). This curvez = f (x , y) is obtained by the intersection of the surface z = f (x , y)with the plane y = y0. df

dxis the slope of the tangent line to the curve

z = f (x , y0).Example Given z = f (x , y) = x2

y3

(a) Find the slope of the tangent lines at (2,5)for

(i) y fixed and (ii) x fixed

(b) Determine whether z = f (x , y) is increasing or decreasing at(2,5) for

(i) y fixed and (ii) x fixed


Solution(ai) When y is fixed means differentiating f w.r.t x whiles y is kept

constant. Thus:df (x ,y)

dx= 2x

y3 = df (2,5)dx

= 2(2)53

= 4125

Hence the slope of the tangent line at (2,5) for fixed y is 4125

(aii) When x is fixed means differentiating f w.r.t y whiles x is keptconstant. Thus:df (x ,y)

dy= −3x2y−4 = −3x2

y4 = df (2,5)dy

= −3(2)254

= −12625


(bi) fx = 4125

> 0⇒ f (x , y)df (2,5)dy

= −12625

< 0 is increasing

(bii) Decreasing


Exercise 31 Given f (xy) = ex2y . Verify the Clairaut’s theorem

2 Find the slopes of the traces to z = 10− 4x2 − y 2 at the point(1,2).

3 Find the vector equations of the tangent lines to the traces toz = 10− 4x2 − y 2 at the point (1,2).

4 Verify Clairaut’s Theorem for the function f (x , y) = xe−x2y2


Jacobian

Implicit Differentiation Using JacobiansConsider the system

F (u, v , x , y) = 0

G (u, v , x , y) = 0

Where u = u(x , y) and v = v(x , y).To find ......... for the above system we proceed as follows:Find expressions for ... and... in a step by step manner and deduceexpressions for the others.Using implicit function differentiation for each of the equations in theabove system we have;

d

dx[F (u, v , x , y)] = Fx + Fuux + Fvvx = 0


Jacobian Cont’d

dx[G (u, v , x , y)] = Gx + Guux + Gvvx = 0


Differentiation of a Vector function of several

variables

Let (a, b) be x − y coordinates of a point on the surface z = f (x , y).Then the equation of the surface as a vector function is

ρ(x , y) =< x , y , z >=< x , y , f (x , y) >

The tangent vector for traces with fixed y is given asρx(x , y) =< 1, 0, fx(x , y) >. The tangent vector for traces with fixedx is given as ρy (x , y) =< 0, 1, fy (x , y) >.NB:Trace- The point at which a line or the curve in which a surfaceintersects a coordinate plane.


Cont’The equation for the tangent line to traces with fixed ρ is then

ρ(t) =< a, b, f (a, b) > +t < 1, 0, fx(a, b) >

Examples Find the vector equation of the tangent lines


The Tangent Vector

ExampleFind a tangent vector at the point where t = 3 on the vectorf (t) = e2t i + (t2 − 1)j + lntk

Solution

f ′(t) = 2e2t i + 2tj +1

tk

f ′(t)|t=3 = 2e6i + 6j +1

3k


Gradient, Divergence and Curls.

GradientThe gradient of a scalar point function f = f (x , y , z) is defined b:

Gradf = ∇f =

(i

d

dx+ j

d

dy+ k

d

dz

)= i

df

dx+ j

df

dy+ k

df

dz

Gradf , ∇f is a vector quantity.

df =df

dxdx +

df

dydy + k

df

dzdz

∇f =

(idf

dx+ j

df

dy+ k

df

dz

)· (idx + jdy + dzk)

= ∇f


= |∇f |cos θ

θ is the angle between ∇f and dr .The value of df is greatest when θ = 0, or cos θ = 1.This gives the gradient of f to ∇f .


Example

If φ = 3x2yy − y 3z2 , find ∇φ at the point (1,-2,-1).Solution.

∇f =

(idφ

dx+ j

dφ

dy+ k

dφ

dz

)= 6xy i + (3x2 − 3y 2z2)j + (−2y 3z)k

φ|(1,−2,1) = −12i− 9j− 16k


DivergenceThe divergence of a vector point function F is denoted by div Fwhere,

divF = ∇·F =

(i

d

dx+ j

d

dy+ k

d

dz

)· (F1i + F2j + F3k)

Clearly, div is a scalar. If div = 0, F is called Solenoidal.ExampleEvaluate div(3x2i + 5xy 2j + xyz3k) at the point (1,2,3).Solution

div(3x2i+5xy 2j+xyz3k) =

(i∂

∂x+ j

∂

∂y+ k

∂

∂z

)· (3x2i+5xy 2j+xyz3k)

= 6x + 10xy + 3xyz2

div(3x2i + 5xy 2j + xyz3k)|(1,2,3) = 6 + 20 + 54 = 80Rev. Dr. W. Obeng Denteh (KNUST) MATH 265(MATHEMATICAL METHODS) November 20, 2015 24 / 20

CurlThe Curl of a vector point function F is defined as: curlF = ∇× F .If F1i + F2j + F3k, then:

curlF = ∇× F

is given as (∂i

∂x+∂j

∂y+∂k

∂z

)(F1i + F2j + F3k)


Exercise1 Find the unit normal to the surface x2 + y 2 − z = 1 at P(1,1,1).

2 Let v = (x + y)i+ (y − 2z)j+ (x +λ)k. Find λ if v is solenoidal.

3 If v = x i+y j+zk√x2+x2+z2

, find λ· v


The Theorems Of Green, Gauss and Stokes.

Green’s TheoremLet C be a positively oriented, piecewise smooth, simple, closed curveand let D be the region enclose by the curve. If P and Q havecontinuous first order partial derivatives on D then,∫

C

Pdx + Qdy =

∫ ∫D

(∂Q

∂x− ∂P

∂y

)dA

There are some alternate notations to be acknowledged beforeworking some examples.When working with a line integral in which the path satisfies thecondition of Green’s Theorem, we will often denote the line integralas, ∫

Pdx + Qdy or

∫Pdx + Qdy


Green’s Theorem Cont’Both of these notations do assume that C satisfies the conditions ofGreen’s Theorem.Also, sometimes the curve C is not thought of as a separate curvebut instead as the boundary of some region D and in these cases Cmay be denoted ad ∂D.

Example

Use Green’s Theorem to evaluate∫C

xydx + x2y 3dy where C is thetriangle with vertices (0,0), (1,0),(1,2).

SolutionThe following inequalities will define the region enclosed.

0 ≤ x ≤ 1 0 ≤ y ≤ 2x

We can identify P and ! from the line integral. Here they are.

P = xy Q = x2y 3Rev. Dr. W. Obeng Denteh (KNUST) MATH 265(MATHEMATICAL METHODS) November 20, 2015 28 / 20

So the Green’s Theorem of line integral becomes,∫C

xydx + x2y 3dy =

∫ ∫D

(2xy 3 − x)dA =

∫ 1

0

∫ 2x

0

(2xy 3 − x)dydx

=

∫ 1

0

(1

2xy 4 − xy

)|2x0 dx


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MATH 265 (MATHEMATICAL METHODS) · 2018. 8. 8. · @x = 2x + y @u @y = x + 2y 2 Find the partial...

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