applications of 1st order diffential e

543

APPLICATIONS OF 1ST ORDER DIFFENTIAL

EQUATIONS

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8.1 INTRODUCTION

Use of differential equations is of great importance in engineering and science because manyphysical laws and relations appear mathematically in the form of differential equations.Among applications of first order differential equations, linear differential equations areused very frequently in solving problems related to electrical circuits, radio active decays, carbondating, population dynamics, mixture problems, Newton cooling etc.

To begin with, let us consider basic physical and geometrical applications that illustratesthe typical steps of modeling i.e. the steps which lead to form the physical situation or reallife problem to a Mathematical formulation and situation, and its interpretation.

8.2 GEOMETRICAL APPLICATIONS (CURVES)

(a) Cartesian Coordinates: Let the tangent and the normal at any point P(x, y) on thecurve f(x, y) = 0 meet the X-axis at T and N respectively. PM is the perpendicular toX-axis (Fig. 8.1), thenSlope of the tangent at P(x, y)

= ψ = = 1tandy

ydx

(i) Equation of the tangent at P is

− = −( )dy

Y y X xdx

...(1)

So that, the X intercept of the tangent,

OT = − dxx ydy

(putting Y = 0 in (1)) .. (2)

the Y intercept of the tangent,

′ = − dyOT y x

dx (putting X = 0 in(1)) ...(3)

Fig. 8.1

A s

BY

O

T´

T

ψ Subtang

Tang

ent

M NSubnormal

y

ψ

Normal

f x y( , ) = 0

P x y( , )

X

Engineering Mathematics through Applications544

(ii) Equation of the normal at P is

− = − −( )dxY y X xdy ...(4)

So that, the length of the tangent,2

cos 1 dxPT y ec ydy

= ψ = +

the length of the normal, 2

sec 1dy

PN y ydx

= ψ = +

the length of the sub-tangent, cot dxTM y ydy

= ψ =

the length of the sub-normal, tandy

MN y ydx

= ψ =

the derivative of the arc length, 22

1 , 1dyds ds dx

dx dx dy dy

= + = +

(b) Polar Coordinates: Let P(r, θ) be any point on the curve, r = f(θ), then

(i) ψ = θ + φ

(ii) θφ =tan drdr

, p = rsinφ

(iii) Polar sub-tangent, θ= 2 dOT rdr

(iv) Polar sub-normal, =θ

drONd

(v) Perpendicular from the pole on the tangent,

2

2 2 41 1 1 drp r r d

= + θ(vi) Derivative of the arc length,

2 2

21 ,ds d ds drr rdr dr d d

θ = + = + θ θ

Example 1: Find the curve whose sub-tangent is twice the abscissa of the point of contactand pass through the point (1, 2).

Solution: Given 2dxy xdy

= implying =2dy dxy x

On integration, 2 logy = logx + logc or y2 = cx ...(1)

Now the curve pass through (1, 2) implying c = 4 ... (2)

Fig. 8.2

N O T

Pr(,

)θNorm

al

φSubnormal

Tangent

Yr

pSubtangent

Aθ

S

ψφ

X

Applications of 1st Order Differential Equations 545

Hence the required curves, y2 = 4xExample 2: Show that the curve in which the portion of the tangent included betweencoordinate axes, bisected at the point of contact is a rectangular hyperbola.

Solution: Let the tangent at some general point P(x, y) cut the axis at T and T' (Fig. 8.3).

Its X intercepts, dxOT x ydy

= − and Y intercept ′ = − dyOT y x

dx;

then the coordinate of T and T’ are , 0 , 0,dydxx y y x

dy dx − −

respectively.

As P is the middle point of TT’, therefore,

0

2

dxx ydy

x

− +

= or − = 2dxx y xdy

Implying xdy + ydx = 0 i.e. d(xy) = 0

On integration, xy = c is the required equation ofrectangular hyperbola.

Example 3: Find the curve whose tangent cut off intercept on coordinate axes, the sum ofwhich is 'a'.

Solution: See fig.8.3, the X intercept, = − ,dxOT x ydy

the Y intercept, ′ = − ,dy

OT y xdx

Given that − + − =( ) ( )dydxx y y x a

dy dxor − + − =y

x y xp ap

, where dy

pdx

=

− = − +( 1)

( 1)p

y p x ap

implying, 1

py px a

p= +

−

Which is a Clairaut’s equation. Hence its solution is = +− 1acy cx

c.

Example 4: Find the curve for which the polar sub-tangent is equal to the polar sub-normal.

Solution: Given condition, θ =θ

2 d drrdr d

⇒ =θ

drrd

On integration, log r = θ + logc i.e. r = ceθ

Example 5: Obtain the curve for which the normal makes equal angle with the radiusvector and the initial line.

P x y( , )

Y

T´

O T XFig. 8.3


Solution: Let PT and PN be the tangent and the normal at some general point P(r, θ) of the

curve (fig.8.4), so that θφ =tan drdr

.

But by the given conditions, ∠ONP = ∠OPN = (90 – φ)∴ θ = ∠PON = (180 – (∠ONP + ∠OPN))°

= (180 – (180 – 2φ))º = 2φ

orθ = φ2

there by implying

θ θ= φ =tan tan2

drdr

or cos

2sin

2

dr dr

θ

= θθ

On integration, log 2log sin log2

r cθ= +

2 1sin (1 cos )2 2

r c cθ= = − θ , the required equation of the cardioid.

Example 6: Determine the curve for which angle between tangent and the radius vector istwice the vectorial angle.

Solution: See the general fig. 8.2, given φ = 2θ

Further, θφ =tan drdr

implying θθ =tan 2 drdr

Rewrite as θ θ=θ

sin 2cos 2

drdr

or θ= θ θ

1 cos222 sin 2

dr dr

On integration, 2 log r = log sin2θ + log c or r2 = a2sin2θ, c = a2

Example 7: Obtain the equation of the curve for which the angle between the radiusvector and the tangent is the supplement of half the vectorial angle.

Solution: See fig. 8.2, as a particular case when the given condition is θφ = π −2

implying, θ θ φ = π − = − tan tan tan

2 2

tan2

drdrθ θ= − or

θ− = θθ

cos2

sin2

dr dr

On integration, log log 2 log sin2

c r θ− =

Fig. 8.4

T φ

P r( , )θ

O

θ

N X

r

θ = 0


θ= 2sin2

cr

i.e. = − θ2 (1 cos )cr

, which is a parabola

Example 8: Find the equation of the curve in which perpendicular from the pole uponthe tangent at any point is λλλλλ times the radius vector of the point.

Solution: Let P(r, θ) be any general point on the curve r = f(θ), then by the givencondition, p = rλ ...(1)

Where p is the perpendicular distance of the tangent at P(r, θ) from the pole.

From the pedal equation, 2

2 2 41 1 1 drp r r d

= + θ ...(2)

Implying 2

2 2 2 41 1 1 dr

r r r d = + λ θ

or22

22

r drrd

− = λ θ

22 2

2

(1 )r drd

− λ = λ θ or21dr d

r− λ= θλ

On integration, − λ= θ +λ

21log logr c or−λ θλ=

21

r ce

Observation: This curve represents an equiangular spiral for λ = sinα.

Miscellaneous Problem

Example 9: Elaborate the shape of a reflector such that light coming from a fixed source isin parallel rays.

Solution: Let the origin be the fixed source of light and the reflected rays; so that X- axisparallel to the reflector will be a surface generated by revolution of the curve f(x, y) = 0about X-axis (Fig. 8.5).From geometry, if TPT’ is the tangent at P(x, y), then the angle of incidence is equal to theangle of reflection viz.

φ = ∠OPT = ∠P'PT’ = ∠OTP = ψ

Further, = ∠ = φtan tan 2 ,dy

XOPdx (as external ∠XOP = φ + ψ = 2φ)

2 2

22 tan1 tan 1

pyx p

φ= =− φ −

or = −2y

x ypp

,

which is solvable for x(y) ...(1)

On differentiating above equation with respect to y, = − − −2

2 1 dp dpyp y

p p p dy dy

Y

T

T´

φ

O

ψX

P´P x y( , )

ψ

Fig. 8.5


1 1 1 0dp

p p yp p p dy

+ + + =

or 1 1 0dpy

pp p dy

+ + =

From above, = −dp dyp y

or logp = log c – logy or = cpy

... (2)

Elimination of p from (1) and (2) results in, y2 = 2cx + c2. This shows that reflector is amember of the family of paraboloids z2 + y2 = 2cx + c2

ASSIGNMENT 1

1. Determine the curve in which the length of the sub-normal is proportional to thesquare of the ordinate.

2. Find the curve which pass through the origin and is such that the area included betweenthe curve, the ordinate and the x-axis is twice the cube of the ordinate.

3. A curve is such that the length of the perpendicular from the origin on the tangent atany point P of the curve is equal to the abscissa of P. Prove that the differential equation

of the curve is − − =2 22 0dy

y xy xdx

, and hence find the curve.

4. A plane curve has the property that the tangents from any point on the y-axis to thecurve are of constant length ‘a’. Find the differential equation of the family to whichthe curve belongs and hence obtain the curve.

5. Find the curve whose(i) Polar sub-tangent is constant.

(ii) Polar sub-normal is proportional to the sine of the vectorial angle.6. Find the curve for which the tangent at any point P on it, bisects the angle between the

ordinate at P and the line joining P to the origin.7. Find the curve for which the tangent, the radius vector and the perpendicular from

the origin on the tangent form a triangle of area kr2.8. Find the curve in which the length of the arc measured from a fixed point A to any

point P is proportional to the square the abscissa of P.

8.3 ORTHOGONAL AND ISOGONAL TRAJECTORIES

Definitions

(a) Family of curves: The equation F(x, y, c) = 0, where c a parameter is called a family ofcurves as for different values of c, we get different curves having the same property.

(b) Trajectory: Trajectory of a family of curves is the curve which intersects every memberof the other family according to a certain rule.

Orthogonal Trajectory: An orthogonal trajectory of a family of curves is the curve whichmeets each member of the other family at right-angles (fig.8.6)

In other words, two families of curves are called orthogonal trajectories of each other if every memberof either family intersects each member of the other family at right-angles.


OX

Y

P

O

Y

X

Fig. 8.6 (a) Fig. 8.6 (b)

Note: When each member of a family cuts every other member of the same family orthogonally, then thegiven system is self orthogonal.

Orthogonal trajectories are of immense importance in the study of certain practical problemsrelated to plane vector fields such as electric, magnetic, fluid flow and heat-flow fields. Forexample, it is observed that the lines of equal potential, due to distribution of steady currentflowing in a homogenous medium, intersects the lines of current flow at right angles. Againin the steady of fluid flow, the lines of flow (stream lines) are perpendicular to the lines ofconstant velocity potential (equipotential lines or contour lines). Thus, with the application oforthogonal trajectories, the lines of level can be obtained from the lines of flow or vice-versa. Likewise, the lines of heat flow for a body are perpendicular to the isothermal curvesor loci of points at the same temperature.

Isogonal Trajectories (Oblique Trajectories): Lines intersecting all the curves of the givenfamily at a constant angle are called isogonal trajectories. In other words, oblique trajectoriesof a family of curves are the curves which cut every member of the given family at certainangle other than 90°.

If = = φtandy

mdx

is the slope of the tangent to a member of the given family, then the

slope m' = tanψ of the trajectory is given by ( ) φ − α′ = ψ = φ − α =+ φ α

tan tantan tan1 tan tan

m or m and

m' are related by the relation tan constant1m m

mm− ′ = α = + ′

Remarks: For trajectories, only those curves which have continuously turning tangent lines are considered. Inthe neighbourhood of any point on such a curve, the curve can either be represented by y = f(x) or x = φ(y)where f(x) and φ(y) are functions with continuous derivatives. For this purpose, a portion of the curves lying ina particular region A : a ≤ x ≤ b, α ≤ y ≤ β is considered.Working Rule: Finding Orthogonal Trajectories of the Family of Curves f (x, y, c) = 0

(i) Form the differential equation of the family by eliminating c and let it be φ =

, , 0dy

x ydx

(ii) In this differential equation, replace dy dxbydx dy

− so that the differential equation of

the desired family is , , 0dxx ydy

φ − =

(iii) Solve this differential equation to find the family of orthogonal trajectories.


Working Rule: Finding Oblique (Isogonal) Trajectories of the Family of Curves f(x, y, c) = 0.

Form the differential equation of the family by eliminating c and let it be , , 0dy

x ydx

φ =

Replace, in this differential equation tan

by1 tan

dydy dx

dydxdx

+ α

− α so that the differential equation

of the desired family is

+ α φ =

− α

tan, , 0

1 tan

dydxx y dy

dxIntegrate the above differential equation to get required equation of oblique or isogonal

trajectories.

Working Rule: For Finding the Orthogonal Trajectories of the Polar Curve f(r, θθθθθ, c) = 0

(i) By eliminating c form the differential equation of the given family as φ θ = θ , , 0drr

d

(ii) Replace in this differential equation, θ−θ

2bydr drd dr

so that the differential equation of

the desired family is θ φ θ − = 2, , 0dr r

dr

Fig. 8.7

As for any pt. P(r, θ) on the given system of curve (see Fig. 8.7), θφ =tan drdr

, where as in the orthogonal system,

( )2 11 1tan tan 90 cot drd r drdr

°φ = φ + = − φ = − = −θ θ Thus, for getting the differential equation of the orthogonal

system, θdrdr

is to be replaced by −θ

1 drr d

or θ

drd

is to be replaced by θ− 2 drdr


(iii) Solve the above differential equation to obtain the equation of orthogonal trajectories.

Example 10: Electricity steams across a flat plate are following the curves of the family ofparabolas y = ax2. Of what nature are the equipotential lines?

[NIT Kurukshetra, 2006; JNTU, 2006, 03; KUK, 2003-2009]

OrFind the orthogonal trajectories of the system of parabolas y = ax2.

Solution: The given equation is y = ax2 … (1)On differentiating it with respect to x, we get

12 or

2dy dy

ax adx x dx

= = … (2)

Substituting this value of the parameter ‘a’ in equation (1), we get

= =

2 21 or2

dy dy yy x

x dx dx x … (3)

which is the differential equation of the given family of system.

Now, for finding the differential equation of the orthogonal system, replace −bydy dxdx dy

in (3).

∴ − = + =2or 2 0

ydx xdx ydydy x

… (4)On integrating (4), we have

+ =22

2

2 1yx c or

( )22

2 21

2

yxcc

+ =

This equation is the desired family of curves representing asystem of ellipses

Thus, the equipotential lines (the lines of level) of the given parabola are the family of ellipses withcentres at the origin, semi-major axis 2c along the x-axis, semi-minor axis c along y-axis and the

eccentricity 12

, as shown in figure.

Note: The graph of the given family, above the initial axis is for positive values of ‘a’, where as below initial axisis for negative values of a.

Example 11: Find the orthogonal trajectories of the family x2 + y2 + 2λλλλλy + c = 2, λλλλλ beingthe parameter. [KUK, 2002]

Solution: Differentiating the curvex2 + y2 + 2λy + c – 2 = 0 …(1)

Fig. 8.8

OX

Y


we get, 2x + 2yy1 + 2λy1 = 0, 1dy

y =dx

i.e.,1

1

( )x yy

y

+λ = − …(2)

Substituting the value of λ in equation (1), we get

− +

+ + + − =12 2

1

( )2 2 0

x yyx y y c

y …(3)

Changing y1 to −1

1y

in (3), we get

+ + − + =

2 2

11

2 0y

x y y x y ky

x2 + y2 + 2xyy1 – 2y2 + k = 0

+ + − =2 2( ) 2 0dy

x k xy ydx

or−

+ + =2

2 2( ) 0

xydy y dxx k

dx

−+

+ =22

2 2

2( )0

xydy y dxx kdx

x xor

+ + =

2

21 0

yk dx dxx

On integrating both sides, we get

− + =

2ykx Kx x

which is the desired orthogonal family.

Example 12: Find the orthogonal trajectories of the confocal conics 22

2 21,

yxa b

+ =+ λ + λ

λλλλλ

being the parameter. [KUK, 2006; NIT Kurukshetra, 2010]

Solution: The given family is 22

2 21.

yxa b

+ =+ λ + λ

…(1)

Differentiating (1) with respect to x, + =+ λ + λ

12 2

22 0yyx

a b x(b2 + λ) + yy1(a2 + λ) = 0

λ(x + yy1) = –(b2x + a2yy1) ⇒2 2

1

1

( )

( )

b x a yy

x yy

+λ = −

+ …(2)

Now, + −

+ λ = − =+ +

2 2 2 212 2

1 1

( ) ( )( )

( ) ( )b x a yy a b x

a ax yy x yy

…(3)


SImilarly −

+ λ = −+

2 212

1

( )( )

( )

a b yyb

x yy…(4)

On using results (3) and (4) in (1), we get

+ +

+ =− − −

1 12 2 2 2

1

( ) ( )1

( ) ( )

x x yy y x yy

a b a b y

i.e. +

− = − 1

2 21

( )1

( )

x yy yx

ya b …(5)

Equation (5) is the differential equation of the given family.

In order to find the differential equation of the desired family, replace y1 by −1

1y

in (5)

i.e. [ ]

−

+ =−

112 2

1( )

yx

yx yy

a b…(6)

Clearly equations (5) and (6) are the same. Hence the given system is self orthogonal.

Example 13: The electric lines of force of two opposite charges of the same strength at(± 1, 0) are circles (through these points) of the form x2 + y2 – ay = 1. Find their equipotentiallines.

Solution: On differentiating the given family x2 + y2 – ay = 1 …(1)

+ − =2 2 0dy dy

x y adx dx

…(2)

On eliminating ‘a’ from (2), we get

2 2(1 )

2 0dy x y dy

xdx y dx

− −+ + = …(3)

Which is the differential equation of the given family.

Replace bydy dxdx dy

− in (3), we obtain

+ −

− + =2 2( 1)

2 2 0x ydx dxx y

dy y dy

i.e., + −

− + =2 2( 1)

2 2 0x ydx dxx y

dy y dy

i.e. 2xydy + (x2 – y2 –1)dx = 0 …(4)

which is comparable to Mdx + Ndy = 0, where M = x2 – y2 – 1, N = 2xy

Here, we see that − − −

= = − =2 2 2 ( )

2y xM N y y

f xN xy x


∴ −∫∫= = =

2( )

2

1. .dxf x dx xI F e e

xNow multiplying throughout by I.F. to equation (4), we get

− − + =

2

2 211 2 0

y ydx dy

xx x

Hence is solutions is

− − = ∫

2

2 211

ydx c

x xi.e.

2 1yx c

x x+ + =

x2 + y2 + cx + 1 = 0 or− − − = +

2 22( )

( 0) 12 4

x c cy …(5)

Hence, equation (5) represents the desired family of circles orthogonal to the given family.

Observations: The equation of the given family may be written like

( )2 22( 0 ) 12 4

a ax y− + − = + …5(a)

i.e. a family of circle with centre ( )0 ,2

a and radius

21

4a+

With x = ± 1, equation 5(a) becomes, y2 – ay – 1 = 0For a = 0, y = ±1. r = 1.

For a = 1, y = 1.615, –0.615,5

2r =

For a = 2, y = 2.414, –0.414, 2r = and so on.Thus, we see that for various value of ‘a’ the circles through

(x = ±1), get their centre shifted with an increases in value ofradius r.

Orthogonal trajectories of polar curves

Examples 14: Find the family orhtogonal to family F1 of cardiods r = a(1 + cosθθθθθ).[NIT Jalandhar, 2004; JNTU, 2006; KUK, 2008]

Solution: Given curve is r = a(1 + cosθ) …(1)Differentiating it, dr + asinθdθ = 0 …(2)On eliminating the parameter 'a' we get,

θ+ θ =+ θsin 0

1 cosrdr d or

1 sin1 cos

drr d

− θ=θ + θ …(3)

This is the differential equation of the direction field D1 for the given family F1.

To find the differential equation of the direction field orthological to D1, we replace θ

drd

by 2 drdrθ− as evident from the geometry.

Fig. 8.9


Where φ2 and Ψ2 are the angles for the orthogonal trajectory corresponding to the angles φ1and Ψ1 for the given family (Fig. 8.7).

Clearly, 2 1 2 1 1, whence tan tan cot2 2π π φ = φ ± φ = φ ± = − φ

2 1

1d drrdr r dθ = − θ

Whence from (3), the differential equation for the orthogonal trajectories becomes

sin tan

1 cos 2drdrθ θ θ= =

+ θ …(4)

which is a case of variable-separable, and on integration gives

log 2log sin log 22

r cθ= +

This is the equation of the orthogonal family F2.Since r = c(1 – cosθ) represents the same curve as r = c(1 + cosθ), the member of F2 with

label c is the same as the member of F1 with label a = –c. Thus the given family is self-orthogonal.

F a1 for > 0F c2 for > 0

Fig. 8.10

Remarks: Family F1 of this example is not a one-parameter family in any rectangle. However, if we restrict theparameter ‘a’ to only positive value, then F1 is a one parameter family in any rectangle R not enclosing pointswith θ = π. The parameter values c in F2 are then also restricted to positive values, and F2 is a one-parameterfamily in any rectangle R not enclosing points with θ = 0.

Example15: Find the orthogonal trajectory of the family of curve 2

cos ,krr

+ θ = α ααααα being the parameter. [NIT Kurukshetra, 2002, 2006]

Solution: r = f(θ), is the given curve 2

cos ,krr

+ θ = α … (1)


Differentiating with respect to θ, we get

2 2

2cos sin 0dr k dr kr

d r d r − θ − + θ = θ θ … (2)

In order to have the differential equation of the desired family of curves, change

2 to dr drd dr

θ−θ

in equations (2)

( ) 22 2 cos sin 0d kr k r

dr rθ − + θ − + θ =

( )

2

2 2cot

krrd dr

r k

+ θ θ =

− + ... (3)

which may be re-written as

( )( )

2 2

2 21cot

k rd dr

r k r

+θ θ =

−or ( )2 2

cos 1 2sin

rd drr k r

θ θ = + θ −

… (4)

(variable separable form)Integrating both sides w.r.t. θ, we get

2 22log sin log rr dr

k rθ = +

−∫ … (5)

Put k2 – r2 = z ⇒ 2rdr = – dz

∴ 1log sin log logr dz cz

θ = + − +∫

⇒ log sin log rcz

θ = or ( )2 2sin rc

k rθ =

−…(6)

which is the desired system of family of curves.

Example 16: Find the orthogonal trajectories of the curves r2 = a2cos2θθθθθ. [VTU, 2003]Solution: Given r2 = a2cos2θ … (1)

On differentiating (1), we have

2 sin 2drr ad

= − θθ … (2)

Eliminate 'a' from (2), using (1)

2 tan 2drr rd

= − θθ

θ π = θ = 0

r a2 = cos22 θ

θ = 45

r c2 = sin22 θ

Fig. 8.11


i.e. tan 2dr rd

= − θθ

… (3)

which is the differential equation of the given family.

To obtain the differential of the trajectories, replace 2 by dr drd dr

θ−θ

in equation (3).

i.e. 2 tan 2 or cot 2d drr r ddr rθ− = − θ θ θ = … (4)

Integration of (4), gives the equation of the desired system.

∴ log sin 2log log

2c r

θ + = i.e. c2sin2θ = r2 … (5)

Problems on Isogonal Trajectories

Example 17: Find the isogonal trajectories to a family of straight lines y = cx that cut thelines of the given family at an angle ααααα, the tangent of which equals k.

Solution: The equation of the family is y = cx … (1)

⇒ or [using(1)]dy dy y

cdx dx x

= = … (2)

which is the differential equation of the given family.In order to obtain the differential equation of the desired family,

replace dydx

by tan

1 tan

dydx

dydx

+ α

− α in (2),

we get,

tan

1 tan

ydy x

ydxx

+ α=

− α … (3)

or 1

ykdy x

ydx kx

+=

−

It is a homogenous equation, take y = vx

( )

( )21

or 1 1

k vdv v k dvx v xdx vk dx kv

++ + = = − −

⇒ ( )( )2

11

kv dxdvk v x

−=

+

Y

X

Fig. 8.12


Integrating both sides, we get

2 21 1 1 2

1 2 1v dxdv dv a

k v v x− = +

+ +∫ ∫ ∫

( )1 21 1tan log 1 log log2

v v x ak

− − + = +

− + − = + =

12 2 2

12

1 tan log log log putting y x y y

x a vk x x x

⇒ 2 21 arc tan logy

a x yk x

= + …(4)

which is the desired family of curves.In order to precisely define what curve equation (4) represents, we change it to polar

coordinates. We take 2 2tan , ,y dy

x y rx dx

= = φ + = equation (4) reduces to

1log log or kr C r Cek

φ= φ + = ... (5)

Whence, the family of isogonal trajectories of y = cx is the family of logarithmic spiral.

Example 18: Find the isogonal trajectories of the family of circles x2 + y2 = a2 whichintersects at 45°.

OR

Determine the 45° trajectories of the family of concentric circles x2 + y2 = a2

Solution: The given equation of the family is x2 + y2 = a2 … (1)

Differentiating it, we get 0dy

x ydx

+ = ... (2)

which, being free from the parameter ‘a’ is the differential equation of the given family.

In order to determine the differential equation of the desired family, replace dydx

by

0

0

tan 45

1 tan 45

dydx

dydx

+

− in equation (2),

i.e.

10 or

1

dydy x ydxx y dy dx x y

dx

+ ++ = = −−

… (3)

Equation (3) is a homogeneous differential equation of 1st order 1st degree.

Put y = vx so that dy dvv xdx dx

= + … (4)


Therefore, equation (3) becomes dv x vxv xdx x vx

++ =−

i.e. 2

21 1or1 1

dv v v dxx dvdx v v x

+ − = = − + (case of variable separable)

On integration, we get 2 21 1 2 1

1 2 1vdv dv dx

v v x− =

+ +∫ ∫ ∫

⇒ ( )1 21tan log 1 log log2

v v x C− − + = + or ( )1 2tan log . 1v Cx v− = +

i.e. ( )1 2 2tan logy

C x yx

− = +

ASSIGNMENT 2

1. Prove that the system of Confocal and Co-axial Parabolas y2 = 4a(x + a) is selforthogonal.

2. Find the orthogonal trajectories of the series of hyperbolas xy = k2 (or xy = c)

3. Find the orthogonal trajectories of the series of hypocycloids, 2 2 23 3 3x y a+ =

4. Find the orthogonal trajectories of the family of confocal conics 22

2 21,

yxa b

+ =+ λ

where λ is the parameter. [KUK, 2003-04, 06]5. Find the family of curves orthogonal to the family of circles

x2 + y2 – cx = 0, centre , 02c

and y = 0 for x = 0 (fig. 8.13).

6. Find the orthogonal trajectories of the cardiod r = a (1 – cosθ).[KUK, 2005; NIT, Jalandhar, 2005]

7. Find the orthogonal trajectories of the family of curvesrn = an sin nθ [KUK, 2001]

8. Find the orthogonal trajectories of the family of confocal

and co-axial parabolas 21 cos

ar =+ θ

[Osmania, 2003]

9. Find the orthogonal trajectories of (i) 1a

rθ =

+ θ (ii) rn sin nθ = an

8.4 SIMPLE ELECTRIC CIRCUIT

The formulation of differential equations in an electric circuit is governed by Kirchhoff’slaws which are being of immense importance.

FIRST LAW: The algebraic sum of the voltage drop in a closed circuit is equal to the resultantelectromotive force in the circuit.

Y

X

Fig. 8.13


SECOND LAW: Algebraic sum of the currents flowing into any node is zero, if q(t) be thecharge at a time t on a condenser in a circuit containing a resistance R, an inductance L andcapacitance C in series, a known e.m.f. (electromotive force) E(t) is impressed across thecircuit. The magnitude of q, E, L, R, C are taken in some physically consistent set of units, letthese units be as Coulombs, Volts, Henrys, Ohms and Farads respectively. Then in differentacquisition of the above mentioned physical invariants, we generally, come across knownfundamental sets of equations. A simple Electric circuit is called L-R circuit, viz.

ordi di R ERi L E idt dt L L

+ = + =

Then comes LC and LCR circuits, which are discussed under oscillatory electric circuits insuccesseding sections.

Example 19: Find the current in the simple circuit with C = ∞∞∞∞∞ (condenser absent) andEt = E0 sin wt (if initially there is no current in the circuit) [KUK, 2010]

Solution: By Kirchhoff’s First Law, we know that the governing equation of the currentflow in an L-R circuit:

diL Ri Edt

+ = or 0 sin ,Edi R i wt

dt L L+ = when E = E0sin wt …(1)

This is a Leibnitz’s linear equation with integrating factor, . .R Rdt tL LI F e e∫= =

Whence solution, 0( ) sinR Rt tL L

Ei t e e wtdt C

L= +∫

( )0 2 2 2sin cosR t

L R wt wL wtE e CR w L

−= ++

or ( ) ( )( )0 2 2 2

sin cos R tL

R wt wL wti t E Ce

R w L

−−= ++

…(2)

when t = 0, i = i0 implying ( )0

0 2 2 2

E wLi C

R w L= − +

+ …(3)

and thus, ( ) ( )( ) ( )

00 02 2 2 2 2 2

sin cos R tL

R wt wL wt E wLi t E i e

R w L R w L− −= + + + + …(4)

Equation (4) can be put in a more useful form as follows:

Let φ be that acute angle for which tan wLR

φ = , … (5)

then 2 2 2

cos RR w L

φ =+

and 2 2 2sin wL

R w Lφ =

+… (6)

And (4) may be written as:

Fig. 8.14

R

L

E


( ) ( )0 00 2 2 22 2 2

sinR tL

E E wLi t wt i e

R w LR w L− = − φ + + ++

… (7)

which gives the current at any time t.

Observations: We see that the current is a sum of two terms:

( ) ( )

( )

02 2 2

00 2 2 2

sinS

R tLT

Ei t wt

R w LE WL

i t i eR w L

−

= − φ +

= + + … (8)

In (7), as t increases indefinitely, the exponential term willapproach zero and after some time the current i(t) will executenearly harmonic oscillations only (see Fig. 8.15).

Thus, it is clear that after sufficiently long span oftime, the second term is very small and negligible in comparison with the first term. We calliS the steady-state and iT the transient current.

If ( ) ( )0 00 2 2 2 2 2 2

sin 0SE wL E

i iR w L R w L

= − = −φ =+ + … (9)

i.e. i0 is simply iS(t) at t = 0, then there is no transient current. Thus, the transient is due to thefact that the initial value that we have assumed for the total current does not agree withsteady-state value. We can see from (8), that there is no transient current, if L = 0, whichcorresponds to idealized condition in which the circuit is inertia-less and can jump from thearbitrary initial value i0 to its steady-state value iS(0). This reflects the fact that if L = 0, then(1) is not a differential equation and assigning an initial value to it is meaningless.

The steady-state current is a pure harmonic oscillation of the same circular frequency ω(no. of oscillations per 2π seconds) as the electromotive force (e.m.f). Its ‘amplitude’ is

02 2 2

ER w L+ , whereas it would be 0E

R if no inductance were

present. Thus, the inductance L adds an effective resistanceto the (ohm) resistance R and this resistance depends on the

frequency ω. The total resistance 2 2 2R w L+ is called the

‘impedance’. The steady-state current and voltage are out ofphase by the phase angle, φ. The e.m.f. has its maximum

‘amplitude’ at 3 5, , , ....

2 2 2t

w w wπ π π= whereas the current has

maxima at

( ) 3 5, , , ,2 2 2

wtw w wπ π π− φ = − − − i.e. when / ,

2t wπ = + φ 3 / ,

2wπ + φ …

In other words the current lags behind the voltage by a time .wφ

The simplest and most important e.m.f. is a simple sine wave, in many circuit problems itis necessary to consider more general e.m.f.’s and even discontinuous one. One of the types,

Exponential term In ( )i t

2π4π

π 3π 5π ωt

i t( )

Fig. 8.15

E

E0

t0 t

Fig. 8.16


for example, could be generated by connecting a battery in the circuit and closing or openingthe switch. If the battery has a constant e.m.f. of E0 and the switch is closed at time t = t0after the initial time t = 0, then the graph of E appears as shown.

Example 20: The equation of an electromotive force in terms of current i for an electric

circuit having resistance R and a condenser of capacity C in series is iE Ri dtC

= + ∫ . Find

the current at any time t, when E = E0 sin wt.[NIT Kurukshetra, 2007; PTU, 2006; KUK, 2002, 2003-04]

Solution: The given equation can be written as

0 siniRi dt E wtC

+ =∫ (³ E = E0 sinwt) … (1)

Differentiating both sides of equation (1) w.r. to t,

0 cosdi iR E w wtdt C

+ = or 0 cos ,wEdi i wt

dt RC R+ = …(2)

The equation (2) is Leibnitz’s form,1

. .tdt

RC RCI F e e∫= =

Therefore, the solution of the equation (2) is

0( ) cos .t t

RC RCwE

i t e e wt dtR

= ∫

0

22 2

1 cos sin ,1

tRCwE e wt w wt k

R RCwR C

= + + + k is constant of integration

or ( )02 2 2

cos sin ,1

tRC

E wCi wt RCw wt k e

R C w−

= + ++

is required current.

Note: If conditions are given, k can be evaluated.

Example 21: In a circuit containing resistance R and inductance L, the voltage E and the

current i are connected by the equation diE Ri Ldt

= + . Given that L = 640, R = 250,

E = 500 and i = 0 when t = 0. Find the time that elapse, before it reaches 90 percent of itsmaximum value.

Solution: The given equation can be written as di R Eidt L L

+ = ... (1)

It is Leibnitz’s linear equation with . .R tR dt

L LI F e e∫= =

∴ R t R t RtL L LE Ei e e dt A e A

L R= + = +∫ (A is an arbitrary constant) ... (2)


Initially, when t = 0, i = 0

From (2), 0 EAR

= + or EAR

= −

∴ . 1Rt RtL LEi e e

R

= −

or 1

RtLEi e

R−

= −

As t → 0, 0 andRtL Ee i

R−

→ → which is the maximum value of i(t).

If t1 is the time required to reach 90% of maximum value,

then 19 1

10

R tLE E e

R R

− = −

or

1 1 ;10

R tLe

−=

1

110 or log 10R t

Le

RteL

= =

Implying 164log 10 log 1025e e

LtR

= =

Example 22: Discuss the growth of current of zero initial value in a single circuit containinga resistance R, an inductance L and an electrostatic force E sinwt and no capacitor. In theabove equation, show that if i = 0 at time t = 0, then

( )2 2 2

( ) sin sinR tLEi t wt e

R w L

− = − φ + φ +

, where tan LwR

φ = [KUK, 2004-05]

Solution: Since there is no capacitor in the circuit, the governing equation is

sindiL Ri E wtdt

+ = i.e. sindi R Ei wtdt L L

+ =

On solving the above linear differential equation, we have

= +∫( ) sinR t R tL LEi t e e wt dt k

L

12 2

1on using sin sin tanax ax be bx dx e bxaa b

− = − + ∫

( )

22

2

sin ,

R tLeE wt k

L R wL

= − φ ++

where tan LwR

φ = and

k is an arbitrary constant.

Initially when t = 0, i = 0


∴ ( )2 2 2 2

22

10 sin sin( )E Ek kL R R w Lw

L

= + −φ = + −φ++

Eliminating k and simplifying, we get the required answer.

Note1: i has the same period π2

w as the periodic impressed e.m.f.

Note2: Maximum current 2 2 2

ER w L

=+

(from the value of i), where 2 2 2R w L+ is called impedance and E

is the maximum e.m.f. Hence maximum Maximum e.m.f.

currentimpedamce.

=

Example 23: A condenser of capacity C is discharged through the inductance L and aresistance R in series and the charge q at any time t satisfies the equation

2

20

d q dq qL R

dt dt C+ + =

Given that L = 0.25 Henry, R = 2.50 Ohms, C = 2 × 10–6 Farads and that when t = 0 the

change q is 0.002 coulombs and the current 0dqdt

= . Obtain the value of q in terms of t.

[KUK, 2002]

Solution: For the given values of L, R and C, we get

2

62

1000 2 10 0.d q dq

qdt dt

+ + × =

The auxiliary equation is 2 61000 2 10 0D D+ + × = implying 500 1323D i= − ±

∴ The solution is ( )500 cos1323 sin 1323tq e A t B t−= + … (1)Initially when t = 0, q = 0.002 and, hence A = 0.002. ... (2)Differentiating (1),

( ) ( )500 500500 cos1323 sin1323 1323 sin 1323 cos1323t tdqe A t B t e A t B t

dt− −= − + + − +

When 0, 0; 0 500 1323 Bdq

t Adt

= = = − + … (3)

Giving B = 0.0008 nearly (on using (2)

Hence ( )500 0.002 cos1323 0.0008 sin1323tq e t t−= +

Example 24: An e.m.f. Esinpt is applied at t = 0 to a circuit containing a condenser C and

inductance L in series. The current x satisfies the equation 1 sin .dxL x dt E ptdt C

+ =∫


If 2 1pLC

= and the initially the current and charge q are zero, show that the current in the

circuit at time t is given by sin ,2E t ptL

where dqx

dt= − [NITK, 2008; KUK, 2005-06]

Solution: Since 2

2;

dq d qdxxdt dt dt

= − = − and x dt q= −∫ … (1)

The equation becomes, 2

21 sin

d qL q E pt

dt C+ = − … (2)

i.e. 2

22

sind q Ep q ptdt L

+ = − 21given pLC

=

The auxiliary equation is 2 2 0;D p D ip+ = = ± ... (3)

∴ 1 2. . cos sinC F c pt c pt= + ... (4)

And 2 2 2 21 1. . sin sinE EP I pt pt

D p L L D p = − = − + +

= cos .2Et ptpL … (5)

∴ 1 2cos sin cos2Etq c pt c pt ptpL

= + + … (6)

Initially t = 0, q = 0 implying c1 = 0

∴ 2 sin cos2Eq c pt t ptpL

= +

Differentiating and putting for ,dq

xdt

−

2 cos cos sin2E

x pc pt pt pt ptpL

− = + − … (7)

When t = 0, x = 0.

∴ 202EpcpL

= + i.e. 2 22Ecp L

= − … (8)

Hence from (7)

cos cos sin2 2E Ex pt pt pt ptpL pL

− = − + −

or sin2

Et ptx

L= Hence the result.

Miscelaneous Problems

Example 25: A resistance R in series with inductance L is shunted by an equal resistanceR in series with capacitor C. An alternating e.m.f. Esinpt produces currents i1 and i2 inthe two branches. If i1 and i2 are zero when t = 0, determine i1 and i2 from the equations

11 sin

diL Ri E pt

dt+ = and 2 2 cos .

i diR pE pt

C dt+ =


Verify that if R2C = L, the total current i1 + i2 will be sin .E ptR

Solution: The first equation is 11 sin

di R Ei ptdt L L

+ =

. .R Rdt tL LI F e e∫= =

Hence the solution is

1 1sin ,R t R t

L LEi e e pt dt cL

= +∫ c1 is an arbitrary constant

122

2

sin cosR tL

R pt p ptE Le c

RL pL

− = + +

… (1)

( )

2 2

sin cosusing sinax ax a bx b bx

e bx dx ea b

−= +

∫

When t = 0, i1 = 0

implying 2

1 2 2 2

( )0

L pEcL R p L

−= +

+ or 1 2 2 2

ELpc

R p L=

+Substituting value of c1 in (1) and simplifying.

1 2 2 2 2 2 2sin cos

R tL

ELp EL Ri e pt p ptR p L R p L L

− = + − + + … (2)

Note: The first term on the r. h. s. is called the transient non periodic current.

The second equation, 2 2 cosi di

R pE ptC dt

+ = or 22

1 cos .pEdi

i ptdt RC R

+ =

1 1

. .dt t

RC RCI F e e∫= =Hence the solution is

2 2 cost t

Rc RCpE

i e c e pt dtR

= + ∫ , c2 is an arbitrary constant

22

2 2

1 cos sin1

tRCpE e

c pt p ptR RCp

R C

= + + + … (3)

( )2 2using cos cos sin

axax ee bx dx a bx b bx

a b = + +

∫

when t = 0, i2 = 0 implying 2

2 2 2 210 .

1pERC

cp R C RC

= ++

or ( )2 2 2 21pEC

cp R C

= −+


Substituting above value of c2 in (3) and simplifying

2

2 2 2 2 2 2 21 cos sin

1 1

tRCpEC e pERC

i pt p ptp R C p R C RC

− = − + + + +

… (4)

Equation (2) and (4) give i1 and i2. By adding both and using the equation R2C = L, we canprove the second part.

Verification: Given 11 sin .

diL Ri E pt

dt+ = and 2 2 cos .

i diR pE pt

C dt+ = Since we can write the

second differential equation as

2

22 cos .

diR i R pE ptL dt

+ = i.e. 22 cos .

di LL Ri pE ptdt R

+ =

On adding the above two equations, we get the joint equation as:

1 2

1 2( )

( ) sin cosd i i LL R i i E pt p pt

dt R+ + + = + … (5)

If we take (i1 + i2) = i, (5) becomes,

sin cosdi R E Li pt p ptdt L L R

+ = + ... (6)

Here . .R Rdt tL LI F e e= =∫

∴ { }( ) sin cosR Rt tL LE Li t e e pt p pt dt

L R

= + ∫

sin cosR Rt tL LE Le pt dt p e pt dt

L R

= +

∫ ∫(Integration by parts in Ist integral and keeping 2nd integral unchanged)

sin cos cos

R Rt t RL L tLE e e Lpt p pt dt p e pt dt AR RL R

L L

= − + +

∫ ∫

( ) sinR tLEi t pt Ae

R−

= + … (7)

Initially i = i1 + i2 = 0 when t = 0. implying A = 0

∴ 1 2sinE pt

i i iR

+ = =

ASSIGNMENT 3

1. Show that the differential equation for the current i in an electrical circuit containinginductance L and a resistance R in series and acted on by an electromotive force Esinwt

satisfies the equation, sin .diL Ri E wtdt

+ = [PTU, 2006, 2007]


2. When a switch is closed in a circuit containing a battery E and an inductance L, the

current i build up at a rate given by diL Ri Edt

+ = . Find i as a function of t. How long

will it be before the current has reached one-half its final value if E = 6 volts, R = 100Ohms, L = 0.1 Henry?

3. When a resistance R ohm is connected in series with an inductance L henries, an e.m.f. of

E volts, the current i in amperes at a time t is given by diL Ri Edt

+ = .

If E = 10 sint volts and i = 0 when t = 0, find i as a function of t.4. An inductance of 1 henry and a resistance of 2 ohms are connected in series with an

e.m.f. of 100e–t volts. If the current is initially zero, what is the maximum current attained.

8.5 PHYSICAL APPLICATIONS

Under physical applications, we broadly discuss Newton’s Law, Planetary and Satellitemotions and Hooks Law,.

(i) Particle under Newton’s law of motion follows:

2

2F =mass × acceleration dv d xma m m

dt dt= = = , for x as displacement.

Further, if the body mass moves along a curve through a distance s in time t, its

velocity and acceleration are given by 2

2,ds d s

dt dt respectively.

(ii) Under Newton’s laws of gravitations:(a) Resistance varies as the velocity of the particle falling under gravity.(b) Resistance varies as the velocity of the particle projected upward under gravity in

resisting medium.

(c) In velocity of escape from the earth, gravitational force is proportional to 21r

, where

r is the distance from the centre to the projectile.

Note: In reaching other planets, it is essential that velocity v of a spaceship be so large that v doesnot become zero in no case to rule out the possibility to pull back of the ship towards the earth.

(iii) Problem following Hook’s Laws: Tension of an elastic string (or a spring) isproportional to the extension of the string (or a spring) beyond its natural length i.e.,

,xTa

= λ x is the extension beyond the natural length a and λ is the modulus of elasticity.

Example 26: Resisted Upward Motion: – A particle of mass m is projected vertically undergravity. The resistance of the air being mk times the velocity. Show that the maximum

height attained by the particle is [ ]2

log(1 )Vg

λ − + λ , where V is the terminal velocity and

λλλλλ is the initial velocity of the particle.


Solution: Equation of the particle at any time t is given by

2

2d x dxm mg mkdt dt

= − − …(1)

where x is the height of the particle at time t from the initial position.

dvmv mg mkvdx

= − − ;2

2d x d dx d dv dx dvv vdt dt dt dt dx dt dx

= = = =

or dvv g kvdx

= − − …(2)

When acceleration is downward, 0dvvdx

= and v = V (the terminal velocity)

From above equation, 0 = g – kV org

kV

= …(3)

∴ ( )g gdvv g v V v

dx V V= − − = − +

gv dv dx

v V V= −

+ or 1gV dv dx

v V V − = − +

On integration log( )g

v V v V x cV

− + = − + , where c is an arbitrary constant …(4)

Initially, for x = 0, v = λV giving, c = λV –V log(V + λV)

∴ ( )log( ) log( )g

v V v V x V V V VV

− + = − + λ − + λ …(5)

If x is the highest height of the particle, then for , 0x x v= =Hence from the above equation,

0 log( 0) log (1 )g

V V x V V VV

− + = − + λ − + λ

log (1 ) logg

x V V V V VV

= − + λ + + λ

log log(1 ) logg

x V V V V VV

= − + + λ + + λ

log(1 )g

x VV

= λ + + λ

2

log(1 )Vxg

= λ + + λ …(6)

Example 27: Problem on Resisted Downward Motion:- A body of mass m is falling undergravity and facing air resistance proportional to the square of the velocity (i.e. k2v). If


after falling through a distance x, it possesses a velocity v at that instant, prove that2

2 22 logx ak

m a v=

−, where k is the constant of proportionality.

Solution: By Newton’s Law of Motion,

2

22

,d xm mg kvdt

= − dxvdt

= , mg is the weight of the object

Here in above equation, 2

2d x d dx d dv dx dvv vdt dt dt dt dx dt dx

= = = =

∴ 2dvmv mg kvdx

= −

Taking, v2 = z so that 2 dv dzvdx dx

= , above equation reduces to

1.2

dzm mg kzdx

= − or 2 ,2

mgm dz z a zk dx k

= − = − when 2mga

k=

Implying 22dz k

a z m=

− (a case of variable-separable)

On integration 2 2log( ) log ka z c xm

− − + = , c is an arbitrary constant and z = v2

Initially, v = 0 at x = 0 ⇒ c = loga2

Hence, 2

22 2

2log ,mga k x a

a v m k = = −

Example 28: Problem on Velocity of Escape:- If 22

2 2

gRd rdt r

= − where R is the radius of the

earth and r( ≥ ≥ ≥ ≥ ≥ R) is the distance of an object at any time t, projected upwards from the

centre of the earth with velocity 0 2v gR= , show that the object will never return to the

earth.

Solution: As 2

2d r d dr d dv dr dvv vdt dt dt dt dr dt dr

= = = =

Therefore from the given relation,

2

2dv Rv gdr r

= − or2

22 2 Rv dv g dr

r= −

Integrating, 2

2 2gRv c

r= +

Given, 0 2v v gR= = when r = R implies 22

2gR

gR cR

= + or c = 0


∴ 2

2 2gRv

r= or

2gv R

r=

Thus, the object will not come to rest as v = 0 only when r = ∞. Meaning thereby that theobject will never come back to the earth.

Example 29: Problem on Motion of ship/boat across a stream:- A ship is moving in waterwith uniform velocity v0. Suddenly the engine is switched off. Prove that the time inwhich the speed comes down to one half of its original value, is equal to

012 2 2

tan2

abvwabg a b v

− + where w is the weight of the ship and g is the acceleration due to

gravity, assuming that the resistance to the movement of the ship is of the form (a2 + b2v2).

Solution: Equation of motion is

2 2 2( ), where is the mass of the ship

gw dv wa b vg dt

= − +

2 2 2( )wdt dv

g a b v= −

+Integrating under given limits,

0

0

2

2 2 20

1( )

vt

v

wdt dvg a b v

= −+∫ ∫

( ) ( )

02

0

12

1. tan

v

a ab b v

w vtgb

−

= −

01 1

0tan tan2vw b b v

abg a a− − = − −

00

1

00

2tan1

2

vb b vw a avb babg v

a a

−−

= −+ ⋅

01

2 2 2tan

2abvw

abg a b v− = − +

Example 30: Problem on Atmospheric pressure:- Find the atmospheric pressure p lb. per ft.at a height h ft. above the sea level, both when the temperature is constant or variable.

Solution: Let a vertical column of air of unit cross-section, with an element of it boundedby two horizontal planes at height h and h + δh above the sea level, and p, p+ δp be thepressure at h, h + δh respectively. Let ρ be the average density of the element.

Here the thin air column δh is in equilibrium under the action of the pressure (p + δp) andthe weight (ρgδh), both download.


Hence p = (p + δp) + ρgδh

Implying δp = – ρgδh orp

gh

δ = −ρδ

In the limiting case, when δh → 0

dp

gdh

= −ρ … (1)

This is the differential equation of atmospheric pressure at height h.

1. When temperature is constant: Here v is the volume of 1 kg ofair, supposed to move about any expansion or contraction in itsvolume will follow Boyle’s law.

pv = k, k is constant.

or 1 p

p k p kv k

= ⇒ = ρ ⇒ ρ =

From (1), dp p

gdh k

= − ordp g

dhp k

= −

On integrating,

log log or ghk

ghp c p ce

k−

− = − = … (2)

At sea level, h = 0, p = p0 (say) so that c = p0 … (3)

Hence (2) becomes 0

ghkp p e

−= … (4)

2. When temperature is variable: As generally temperature of atmosphere varies greatlywith the change of altitude, in this case we take the relation

pvn = k, n ≠ 1

Implying

1

1 or , 1nn

n

pp k p v k n

v k = = ⇒ ρ = ≠ … (5)

From (1), 1/ndp p

gdh k

= − or

( ) ( )1 1n n

dp gdh

kp= − … (6)

On integration,

11 1

111

nn

pghk C

n

−−

= +−

or 1 11

11n nn p ghk C

n− −

= +−

… (7)

At the sea level h = 0, p = p0 (say)

∴ 11

1 01nnC p

n−

=−

O Sea level

g zρδp z

p p + δ z z + δ

Fig. 8.17


Hence Eq. (7) becomes, 1 1 11 1

01n n nn p p ghk

n

− − − − = −

This is the desired relation between p and h when the temperature is variable.

Example 31: Problem on Rotating Cylinders:- A cylindrical tank of radius r filled withwater to a depth h is rotated with angular velocity ωωωωω about its axis, centrifugal force tendsto drive water outwards from the centre of the tank. Under the steady conditions of uniformrotation, show that the section of the free surface of the water by a plane through the axis,is the curve

2 2

2

2 2ry x h

g ω= − +

Solution: Forces acting on a particle of mass m cut out from the free surface of water at P(x, y)on the curve are the weight of the particle acting vertically downwards and the centrifugalforce acting upwards (Fig. 8.18 (b)).

In case of steady motion, the particle P(x, y) will move on the free surface of the water andhence no tangential force will be acting there upon, as shown in figure 8.18 (a).

Yr

h

O

Hy

mω2

Y

r

x P

X

AT

mg

Rψ

O

Steady state Rotating state

Fig. 18 (a) Fig. 8.18 (b)

∴ Rcosφ = mg and Rsinφ = mω2x

Implying 2 2

tandy m x xdx mg g

ω ω= φ = = … (1)

which is the differential equation of the surface of rotating fluid mass. On integration,

2

dy xdx Cg

ω= +∫ ∫ or2 2

2xy Cg

ω= + … (2)

From (2); when

2 2

0, (=OA) C ( )

, ( ) ...( )2

x y irx r y H C iig

= = … ω = = = +

… (3)


∴ y = 2 2 2 2

2 2w x w r

g g−

Volume of the fluid mass in the non rotating state = πr2h … (4)

Whereas in the rotating state = 2 2H

OAr H x dyπ − π∫

2

2

2( )

H

C

gr H y C dy

π= π − −ω ∫ (using (2))

2 2

2( )

gr H H C

π= π − −ω

22 2 2 2

222 2gr rr C

g g

πω ω= π + − ω (using 3(ii))

2 4 4 4

22 22 4gr rr C

g gπω ω= π + π −ω

2 2

2

4rr Cg

ω= π + … (5)

But the volume in the two states should be equal viz. expression (4) and (5) must equate,

2 2

2 2

4rr h r Cg

ωπ = π +

Implying 2 2

4rC hg

ω= − …(6)

Therefore Eq. (2) gives, 2 2 2 2

2 4x ry hg g

ω ω= + − or2 2

2

2 2ry x h

gω = − +

which is the desired equation of the curve.

ASSIGNMENT 4

1. A particle is projected vertically upwards in the gravitational field under a resistanceequal to k times the square of its velocity per unit mass. If V0 is the initial velocity, findthe maximum height attained.

2. The differential equation for the motion of a particle moving in a central orbit is

2 4

2 3d x axdt x

= −µ + If it starts from rest at a distance ‘a’, find the time it will take to arrive at the origin.

3. A particle of mass m moves under gravity in a medium whose resistance is k times itsvelocity, where k is a constant. If the particle is projected vertically upwards with avelocity v, show that the time to reach the highest point is

log 1em kvk mg

+


4. A body of mass m falls through a medium that opposes its fall with a force proportionalto the square of its velocity so that its equation of motion is

2

22

,d xm mg kv v xdt

= − =

where x is the distance of fall and v is the velocity. If at t = 0, x = 0 and v = 0, find thevelocity of the body and the distance it has fallen in t seconds.

5. A particle is projected with velocity v along a smooth horizontal plane in the mediumwhose resistance per unit mass is µ times the cube of the velocity. Show that the distanceit has described in time t is

( )21 1 2 1v tv

+ µ −µ

6. When a bullet is fired into a sand tank, its retardation is proportional to the square rootof its velocity. How long will it take to come to rest if it enters into the sand tank withvelocity v0?

7. Show that a particle projected from the earth’s surface with velocity of 7 miles/sec willnot return to the earth. (Take earth’s radius = 3960 miles and g = 32.17 ft / sec2)

8. A chain coiled up near the edge of a smooth table just starts to fall over the edge. Thevelocity v when a length x has fallen is given

2dvxv v gxdx

+ = . Show that 8 3 /sec.v ftx

=

9. Upto a certain height in the atmosphere, it is found that the pressure p and the densityρ are connected by the relation p = kρn, (n > 1). If this relation continued to hold uptoany height, show that the density would vanish at a finite height.

10. A canonical cistern of height h and semi vertical angle α is filled with water and is heldin vertical position with vertex downwards. Water leaks out from the bottle at the rateof kx2 cubic constant per second, k is a constant and x is the height of water level from

the vertex. Prove that the cistern will be empty in 2tanh

k π α seconds.

11. The rate at which water flows from a small hole at the bottom of a tank is proportionalto the square root of the depth of the water. If half the water flows from a cylindricaltank (with vertical axis) in 5 minutes, find the time required to empty the tank.

ANSWERS

Assignment 1

1. y = aecx 2. x = 3y2

3. x2 + y2 = cx 4.2 2

2 2 log a a xy a x a cx

− −= − + + 5. (i) r(θ – α) = c (ii) r = a + bcosθ

AS


6. c2x2 = 2cy + 1 7. r = a eθcotα

8.2

1 2sin ,4

x ky c a ax x aa

−+ = + − =

Assignment 2

2. x2 – y2 = a 3. x4/3 – y4/3 = C4. x2 + y2 = 2a2logx + c 5. x2 + y2 – cy = 0

6. a(1 + cosθ) 7. rn = bncosnθ

8 .8 .8 .8 .8 .2

1 cosbr =

− θ

9. (i) 2 31 1ln . ,2 3

r d+ θ + θ = (ii) rncosnθ = d

Assignment 3

2. 0.0006931 seconds 3. 2 20.1 sin cos

R tLR t L t Le

L R−

− + + 4. 25 Amp.

Assignment 4

1.2

01 log 12

kVk g

+

2.

4 44

3

( )Hint: with solution

2 Retain the -ve sign as the motion is down towards origin

a xdv a dxv x vdx x x dt

µ − π = −µ + = = µ

4. tan and logmg gk gkmh t tk m k m

6. 02 v

k11. 17 minutes 4 seconds

applications of 1st order diffential e

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