UPSC Civil Services Main 1998 - MathematicsCalculus

Question 1(a) Find the asymptotes of the curve

(2x− 3y + 1)2(x+ y)− 8x+ 2y − 9 = 0

and show that they intersect the curve again in three points which lie on a straight line.

Solution. The given equation is

(2x− 3y)2(x+ y) + 2(2x− 3y)(x+ y) + x+ y − 8x+ 2y − 9 = 0

Now x3φ3(yx) = homogeneous terms of degree 3 = (2x − 3y)2(x + y). If y = mx + c is an

asymptote, then m is a root of φ3(m) = (2− 3m)2(1 + m) = 0⇒ m = −1,m = 23, 2

3, so we

may have two parallel asymptotes.For m = −1, c = −φ2(−1)

φ′3(−1)= 0

φ′3(−1)= 0, as φ′3(−1) 6= 0. Thus x+ y = 0 is one asymptote.

For parallel asymptotes y = 23x+ c, c is the root of

c2

2!φ′′3(

2

3) + cφ′2(

2

3) + φ1(

2

3) = 0

φ3(m) = (2− 3m)2(1 +m)

φ′3(m) = (2− 3m)2 − 6(2− 3m)(1 +m)

φ′′3(m) = −12(2− 3m) + 18(1 +m)− 6(2− 3m)⇒ φ′′3

(2

3

)= 30

φ2(m) = 2(1 +m)(2− 3m)

φ′2(m) = −6(1 +m) + 2(2− 3m)⇒ φ′2

(2

3

)= −10

φ1(m) = −7 + 3m⇒ φ1

(2

3

)= −5

1

Brij BhooshanAsst. Professor

B.S.A. College of Engg & TechnologyMathura

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Thus 15c2 − 10c − 5 = 0 ⇒ c = 1,−13. So the asymptotes are y = 2

3x + 1, y = 2

3x − 1

3or

2x− 3y + 3 = 0, 2x− 3y − 1 = 0.The joint equation of the asymptotes is Pn = (x + y)(2x − 3y + 3)(2x − 3y − 1) = 0

or Pn = (2x − 3y)2(x + y) + 2(2x − 3y)(x + y) − 3(x + y) = 0. The equation of the curveis f(x, y) = Pn − 4x + 6y − 9 = 0. This implies that the points of intersection of f(x, y)and Pn lie on 4x − 6y + 9 = 0, which is a straight line. But this line is parallel to thetwo parallel asymptotes, which means that the two aymptotes do not cut the curve — thiscan be verified by solving the simultaneous equations f(x, y) = 0, 2x − 3y + 3 = 0 andf(x, y) = 0, 2x − 3y − 1 = 0. The asymptote x + y = 0 cuts the curve at (− 9

10, 9

10), which

lies on the line 4x− 6y + 9 = 0.(It seems that there is an error in the question. The following statement however is true

— if there are 3 points at which the asymptotes of a cubic curve intersect it, then these mustlie in a straight line.)

Question 1(b) A thin closed rectangular box is to have one edge n times another edge,and the volume of the box is given to be V . Prove that the least surface S is given bynS3 = 54(n+ 1)2V 2.

Solution. Let the edges be given by x, nx, y, so that V = nx2y ⇒ y = Vnx2 . Then

S = 2(nx2 + xy + nxy) = 2

(nx2 +

(n+ 1)V

nx

). For critical points,

dS

dx= 2

(2nx− (n+ 1)V

nx2

)= 0

⇒ 2n2x3 − (n+ 1)V = 0⇒ x3 =(n+ 1)V

2n2

d2S

dx2= 2

(2n+

2(n+ 1)V

nx3

)= 2(2n+ 4n) > 0(when x3 =

(n+ 1)V

2n2)

Thus S is minimum when x3 =(n+ 1)V

2n2.

S =2

nx(n2x3 + (n+ 1)V ) =

3(n+ 1)V

nx, so n3x3S3 = 27(n+ 1)3V 3 ⇒ n(n+ 1)V S3/2 =

27(n+ 1)3V 3 ⇒ nS3 = 54(n+ 1)2V 2 as required.

Question 1(c) If x+ y = 1, prove that

dn

dxn(xnyn) = n!

[yn −

(n

1

)2

yn−1x+

(n

2

)2

yn−2x2 + . . .+ (−1)nxn]

Solution. By the Leibnitz formula

dn

dxn(uv) =

n∑r=0

(n

r

)un−rvr

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Let u = xn, v = yn = (1− x)n, then

un−r = n(n− 1) . . . (n− (n− r) + 1)xn−(n−r) =n!

r!xr

vr = (−1)rn(n− 1) . . . (n− r + 1)(1− x)n−r = (−1)rn!

(n− r)!yn−r(

n

r

)un−rvr =

(n

r

)n!

r!xr(−1)r

n!

(n− r)!yn−r = (−1)rn!

(n

r

)2

xryn−r

⇒ dn

dxn((xnyn) = n!

n∑r=0

(−1)r(n

r

)2

xryn−r

= n!

[yn −

(n

1

)2

yn−1x+

(n

2

)2

yn−2x2 + . . .+ (−1)nxn]

Question 2(a) Show that ∫ ∞0

xp−1

(1 + x)p+qdx = B(p, q)

Solution. By definition, B(p, q) =

∫ 1

0

xq−1(1− x)p−1 dx. Let x =1

1 + y, dx =

−dy(1 + y)2

.

B(p, q) =

∫ 0

∞

1

(1 + y)q−1

(1− 1

1 + y

)p−1 −dy(1 + y)2

=

∫ ∞0

yp−1

(1 + y)p+qdy

as required.

Question 2(b) Show that

∫∫∫dx dy dz√

1− x2 − y2 − z2=

π2

8where the integral is over all

positive values of x, y, z for which the expression is real.

Solution. Switching over to polar coordinates, x = r sin θ cosφ, y = r sin θ cosφ, z =r cosφ, ∂(x,y,z)

∂(r,θ,φ), we get

I =

∫ π2

0

∫ π2

0

∫ 1

0

(1− r2)−12 r2 sin θ dr dθ dφ

=π

2

∫ 1

0

(1− r2)−12 r2 dr

Put r2 = t

=π

2

∫ 1

0

(1− t)−12t

12

2dt

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=π

4

∫ 1

0

(1− t)12−1t

32−1 dt

=π

4B(3

2,1

2

)=π

4

Γ(32)Γ(1

2)

Γ(2)=π

4

1

2

(Γ(1

2

))2

=π2

8∵ Γ(1

2

)=√π

Question 2(c) The ellipse b2x2 + a2y2 = a2b2 is divided into two parts by the line x = a/2and the smaller part is rotated through four right angles about this line. Prove that the volumegenerated is

πa2b

[3√

3

4− π

3

]Solution.

The part rotated is LL′AL, the coordi-nates of L are (a

2, y) where y is given by

a2

4a2 + y2

b2= 1. Thus L is the point (a

2,√

32b)

and L′ is the point (a2,−√

32b). P is the generic

point (x, y) so that OM = x,NM = x− a2

=PM ′. The required volume

V =

∫π(PM ′)2 d(NM ′) = 2

∫ √3

2b

0

π(x−a

2

)2

dy

A′ O N M A

L′

LP

M ′

x = a2

Put x = a cos θ, y = b sin θ, so that θ varies from 0 to π3.

V = 2π

∫ π3

0

(a cos θ − a

2

)2

b cos θ dθ

=2πa2b

4

∫ π3

0

(2 cos θ − 1)2 cos θ dθ

=πa2b

2

∫ π3

0

(4 cos θ(1− sin2 θ)− 4 cos2 θ + cos θ

)dθ

=πa2b

2

[5 sin θ − 4 sin3 θ

3

]π3

0

− 4πa2b

2

∫ π3

0

1 + cos 2θ

2dθ

=πa2b

2

[5 sin θ − 4 sin3 θ

3− 2θ − sin 2θ

]π3

0

=πa2b

2

[5

√3

2− 4

3

3√

3

8− 2

π

3−√

3

2

]=

πa2b

2

[3√

3

2− 2

π

3

]= πa2b

[3√

3

4− π

3

]4

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Paper II

Question 3(a) Show that the function f(x, y) = 2x4 − 3x2y + y2 has (0, 0) as the onlycritical point but the function has neither a minimum nor a maximum at (0, 0).

Solution. For critical points, ∂f∂x

= 8x3− 6xy = 0, ∂f∂y

= −3x2 + 2y = 0. ∂f∂x

= 0⇒ x = 0 or

6y = 8x2. But 6y = 8x2 is not compatible with the second equation 2y = 3x2. Thus x = 0,which implies y = 0, hence (0, 0) is the only critical point.

Now f(0, 0) = 0, f(δ, 0) = 2δ4 > 0. Let us take y =√

3x2, f(x, y) = 2x4− 3√

3x4 + 3x4 <0. Now whatever neighborhood of (0, 0) we take, it has points of the form (δ, 0) for suitableδ > 0, as well as points that lie on the parabola y =

√3x2, thus f(x, y) takes positive as well

as negative values in any neighborhood of (0, 0), hence has neither maximum nor minimumat (0, 0).

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