Engineering mathematics-1
Previous year question paper with solutions for Engineering mathematics-1 May-2018
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Question paper 1
Section A
1. a) Find asymptotes, parallel to axes, of the curve: y =
Answer:
(a) yx2 - y = x2 + 1
|| to x axis
X2=0
X = 0 No asymptote || to x axis
|| to y axis
X2-1 = 0
X2 = 1
X = 1
b) Write a formula to find the volume of the solid generated by the revolution, about
x-axis, of the area bounded by the curve y = f(x), the x-axis and the ordinates
x = a and x = b.Answer:
(b) Volume will be given by integrating the terms
=
=
c) Find the value of
, where x = rcosθ & y = rsinθ .Answer:
d) If an error of 1% is made in measuring the major and minor axes of an ellipse, what is the
percentage error in its area?Answer:
(d) let x and y be semi major and semi minor axes of an ellipse.
Area of ellipse = A=
Log A = Logπ + Log x +Log y
1+1=2Error in area = 2 %
e) Is the function
? If yes, what is its degree?Answer:
Yes, Degree = 2
f) What is the value of
over the positive quadrant of the circle x2 + y2 = 1?Answer:
=
g) Give geometrical interpretation of
Answer:
=
=
=
= 2
h) Show that for the vector field
(x2 - y2 + x) 
— (2xy + y) 
= 0.Answer:
=
=
= -
2y+2y[
]i) Show that the vector field
= (-x2 + yz) + (4y — z2x) + (2xz — 4z) 
is solenoidal.Answer:
For solenoidal function
div
div f =
= -
2x + 4+2x -4= 0
So it is solenoidal
j) State Green’s theorem in plane.
Answer:
Green’s theorem in plane:-
Let R bed closed region of x-y plane bounded by a simple closed cense c and let M and N be continuous function of x and y having continuous partial derivative
and 
in R then: 
Section B
2. Trace the following curves by giving their salient feature:
a) y2 (a — x) = x2 (a + x).
b) r = a (1 — cosθ)Answer:
(a) Symmetry: Symmetrical about x-axis
Origin: passes through origin and tangent at origin
y= x and y =-x therefore origin is a node.
Asymptotes x = a
Points: it crones the Ares at (0,0) and (-a,0 )
When x >a or <-a y is imaginary
Shape of curve is strophoid
(b) Given figure is a cardioid and symmetrical about q = 0 i.e. initial line.
q varies from (0-π/2 ) or (0-180).
3. a) Find the whole length of the curve x2/3 + y2/3 = a2/3 .
b) Use definite integral to find the area of ellipse
Answer:
(a) Given curve is asteroid. Which is symmetrical about x-axis and y-axis therefore entries length of curve is four times the length of one part.
Diff given curve w.r.t x
2/3x-1/3+ 2/3y-1/3dy/dx =0
2/3y-1/3dy/dx = -2/3x-1/3
dy/dx=-(y/x)1/3
total length = 4
= 4 
= 4
= 6a
(b)
Therefore area of ellipse = 4x area of ellipse in I quadrant.
=
=
=
Therefore required area = area of region PQRS
= 4 x area of region OFPB in the first quadrant
=
=
=
4. a) lf u = log (x3 +y3 +z3 —3xyz), show that
= -9(x+y+z)-2b) State Euler’s theorem for homogeneous functions and apply it to show that
, where sin u = 
Answer:
(a)
=
=
=
=
(b) Euler’s theorem
If H= f (x, y, z)is ahomogeneous function of x, y and z of degree n, then
U= sin-1(x2+y2/x+y)
Sin u = x2+y2/x+y
F = x2+y2/x+y
Which is homogeneous of degree 1.
By Euler’s theorem
=
5. a) The temperature T at any point (x, y, z) in space is T = 400xyz2 . Find the highest
temperature on the surface of the unit sphere x2 + y2 + z2= 1.b) If f(x,y) = tan-1 xy, compute f(0.9, –1.2) approximately.
Answer:
(a) Temperature at any point on sphere is given by
T= 400 xy (1-x2-y2)
= 400y -400y3-1200x2y
400x – 1200xy2-400x3For critical points
400y-400y3-1200x2y
400x(1-3y2-x2)=0
critical points after solution above equation are
(0,0),(±1,0),(0, ±1),( ±1/2, ±1/2) out of which first three given the value of T to be zero. For other four point (±1/2, ±1/2). Now A =
= 2400 xy, B = 
= 400-1200y2-1200x2, C=
= -2400xyAt (1/2,1/2,) A= -600 < 0, B= -200, C = -600 so that AC-B2= 320000> 0
∴
T is maximum at(1/2,1/2)Whose maximum value will be given by
= 400 ¼(1 -1/4 - 1/4) = 100 x ½ =50
(b) f(x,y) = tan-1xy
Let x=0.9 , y=-1
X+δx= 0.9 1+δx=0.9 =δx=-1 =
-y+ δy=-1.2 δy=-1.2+1 = δy = -0.2
df=
= y/1+x2y2 dx + x/1+x2y2dy = ydx+xdy/1+x2y2When x =1
Y = -1
df =
=
= 
= 
Section C
6. a) Evaluate the following integral by changing the order of integration:
b) Evaluate the triple integral
Answer:
(a) Given region of integration
0
it is a circle of radius 1
we will introduce a vertical strip which will vary from
order to change the order into integration
= 2
(b)
==
=
=
=
=
=
7. a) Find a unit vector normal to the surface x2+y2+z2= 9 at the point (2, -1, 2) .
b) If u = x2+y2+z2 &
Show that Ñ. 
= 5uAnswer:
(a) f(x,y,z) = x2+y2+z2 = 9
=2x
= 2y
= 2zVector =
A point (2,-1,2)
=
=
Unit vector normal to it =
(b) u= x2+y2+z2
(by identity)
)=
[
] 8. a) If
evaluate, 
where C is the curve in the xy-plane y = 2x2
from (0,0) to (1,2).(b) Compute
, where 
and S is the triangular surface with vertices (2, 0,0), (0, 2, 0) and (0, 0, 4).Answer:
(a)
where 
y =
and 
(b)
Equation of given plane x/2+y/z+z/4=1
Let R be orthogonal projection of x+y +z/2=1in XoY plane i.e. z=0 given by x+y=1
Therefore R=0{(x.y):0≤ y≤ 1-x;0≤ x≤ 1}
=x +yx 
=(x
) 
ds =
[Put x=2,y=2,z=neglected ]
9. State Gauss Divergence theorem and verify it for
taken over
the cube bounded by x = 0, x = 1, y = 0, y = 1, z = 0, z = 1.Answer:
Surface contains six faces
(i) OABC faces Z = 0,
.
(ii) On face DEFG Z = 1,
(iii) On face OAFG y = 0,
(iv) On face DEBC y = 0,
(v) On face ABEF x = 1,
(vi) On face OCDG x = 0,
Gauss Divergence theorem
If
be a vector point function having continuous first partial derivative in reason v bounded by surface (s).Then
