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SK Engineering Academy
Mathematics – I (181101) Question Bank (2011- 2012)
Part A (Two Marks) Questions and Answers & Part B Questions
I-Year B.E/ B.Tech. Common to all Branches
UNIT – I MATRICES PART – A
⎡ 1
1. Find the sum and product of the eigen values of the matrix ⎢ ⎢ ⎢⎣− 2
2 − 2⎤
⎥
0 ⎥
− 1 3 ⎥⎦
Sum of the eigen values = Sum of the main diagonal elements = 1+0+3 = 4
Product of the eigen values = │A│ = -13
2. If 3 and 15 are the two eigen values of
⎡ 8 − 6
A = ⎢− 6 7
2 ⎤
− 4⎥ , find A , without expanding the
determinant.
2 − 4 3
If λ is the third eigen values of A, then 3 + 15 + λ = 8 + 7 + 3 => λ = 0
We know that , A = product of eigen values = (0)(3)(15) = 0
3. The product of two eigen values of the matrix
⎡ 6 ⎢− 2 ⎢⎣ 2
− 2 2 ⎤
3 − 1⎥
− 1 3 ⎥⎦
is 16. Find the third eigen value.
Let λ1 , λ2 , λ3 be the eigen values of the given matrix , then λ1 λ2 λ3 = A
(16) λ3 = 6(9-1)+2(-6 +2)+2(2-6) [ since product of two eigen values is 16]
= 32 ∴λ3 = 2
⎡7
4. One of the eigen values of ⎢4
⎢⎣4
4 4 ⎤
− 8 − 1⎥
− 1 − 8⎥⎦
is -9, Find the other two eigen values.
If λ1 , λ2 are the other two eigen values , then
λ1 + λ2 - 9 = 7 - 8 – 8 = -9 (since sum of the eigen values = sum of the leading diagonals)
λ1 + λ2 = 0 => λ1 = – λ2 …(1)
–9λ1 λ2 = │A│ = 441 ( since product of the eigen values = A )
λ1 λ2 = - 49 =>
λ1 =
,− 49
λ
…… (2)
substitute in (1) weget,
2
λ = ± 7
(1) => λ =
m 7 . Hence the other two eigen values are 7 and -7.
5. Find the eigen values of A2 , if
⎡3
A = ⎢0
⎢⎣0
1 4⎤
2 ⎥
0 5⎥⎦
In a triangular matrix, the main diagonal values are the eigen values of the matrix.
Here 3, 2, 5 are the eigen values of A. Hence the eigen values of A2 = 32, 22, 52 = 9,4,25.
1
6. If -2,3,6 are the eigen values of a 3 × 3 matrix A, then what are the eigen values of 6A −1 and A T .
Eigen values of A-1 = -1 , 1 , 1
, ∴Eigen values of 6A-1 = - 6 , 6 , 6 = − 3,2,1.
2 3 6
2 3 6
Eigen values of AT = Eigen values of A = − 2,3,6
7. State Cayley Hamilton theorem.
Every square matrix satisfies its own characteristics equation.
⎡1 2⎤ -1
8. Given A = ⎢
⎣
⎥ , Find A
⎦
using Cayley – Hamilton theorem.
The characteristics equation is λ2 – 4 λ – 5 = 0. By Cayley – Hamilton theorem A2 – 4 A – 5I = 0.
Multiply by A-1 we get A – 4 I – 5 A-1 = 0 ∴
A -1 =
⎡ − 3
1 [A − 4I] = ⎢ 5
5 ⎢ 4
⎣ 5
2 ⎤
5 ⎥
−1⎥ ⎥
5 ⎦
⎛ 1 ⎞
⎡− 2
2 − 3⎤
⎜ ⎟ ⎢ ⎥
9. If ⎜ 2 ⎟ is an eigen vector of ⎢ 2
1 − 6⎥
, find the corresponding eigen value.
⎜ ⎟
⎝ − 1⎠
− 1 − 2 0
⎛ − 2 − λ 2
− 3 ⎞ ⎛ x1 ⎞
⎛ 0 ⎞
⎛ − 2 − λ 2
− 3 ⎞ ⎛ 1 ⎞
⎛ 0 ⎞
⎜ ⎟ ⎜
⎟ ⎜ ⎟ ⎜
⎟ ⎜ ⎟ ⎜ ⎟
(A - λI)X = 0 ⇒ ⎜ 2
1 − λ
− 6 ⎟ ⎜ x 2 ⎟ = ⎜ 0 ⎟ ⇒ ⎜ 2
1 − λ
− 6 ⎟ ⎜ 2 ⎟ = ⎜ 0 ⎟
⎜ ⎟ ⎜
⎟ ⎜ ⎟ ⎜
⎟ ⎜ ⎟ ⎜ ⎟
⎝ −1
− 2 − λ ⎠ ⎝ x 3 ⎠
⎝ 0 ⎠
⎝ −1
− 2 − λ ⎠ ⎝ −1⎠
⎝ 0 ⎠
(–2–λ)(1)+2(2)+(-3)(-1) = 0 ⇒ λ = 5.
⎛ 1 2 1 ⎞ ⎜ ⎟
10. The matrix A = ⎜ 2 0
⎜
− 2 ⎟
⎟
is singular and one of its eigen values is 2. Find the other two eigen
values.
⎝ 1 2 1 ⎠
Since A is singular then one of the eigen value is zero
Let λ1 = 0 & λ2 = 2. Trace of the matrix = Sum of the eigen values
1 + 0 +1 = λ1 + λ2 +λ3
2 = 0 + 2 + λ3
∴ λ3 = 0
⎛ 2 0
⎜
11. Find the eigen vector of ⎜ 0 2
⎝ 1 0
1 ⎞
⎟
0 ⎟ corresponding to eigen value 2.
⎟
⎠
The eigen vectors are given by (A - λI)X = 0
⎛ 2 − λ 0
1 ⎞⎛ x1 ⎞
⎛ 0 ⎞
⎜ ⎟⎜
⎟ ⎜ ⎟
⇒ ⎜ 0
2 − λ
0 ⎟⎜ x 2 ⎟ = ⎜ 0 ⎟
⎜ ⎟⎜
⎟ ⎜ ⎟
⎛ 0 0
⎝
1 ⎞⎛ x1 ⎞
1 0
⎛ 0 ⎞
2 − λ ⎠⎝ x 3 ⎠
⎝ 0 ⎠
⎛ 0 ⎞
⎜ ⎟⎜
⎟ ⎜ ⎟ ⎜ ⎟
When λ = 2, ⎜ 0 0
0 ⎟⎜ x 2 ⎟ = ⎜ 0 ⎟
⇒ x3 = 0 & x1 = 0 . Therefore the eigenvector is ⎜ 1 ⎟
⎜ ⎟⎜
⎟ ⎜ ⎟ ⎜ ⎟
⎝ 1 0
0 ⎠⎝ x 3 ⎠
⎝ 0 ⎠
⎝ 0 ⎠
2
⎛a
12. Find the constants a and c such that the matrix ⎜ ⎝1
4⎞
⎟ has 3 &–2 as eigen values.
c⎠
Sum of the eigen values = Trace of the matrix ⇒ a + c = 3-2 = 1-----(1) Product of the eigen values = Determinant of the matrix
ac- 4 = (3)(-2) = -6 ⇒ ac = -2 ∴ c = -2/a
sub c in (1) a + c = 1 ⇒ a + (-2/a) =1 ⇒ a = -1, 2 ⇒ c = 2,-1
13. Determine λ so that λ (x2 + y2 +z2) + 2xy – 2xz + 2zy is positive definite.
⎛ λ 1
⎜
The matrix of the given quadratic form is A = ⎜ 1 λ
⎝ −1 1
−1⎞
⎟
1 ⎟
⎟
⎠
The principal sub determinants are given by
λ
D1 = λ, D2 =
1
1 = λ2
λ
− 1 = (λ + 1)(λ − 1)
& D3 = |A| = (λ+1)2(λ -2)
The Quadratic form is +ve definite if D1, D2 & D3 > 0 ⇒ λ > 2
14. If λ1 , λ2 , λ3 , … , λn are the eigen values of an nxn matrix A, then show that
λ13, λ 3
, λ 3
,… , λ 3
are the eigen values of A3.
Let λr be the eigen value of A with the eigen vector Xr , then AXr = λr Xr
Consider, A3 Xr = A2 (AXr)
= A2 (λr Xr)
= λr A(AXr)
= λr A(λr Xr) = λr 2(Ar Xr) = λ 3X
2 2
15. Determine the nature of the following quadratic form : f(x1,x2,x3) = x1
+ 2x2 .
⎛ 1
⎜
The matrix of the given quadratic form is A = ⎜ 0
⎝ 0
0 0 ⎞
⎟
2 0 ⎟
0 ⎠
1 0
The principal sub determinants are given by D1 = 1, D2 =
0
= 2 & D3 = |A| = 0
2
Therefore it is +ve semi definite (Since D1>0, D2 >0& D3 = 0)
16. If 1 & 2 are the eigen values of a 2x2 matrix A, what are the eigen values of A2, adj A and A+7I.
The eigen values of A2 are 12 , 22 = 1, 4
The eigen values of A+7I are 1+7 , 2+7 = 8, 9.
The eigen values of adj A are
A A 2 2
, = ,
1 2 1 2
= 2,1 [since
A −1 =
1 adjA ⇒ adjA = A A −1 and |A| = 2 ]
A
17. Find the characteristics equation of
⎡ 6 − 2
A = ⎢− 2 3
2 ⎤
− 1⎥
2 − 1 3
The characteristics equation is λ3 – S1 λ2 + S2 λ – S3 = 0.
S1 = Sum of the main diagonal elts = 12; S2 = Sum of the minors of the main diagonal elts = 36
S3 = A = 32. Therefore the char. Eqn is λ3 – 12 λ2 + 36λ – 32 = 0.
18. Find the Rank, index and signature of the Quadratic form whose Canonical form is
2 2 2
x1 + 2x 2 − 3x 3
Rank (r) = Number of terms in the C.F = 3 , Index (p) = Number of Positive terms in the C.F = 2
Signature (s) = 2p – r =1
3
19. Show that
⎡ cos θ
A = ⎢
⎣− sin θ
sin θ ⎤
cos θ⎥
is orthogonal.
⎡ cos θ
sin θ ⎤
⎡cos θ
- sin θ⎤
⎡1 0⎤
AA T = ⎢
⎥ ⎢ ⎥ = ⎢
⎥ = I
⎣− sin θ
= ⎡cos θ
cos θ⎦
- sin θ⎤
⎣ sin θ
⎡ cos θ
cos θ ⎦
sin θ ⎤
⎣0 1⎦
⎡1 0⎤
A T A ⎢ ⎥ ⎢
⎥ = ⎢
⎥ = I
⎣ sin θ
cos θ ⎦
⎣- sin θ
cos θ⎦
⎣0 1⎦
∴ A is orthogonal.
20. If A is an orthogonal matrix prove that A = ±1
Since A is orthogonal, then AT =A-1. ∴ AAT = 1 ⇒
PART – B
⎡ 3
1. Find the eigenvalues and eigenvectors of the matrix ⎢− 2
A AT =1
10 5 ⎤
− 3 − 4⎥ .
⇒ A 2 =1 ⇒ A
= ± 1
3 5 7
⎡ 13
⎢
− 3 5 ⎤
⎥ 4
2. Verify Cayley Hamilton theorem and find the inverse of the matrix
A = ⎢ 0
4 0 ⎥ .Also find A .
⎢⎣− 15 9 − 7
3. Reduce 8x2 + 7y2 + 3z2 –12xy–8yz+4xz into a canonical form by an orthogonal reduction and find the rank, signature, index and the nature of the quadratic form. Also give a nonzero set of values x,y,z which makes this Q.F zero.
⎡3
4. Diagonalise the matrix A = ⎢1
⎢⎣1
1 1 ⎤
3 −1⎥
−1 3 ⎥⎦
by means of an orthogonal transformation.
5. Find the matrix A whose eigen values are 2,3 and 6 and the eigen vectors are (1,0,-1)T, (1,1,1)T(1,-2,1)T.
UNIT – II THREE DIMENSIONAL ANALYTICAL GEOMETRY PART – A
1. Find the equation of the sphere x 2 + y 2 + z 2 + 16x - 17y + 21z + k = 0 which passing through the point
(1,1,0)
The sphere passes through the point (1,1,0) ⇒ 12 +12 +02 +16(1) – 17(1) +21(0) + k = 0 ⇒ k = -1
x 2 + y 2 + z 2 + 16x - 17y + 21z - 1 = 0
2. Find the centre and radius of the sphere 3(x 2 + y 2 + z 2 ) + 6x + 12y + 2z + 15 = 0 .
The equation of the sphere can be written as
⇒ u = 1; v = 2; w =1/3; d = 5
x2 + y 2 + z 2 + 2 x + 4 y + ( 2
) z + 5 = 0
Centre is (-u,-v,-w) ⇒ centre is (-1,-2,-1/3) and radius is
3. Define the right circular cone
u 2 + v2 + w2 − d =
1 + 4 + ( 19
) − 5 = 1
3
A right circular cone is the surface generated by a straight line (axis of the cone) which passes through a fixed point (vertex) and makes a constant angle (semi-vertical angle) with a fixed line through
the fixed point.
4
4. Find the equation of the sphere through the circle the point (1,-2,3) .
The equation of the sphere through the given circle is
x 2 + y 2 + z 2 + 2x + 3y + 6 = 0, x - 2y + 4z = 9 and
x2 + y2 + z 2 + 2x + 3 y + 6 + λ( x − 2 y + 4z − 9) = 0
The sphere passes through the point (1,-2, 3) (substituting in the above equation) ⇒ λ = -2.
⇒ x2 + y2 + z 2 + 2x + 3 y + 6 + (−2)( x − 2 y + 4 z − 9) = 0
⇒ x2 + y2 + z 2 + 7 y − 8z + 24 = 0 .
5. Find the equation of the right circular cone whose vertex is origin, axis is the y-axis and semi vertical angel is 30°.
The equation of right circular cone whose vertex is origin, axis is the y-axis and semi vertical angel is α°
is x2 +z2 = y2 tan2 α ⇒ x2 +z2 = y2 tan2 30 ⇒ is 3(x2 +z2) = y2.
6. Find the equation of the sphere concentric with x 2 + y 2 + z 2 − 4x + 6y − 8z + 4 = 0
through the point (1,2,3).
Centre of the sphere (2,-3, 4) and
radius of the sphere = distance between the centre and the point (1,2,3)
and passing
= (2 −1)2 + (−3 − 2)2 + (4 − 3)2 =
1 + 25 + 1 = 26
Equation of the sphere is (x-2)2 +(y + 3)2 + (z -4)2 = 26.
7. Find the equation of the cone whose vertex is the origin and guiding curve is
x + y
+ z 2 = 1 ,
4 9
x + y + z = 1 .
Since the vertex is origin , homogenizing 9x2 + 4y2 + 36z2 = 36(12) = 36(x + y + z)2
⇒ 27x2 + 32y2 + 72(xy + yz + zx) = 0
8. Find the equation of the tangent plane at the point (1,-1,2) to the sphere
x 2 + y 2 + z 2 − 2x + 4y + 6z - 12 = 0 .
The equation of the tangent plane at the point (x1,y1,z1) is
xx1 +yy1 + zz1 + u(x+ x1) + v(y + y1) + w(z + z1) + d = 0. x(1) +y(-1) +z(2) -1(x +1) +2(y – 1) +3(z + 2) -12 = 0
y + 5z -9 = 0.
9. Find the equation of the sphere having the points (2,-3,4) and (-1,5,7) as the ends of a diameter.
Equation of the sphere is (x-x1)(x-x2) + (y-y1)(y-y2) +(z-z1)(z-z2) = 0
⇒ (x-2)(x+1) + (y+3)(y-5) +(z-4)(z-7) = 0 ⇒ x2 + y2 + z2 – x -2y -11z +11 = 0.
10. Find the equation of the sphere passing through the points (0,0,0) ,(1,0,0), (0,1,0) and (0,0,1).
The equation of the sphere is x2 +y2+z2 +2ux +2vy +2wz +d = 0. The sphere is passing through (0,0,0) ⇒ d = 0,
5
The sphere is passing through (1,0,0) ⇒ u = -1/2, The sphere is passing through (0,1,0) ⇒ v = -1/2, The sphere is passing through (0,0,1) ⇒ w = -1/2, The equation of the sphere is x2 +y2+z2 - x - y - z = 0.
11. Write the equation of the tangent plane to the sphere (x - 2) 2 + (y - 3) 2 + (z - 4) 2 = 14
(1,5,7).
The equation of the sphere is x2 +y2+z2 - 4x -6y -8z +15 = 0. The equation of the tangent plane at the point (x1,y1,z1) is
xx1 +yy1 + zz1 + u(x+ x1) + v(y + y1) + w(z + z1) + d = 0.
at the point
⇒ x(1) +y(5) +z(7) -2(x +1) -3(y +5) -4(z + 7) + 15 = 0 ⇒x + 2y + 3z -23 = 0.
12. Write the equation of right circular cylinder whose axis is the line
radius is r.
x − α
l
= y − β
m
= z − γ
n
and the
(x − α )2
+ (y − β)2
+ (z − γ)2
[ (x − α) + m( y − β) + n(z − γ)] 2
−
l + m + n
= r 2
13. Show that the spheres x 2 + y 2 + z 2 + 6y + 2z + 8 = 0
orthogonally
The condition is 2(u1u2 + v1v2 + w1w2 ) = d1 + d2 .
& x2 + y 2 + z 2 + 6x + 8y + 4z + 20 = 0 cut
Here, 2[ (0)(3) + (3)(4) + (1)(2) ] = 8 + 20 ⇒ 2[14] = 28 . The condition is satisfied. Thus the spheres cut orthogonally.
14. Find the radius of the sphere whose centre is (1,2,3), which touches the plane 2x + 2y –z = 2
Radius of the sphere = Perpendicular distance from the centre to the tangent plane.
= 2(1) + 2(2) − 3 − 2 =
2 + 4 − 5 = 1
4 + 4 + 1 9 3
15. What do you mean by Enveloping cone?
The locus of the tangent lines from a given point to a sphere is a cone called the enveloping cone from the point to the sphere.
16. Find the equation of the right circular cylinder whose axis is the z-axis and radius a.
Let P(x,y,z) be any point on the cylinder. Draw PM ⊥Z-axis, so that OM = z. Then the point M is (0,0,z)
Also, MP = Radius of the cylinder ⇒
( x − 0)2 + ( y − 0)2 + ( z − z)2 = a2 ⇒ x2 + y2 = a2
is the equation
of the right circular cylinder.
6
17. Define right circular cylinder.
A right circular cylinder is the surface generated by a straight line which is parallel to a fixed line and is at a constant distance from it.
18. Define Great circle.
The section of a sphere by a plane passing through its centre is called a great circle. The centre and radius of the great circle are the same as the centre and radius of the sphere.
19. Find the equation of the cone whose vertex is the origin and base the circle x = a; y2 +z2 = b2.
Homogenising, y2 +z2 = b2(12) = b2(x/a)2 ⇒ a2(y2 +z2 ) = x2 b2.
20. Define cone.
A cone is the surface generated by a straight line which passes through a fixed point and is subjected to one more condition (i.e) it may intersect a given curve, touch a given surface or make a given angle with
a line through the fixed point.
PART – B
1. Find the equation of the cone with vertex at (1,1,1) and passing through the curve of intersection of
x2 + y2 + z 2 = 1
and x + y + z = 1.
2. Find the two tangent planes to the sphere
x + 4y + 8z = 0. Find their point contact.
x2 + y2 + z 2 − 4 x − 2 y − 6z + 5 = 0
which are parallel to the plane
3. Find the equation of the cone formed by rotating the line 2x + 3y = 5; z = 0 about the y – axis.
4. Find the equation of the right circular cylinder of radius 3 and axis
x −1 = y − 3 = z − 5 . .
2 2 −1
5. Find the equation of the right circular cone whose vertex is at the origin and axis is the line having semi vertical angle of 30°.
UNIT – III DIFFERENTIAL CALCULS PART – A
1. Find the radius of curvature at x=90o on the curve y=4 sin x.
y = 4sin x
Differentiating w.r.t x , y1 = 4 cos x ,y2= -4 sin x
At x=90o , y1= 0 , and y2= -4
x = y = z .
1 2 3
3
[1 + y 2 ] 2
The radius of curvature is ρ = 1
y 2
3
= [1 + 0] 2
− 4
= -1/4 ∴ ρ = 1/4
2. Find the radius of curvature y=ex at the point where the curve cuts the y axis.
The curve y =ex cuts the y axis at x=0 y1=ex, and y2= ex
At x=0 , y1= 1 and y2= 1
7
The radius of curvature is ρ =
,3
[1 + y 2 ] 2
1
y2
3
[1 + 1] 2
=
1
= 23/2 = 2√2
3. What is the curvature of the curve x2+y2 –2x–4y-4=0?
The given curve is a circle 2g = -2, 2f = -4, c= - 4 ∴g = -1 , f = -2
radius, r =
12 + 2 2 + 4 = 3
∴ the radius of curvature = Radius of the circle = 3
Hence the curvature = 1/r = 1/3
4. Write down the formula for the radius of curvature x=f(t),y=g(t) given in parametric coordinate at any point t.
The radius of curvature ρ =
[ x '2
3
+ y '2 ] 2
x′y′ − x′ y′
5. Find the radius of curvature at any point on y = c log sec (x/c).
1 ⎛ x ⎞
⎛ x ⎞ 1
⎛ x ⎞
, y = ,1
2 ⎛ x ⎞
y1 = c.
sec⎜
⎟ tan⎜ ⎟
= tan⎜ ⎟
2 Sec ⎜ ⎟
⎛ x ⎞ c c c c
c ⎝ c ⎠
sec⎜ ⎟
⎝ ⎠ ⎝ ⎠ ⎝ ⎠
⎝ c ⎠
2 ⎛ x ⎞ 3
[1 + tan ⎜
⎟] 2
ρ = ⎝ c ⎠ .
⎛ x ⎞
= c sec ⎛ x ⎞ .
c
1 sec 2 ⎜ ⎟ ⎝ ⎠
c ⎝ c ⎠
6. Find ρ at any point on the curve r = eθ
r= eθ
Differentiating w.r.t θ, we get r1 = eθ and r2 = eθ
2 2 3
2θ 2θ 3
R.O.C in Polar Coordinate is ρ =
[r + r1 ] 2 = [e
2 2 2θ
+ e ] 2
2θ 2θ
2e 2θ
=√2 eθ =√2 r
r + 2r1
− rr2
e + 2e − e
7. State the formula for the co-ordinates of centre of curvature.
2
X = x –
y1[1 + y1 ]
y 2
, Y = y + [1 + y1 ] .
y 2
8. Define the curvature of a plane curve and what is the curvature of a straight line?
The rate of bending of a curve is called curvature .Curvature of straight line is zero.
9. Write down the equation of circle of curvature .
− −
Circle of Curvature is (x − x) + (y − y) = ρ 2
10. Find the curvature of the curve 2x2+2y2 +5x-2y+1=0
Radius of the circle =
g 2 + f 2 − d =
25 + 1 − 1 = , 21 = radius of curvature ∴Curvature = 1 = 4
11. Define Evolute.
16 4 2 4
ρ 21
The locus of centre of curvature of a curve is called Evolute.
12. State any two properties of evolute.
(i) The normals drawn to a curve become tangents to its evolute.
(ii)The difference between the radii of curvature at two points of a curve is equal to the length of the evolute between the two coressonding points
8
13. Find the co-ordinates of the centre of curvataure of the curve y = x2 at the point (1,-1)
x = x − 1 (1 + y 2 ) = − 4
2
(1 + y 2 ) 3
y = y + 1 =
y 2 2
∴centre is ⎛ − 4, 3 ⎞ ⎝ 2 ⎠
14. Find the evolute of the curve whose the coordinates of the centre of curvature of the curve is
x = 2a + 2
3
, y = − 3
2
⎛ x −
⎜
2a ⎞
⎟
⎛ y ⎞
= t 6 = ⎜ ⎟
2 3
⇒ 27a y = 4 x − 2a
⎜ 3a ⎟
⎜ 2a ⎟
⎝ ⎠ ⎝ ⎠
15. Find the curvature at (3,-4) for the curve x2+y2=25
Radius = 5 therefore radius of curvature at any point is 1/5.
16. Find the envelope of the family of lines x/t +yt = 2c, t being the parameter.
x/t +yt = 2c--------(1)
Diif. w.r.to ‘t’ partially, we get
-x/t2 +y = 0 t2 = x/y
t = √x/y--------(2)
Using (2) in (1), we get xy = c2
17. Find the envelope of (x - α)2 + y2 = 4α, α being the parameter.
(x - α)2 + y2 = 4α ---------------(1)
Differentiating partially w.r.to α, we get
2(x - α)(-1) = 4 ⇒ α = x + 2
Substituting α in (1), we get
(-2)2 + y2 = 4(x+2) ⇒ y2 = 4(x+1) is the required envelope.
18. Find the envelope of the family of straight lines xcosα + ysinα = a secα, α being the parameter.
xcosα + ysinα = a secα-----------------(1)
÷ cosα ⇒ x + ytanα = a( 1+tan2α)
atan2α - ytanα + a – x = 0 which is a quadratic in tanα.
∴The envelope is B2 – 4AC = 0.
(-y)2 – 4a(a – x) = 0 ⇒ y2 = 4a(a – x).
19. Find the envelope of the family of lines (x/a) cosθ + (y/b) sinθ =1, θ being the parameter.
(x/a) cosθ + (y/b) sinθ =1 --------(1)
Diif. w.r.to ‘θ’ partially, we get
-(x/a) sinθ + (y/b) cosθ =1 --------(2)
(1)2 + (2)2 ⇒ x a2
(cos2θ+sin2θ) +
y2
2 (sin2θ+cos2θ) = 1 ⇒
b
x 2 y 2
+ = 1.
a 2 b 2
20. Find the envelope of the family of straight line y = mx + a/m , m being the parameter
Given y = mx + a/m--------(1)
my = m2x + a
m2x -my + a = 0 which is quadratic in m
∴ The envelope of (1) is B2 – 4AC = 0 i.e ) y2 = 4ax
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PART – B
1. Find the equation of the circle of curvature of the curve x +
y = a at (a 4 , a 4) .
2. Find the equation of the circle of curvature of the curve x 3 + y3 = 3axy at ⎛ 3a
⎝ 2
3a ⎞
, ⎟
2 ⎠
3. Find the evolute of the curve
x + y = 1
,treating it as the envelope of normals.
a 2 b 2
4. Find the evolute of the rectangular hyperbola xy = c2.
5. Find the envelope of the straight line
x + y = 1 where a and b are parameters that are connected by the
a b
relation a + b = c.
UNIT – IV FUNCTIONS OF SEVERAL VARIABLES PART – A
1. Write the sufficient conditions for f(x,y) to have a maximum value at (a,b).
(i) fx(a,b) = 0, fy(a,b) = 0 .
(ii)If A = fxx(a,b) , B = fxy(a,b), C = fyy(a,b) and ∆=AC-B2, then f(x,y) is maximum at (a,b) if ∆ > 0 and
A 0 , A> 0 .
∴f is minimum at (-2,0) and the minimum value is f(-2,0) = 24.
5. Find the stationary points of f(x ,y) = x2 – x y + y2 -2x +y.
fx = 2x – y -2 = 0 , fy = -x +2y +1 = 0 .
2x – y = 2, -2x + 4y = -2. ⇒ y = 0, x = 1.
∴ The stationary point is (1 ,0).
6. Find Taylor’s series expansion of x y near the point (1,1) up to first degree terms.
f(x ,y) = f(a ,b) + [(x-a) f x(a ,b) + (y-b) f y(a ,b)]. f(x ,y) = x y , f(1,1) = 1
f x(x , y)=yxy-1, f x(1,1) = 1. f y (x ,y) = x y log x , f y(1,1)= 0,.
f (x ,y) = 1 + [ (x-1)(1) + (y-1) (0)]. f(x ,y) = x.
7. Find Taylor’s series expansion of ex sin y near the point (-1, ) up to first degree terms.
f (x , y) = ex sin y , f x(x , y) = ex sin y, fy (x ,y) = ex cos y
1 ⎡ + ( x + 1) + ⎛ y − π ⎞⎤ .
f(-1, π ) =
4
1 , fx(-1, π ) =
e 2 4
1 , fy(-1, π )= 1
e 2 4 e 2
. ∴ ex sin y =
⎢1
e 2 ⎣
⎜ ⎟⎥ ⎝ ⎠⎦
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8. Expand ex cosy in Taylor’s series in powers of x and y up to terms of first degree.
f(x,y) = ex cosy , f(0,0) = 1 fx(x,y) = ex cosy , fx(0,0) = 1 fy(x,y) = - ex siny, fy(0,0) =0. ex cosy = 1 + x .
9. State sufficient conditions for a function of two variables to have an extreme at a point .
The function f(x,y) has an extreme value at a point (a,b) if fx(a,b) = 0, fy(a,b) = 0 .
and AC – B2 > 0 and A>0 , A
How to do Cartesian plane equations? In order to solve the problem you need to: Choose your x-values to input into the formula. Usually you’ll want to choose integers (whole numbers plus zero and negatives like -10, 3, 1, 7…) because they are easier to ... Create a few (x, y) coordinates. Plot the points on a Cartesian plane. Draw the curve or line of the graph.
