# Spring 2008_MTH301_2_SOLved

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Solution
Assignment No: 2

Q 1.       Consider a point P(1, 2, 2) in rectangular co-ordinate system.
a. What is the position vector of this point?
b. Find the unit vector in the direction of this position vector?
Solution:
→
a)       The position vector r of the point (1, 2, 2) is

OM
→
r = i + 2 j + 2k

C
b)       Since any vector can be expressed as product of its magnitude and unit

S.
vector so we can write
→         ^
r= r r
→
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^     r
r=
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r

r = (1) 2 + (2) 2 + (2) 2
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= 9
=3
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^    i + 2 j + 2k
So,       r=
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3
1     2     2
= i+ j+ k
3 3         3
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→              →
Q 2.       Find whether the vectors a = i − 3 j + 7k ,    b = 8i − 2 j − 2k are
orthogonal to each other or not?
Solution:
As we know, if two vectors are orthogonal their dot product is zero. So let us find
the dot product of given vectors.

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→       →
a . b = (i − 3 j + 7 k ) . (8i − 2 j − 2k )
= 8(i . i ) + 6( j. j ) − 14(k .k )
= 8 + 6 − 14
=0
Thus, the two vectors are orthogonal or perpendicular to each other.

Q 3.            Find the area of the parallelogram determined by the vectors
→                               →
a = i − j + 2k and b = 3 j + k

OM
Solution:
→   →       →   →
Area of a parallelogram = magnitude of a × b = a × b

C
S.
→       →
a × b = (i − j + 2k ) × (0i + 3 j + k )
i     j     k               N
= 1 −1 2
0 3 1
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= (−1 − 6)i + (0 − 1) j + (3 − 0)k
= − 7i − j + 3k
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→    →
a × b = (−7) 2 + (−1) 2 + (3) 2
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= 49 + 1 + 9
= 59
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= 7.68
Thus area of parallelogram is 7.68 square unit
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Q 4.            Find the directional derivative of
f ( x, y ) = y 2 ln x
→
at the point (1, 4) in the direction of the vector u = − 3i + 3 j
Solution:

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f ( x, y ) = y 2 ln x
y2
f x ( x, y ) =   , f y ( x, y ) = 2 y ln x
x
f x (1, 4) = 16 , f y (1, 4) = 0
→
u = − 3i + 3 j
→              ^
u= uu
→
^          u
u=
u

OM
u = (−3) 2 + (3) 2

C
=       9+9

S.
= 18 = 3 2
^          −3i + 3 j
u=
3 2
N
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1     1
= −     i+    j
2     2
⎡                  ^
⎤              ⎡                  ^
⎤
AL

Du f (1, 4) = f x (1, 4) ⎢ x − component of u ⎥ + f y (1, 4) ⎢ y − component of u ⎥
⎣                    ⎦              ⎣                    ⎦
⎛ −1 ⎞ ⎛ 1 ⎞
= 16 ⎜      ⎟ + 0⎜     ⎟
U

⎝ 2⎠ ⎝ 2⎠
−16
RT

=
2
Q 5.              Find a unit vector in the direction in which the function
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f ( x, y ) = 4 x3 y 2 increases most rapidly at the point P(-1, 1).
Solution:
f ( x, y ) = 4 x 3 y 2
f x ( x, y ) = 12 x 2 y 2 ,     f y ( x, y ) = 8 x 3 y
∇f ( x, y ) = f x ( x, y )i + f y ( x, y ) j
= 12 x 2 y 2 i + 8 x3 y j
f x (−1, 1) =12 ,          f y (−1, 1) = −8

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∇f (−1, 1) = f x (−1, 1)i + f y (−1, 1) j
= 12 i − 8 j
^
A unit vector u in the direction of ∇f (−1, 1) is
^       ∇f (−1, 1)
u=
∇f (−1, 1)

∇f (−1, 1) = (12) 2 + (−8) 2
= 144 + 64
= 4 13

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^       ∇f (−1, 1)   12 i − 8 j
u=                 =
∇f (−1, 1)    4 13

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3     2
=        i−    j

S.
13    13

Q 6.
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Write down the general equation in parametric form for the following.
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a. Straight line in two dimensional space.
Let (x0, y0) is any fixed point on the line and the line is parallel to the
vector ai + bj then parametric form of straight line in two dimensional
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space in terms of parameter t is
x = x0 +at , y = y0 + bt
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b. Straight line in three dimensional space.
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Parametric equation of a line in three dimensional space passing through
the point (x0, y0, z0) and parallel to the vector ai + bj + ck is given by
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x = x0 +at , y = y0 + bt         ,   z = z0 +ct
c. Parametric form of curve in two dimensional space is
x = f(t) , y = g(t) , where t is parameter

Q 7.          Find the equation of
a. Tangent plane
b. Normal line
of the surface f ( x, y, z ) = x 2 y − 4 z 2 + 7 at the point (-3, 1, -2)

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Solution:
f ( x, y , z ) = x 2 y − 4 z 2 + 7
f x ( x, y, z ) = 2 xy
f y ( x, y , z ) = x 2
f z ( x, y, z ) = −8 z

f x (−3, 1, − 2) = 2(−3)(1) = − 6
f y (−3, 1, − 2) = (−3) 2 = 9
f z (−3, 1, − 2) = − 8(−2) = 16

OM
Equation of the tangent plane to the surface at (x0, y0, z0) = (-3, 1, -2) is
f x ( x0 , y0 , z0 ) ( x − x0 ) + f y ( x0 , y0 , z0 ) ( y − y0 ) + f z ( x0 , y0 , z0 ) ( z − z0 ) = 0

C
−6( x + 3) + 9( y − 1) + 16( z + 2) = 0

S.
−6 x − 18 + 9 y − 9 + 16 z + 32 = 0
−6 x + 9 y + 16 z + 5 = 0
N
Equation of the normal line of the surface through (x0, y0, z0) = (-3, 1, -2) in
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SYMMETRIC FORM is
x − x0               y − y0               z − z0
=                    =
f x ( x0 , y0 , z0 ) f y ( x0 , y0 , z0 ) f z ( x0 , y0 , z0 )
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x + 3 y −1 z + 2
=    =
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−6    9    16
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Equation of the normal line of the surface through (x0, y0, z0) = (-3, 1, -2) in
PARAMETRIC FORM is
x = x0 + f x ( x0 , y0 , z0 )
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y = y0 + f y ( x0 , y0 , z0 )
z = z0 + f z ( x0 , y0 , z0 )
Thus
x = − 3 − 6t
y = 1 + 9t
z = −2 + 16t

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Q 8.            Show that the sphere x 2 + y 2 + z 2 = a 2 and the cone z 2 = x 2 + y 2 are
orthogonal at every point of intersection.
Solution:
f ( x, y, z ) = x 2 + y 2 + z 2 − a 2 = 0 − − − − − − − − − − − − − −(1)
g ( x, y, z ) = x 2 + y 2 − z 2 = 0 − − − − − − − − − − − − − − − −(2)
f x ( x, y , z ) = 2 x
f y ( x, y , z ) = 2 y
f z ( x, y , z ) = 2 z

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g x ( x, y , z ) = 2 x
g y ( x, y , z ) = 2 y
g z ( x, y , z ) = − 2 z

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f x g x + f y g y + f z g z = (2 x)(2 x) + (2 y )(2 y ) + (2 z )(−2 z )

S.
= 4 x2 + 4 y 2 − 4 z 2
= 4( x 2 + y 2 − z 2 )
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Since from equation (2) , x 2 + y 2 − z 2 = 0 So
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f x g x + f y g y + f z g z = 4( x 2 + y 2 − z 2 ) = 0

Thus, the sphere x 2 + y 2 + z 2 = a 2 and the cone z 2 = x 2 + y 2 are orthogonal at
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every point of intersection.
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Q 9.            For evaluating RELATIVE EXTREME VALUES of a function at some
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point in the domain, what we consider as a neighborhood of a point if it lies in
a. Two dimensional space
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b. Three dimensional space
Solution:
a. For evaluating RELATIVE EXTREME VALUES of a function at some
point (x0, y0) in the domain D, neighborhood of that point is an open disc.
Open disc K ∈ D, centered at point (x0, y0) and of radius r is defined as

K = {( x, y ) ∈ R 2 : ( x − x0 ) 2 + ( y − y0 ) 2 < r 2 }

b. For evaluating RELATIVE EXTREME VALUES of a function at some
point (x0, y0, z0) in the domain D, neighborhood of that point is an open

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sphere. Open sphere S ∈ D, centered at point (x0, y0, Z0) and of radius r is

defined as S = {( x, y , z ) ∈ R 3 : ( x − x0 ) 2 + ( y − y0 ) 2 + ( z − z0 ) 2 < r 2 }

Q 10.       Locate all relative maxima, relative minima and saddle point of
f ( x, y ) = x 2 + 2 y 2 − x 2 y
Solution:
f ( x, y ) = x 2 + 2 y 2 − x 2 y
f x ( x, y ) = 2 x − 2 xy , f y ( x, y ) = 4 y − x 2

OM
The critical points of f satisfy the equations
2 x − 2 xy = 0 − − − − − − − − − −(1)

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4 y − x 2 = 0 − − − − − − − − − −(2)

S.
From equation (1)
2 x(1 − y ) = 0                              N
x(1 − y ) = 0
x = 0 , 1− y = 0 ⇒ y = 1
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Use this value of x and y in equation (2),
If x = 0 then
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4 y − x2 = 0 ⇒ 4 y = 0
⇒y=0
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If y = 1 then
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4 y − x2 = 0 ⇒ 4 − x2 = 0
x2 = 4
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x = ±2
So (0, 0), (2, 1) and (-2, 1) are the critical points
Now find second-order partial derivatives
f xx ( x, y ) = 2 − 2 y , f yy ( x, y ) = 4 , f xy ( x, y ) = − 2 x

For critical point (0, 0)
f xx (0, 0) = 2 , f yy (0, 0) = 4 , f xy (0, 0) = 0
D = f xx (0, 0) f yy (0, 0) − f xy 2 (0, 0)
= (2)(4) − 0 = 8

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D > 0 and f xx (0, 0) > 0 So f has relative minimum at (0, 0)
For critical point (2, 1)
f xx (2, 1) = 0 , f yy (2, 1) = 4 , f xy (2, 1) = − 4
D = f xx (2, 1) f yy (2, 1) − f xy 2 (2, 1)
= (0)(4) − (−4) 2
= −16
D < 0 so f has a saddle point at (2, 1)
For critical point (-2, 1)

OM
f xx ( x, y ) = 2 − 2 y , f yy ( x, y ) = 4 , f xy ( x, y ) = − 2 x
f xx (−2, 1) = 0 , f yy (−2, 1) = 4 , f xy (−2, 1) = 4

C
D = f xx (−2, 1) f yy (−2, 1) − f xy 2 (−2, 1)

S.
= (0)(4) − (4) 2
= −16
D < 0 so f has a saddle point at (-2, 1)
N
IA
U            AL
RT
VI

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