Subject: Mathematics-II for EC Stream

Subject code: BMATE201

Department of Mathematics

Acharya Institute of Technology

Bangalore - 560107

May 5, 2023
Department of Mathematics (AIT) Mathematics-II for EC stream ( 1/32) May 5, 2023 1 / 32
 Module-1
VECTOR CALCULUS

VECTOR DIFFERENTIATION:
Definition
Vector is a quantity having both magnitude and direction. Vector
quantities like force, velocity, acceleration etc.

Basic concepts: Let the position vector of a point P(x,y,z) in space

be ⃗r = xi + yj + zk. If x, y, z are all functions of a single parameter t,
then ⃗r is said to be a vector function of t which is also referred to as
a vector point function usually denoted as ⃗r = ⃗r (t) .

Department of Mathematics (AIT) Mathematics-II for EC stream ( 2/32) May 5, 2023 2 / 32

As the parameter t varies, the point P traces a curve in space.
Therefore ⃗r (t) = x(t)i + y (t)j + z(t)k is called as the vector
equation of the curve.
d⃗r
dt
= ⃗r ′ (t) = dx
dt
i + dy
dt
j + dz
dt
k is a vector along the tangent to the
curve at P.
If t is the time variable, ⃗v = d⃗ r
dt
gives the velocity of the particle at
time t.

Department of Mathematics (AIT) Mathematics-II for EC stream ( 3/32) May 5, 2023 3 / 32

 2
Further ⃗a = d⃗ v
dt
= ddt 2⃗r represents the rate of change of velocity ⃗v and
is called the acceleration of the particle at time t.
Properties:
1 d
[c r⃗ (t) ± c2 r⃗2 (t)] = [c1 r⃗1 ′ (t) ± c2 r⃗2 ′ (t)] where c1 andc2 are
dt 1 1
constants
d ⃗ ⃗ ⃗ ⃗
2
dt
(F .G ) = F⃗ . ddtG + ddtF .G⃗
d ⃗ ⃗ ⃗
3
dt
(F × G⃗ ) = F⃗ × ddtG + ddtF × G⃗
where F⃗ = F⃗ (t)and G⃗ = G⃗ (t)

Department of Mathematics (AIT) Mathematics-II for EC stream ( 4/32) May 5, 2023 4 / 32

Partial Derivatives of vectors:
If F⃗ is a vector depending on more than one scalar variable say, x,y,z,
for eg, then we write F⃗ = F⃗ (x, y , z). The Partial derivative of F⃗ with
respect to x is defined as

∂ F⃗ F⃗ (x + ∆x, y , z) − F⃗ (x, y , z)
= lim (1)
∂x ∆x→0 ∆x
∂F ⃗ F (x, y + ∆y , z) − F⃗ (x, y , z)
⃗
= lim (2)
∂y ∆y →0 ∆y
∂F⃗ F (x, y , z + ∆z) − F⃗ (x, y , z)
⃗
= lim , (3)
∂z ∆z→0 ∆z
are the partial derivatives with respect to x, y and z respectively, if
these limit exists.
Department of Mathematics (AIT) Mathematics-II for EC stream ( 5/32) May 5, 2023 5 / 32
The higher derivatives can be defined as
∂ 2 F⃗ ∂ ∂ F⃗
∂x 2
= ∂x ( ∂x )
2
∂ F ⃗ ∂ ∂ F⃗
∂y 2
= ∂y ( ∂y )
∂ 2 F⃗ ∂ ∂F ⃗
∂z 2
= ∂z ( ∂z )
∂ 2 F⃗ ∂ ∂ F⃗ ∂ 2 F⃗ ∂ ∂ F⃗
∂x∂y
= ∂x ( ∂y ) and ∂y ∂x
= ∂y ( ∂x )
Vector differential operator:
The Vector differential operator Del or Nabla, denoted by ∇ is
defined as
∂ ∂ ∂
∇ = ∂x i + ∂y j + ∂z k
Gradient of a scalar function: If ϕ(x, y , z) is a continuously
differentiable scalar function then the gradient of ϕ is defined as
gradϕ = ∇ϕ = ∂ϕ ∂x
i + ∂ϕ∂y
j + ∂ϕ
∂z
k

Department of Mathematics (AIT) Mathematics-II for EC stream ( 6/32) May 5, 2023 6 / 32

Directional Derivative:
If ϕ(x, y , z) is a scalar point function and d⃗ is a vector then the
directional derivative of ϕ in the direction of d⃗ is given by ∇ϕ.n̂
⃗
where n̂ = |dd⃗ | .

Theorem
If ϕ(x, y , z) is a scalar function then gradϕ is a vector normal to the
surface ϕ(x, y , z) = c.

Proof: Let ⃗r = x(t)i + y (t)j + z(t)k be the position vector of any

point P(x,y,z) on the surface ϕ(x, y , z) = c.
Then d⃗r
dt
= dxdt
i + dy
dt
j + dz
dt
k is tangent to the surface at P.
∂ϕ
We have ∇ϕ = ∂x i + ∂y j + ∂ϕ
∂ϕ
∂z
k.
Therefore ∇ϕ. dt = ∂x dt + ∂y dt + ∂ϕ
d⃗
r ∂ϕ dx ∂ϕ dy dz
∂z dt

Department of Mathematics (AIT) Mathematics-II for EC stream ( 7/32) May 5, 2023 7 / 32

Since ϕ(x, y , z) = c where x=x(t), y=y(t), z=z(t) and c is constant.
On differentiating w.r.t. t both sides we have
∂ϕ dx ∂ϕ dy ∂ϕ dz
+ + =0
∂x dt ∂y dt ∂z dt
d⃗r
∴ ∇ϕ. =0
dt
That implies ∇ϕ is perpendicular to d⃗ r
dt
Hence d⃗r
dt
is a vector tangential to the surface at P. We conclude that
∇ϕ is a normal vector to the surface ϕ(x, y , z) = c at P.
Note: the directional derivative of a scalar function ϕ at any point is
maximum along ∇ϕ.

Department of Mathematics (AIT) Mathematics-II for EC stream ( 8/32) May 5, 2023 8 / 32

PROBLEMS:

1. Find the unit vector normal to the surface x 2 y + 2xz = 4 at

(2, −2, 3)
Solution: Let ϕ = x 2 y + 2xz.
∇ϕ is a vector normal to the surface
∂ϕ ∂ϕ ∂ϕ
We have ∇ϕ = ∂x
i + ∂y
j + ∂z
k
∂(x 2 y +2xz) ∂(x 2 y +2xz) ∂(x 2 y +2xz)
∇ϕ = ∂x
i + ∂y
j + ∂z
k
therefore ∇ϕ = (2xy + 2z)i + x 2 j + 2xk
[∇ϕ](2,−2,3) = −2i + 4j + 4k
(∇ϕ) (−2i+4j+4k) (−i+2j+2k)
The required unit normal vector n̂ = |∇ϕ|
=√ = 3
.
(4+16+16)

Department of Mathematics (AIT) Mathematics-II for EC stream ( 9/32) May 5, 2023 9 / 32

2. Find the unit vector normal to the surface
x 3 + y 3 + z 3 + 3xyz = 3 at (1, 2, −1). (Jan-2018)
Solution: Let ϕ = x 3 + y 3 + 3xyz.
∇ϕ is a vector normal to the surface
∂ϕ ∂ϕ ∂ϕ
We have ∇ϕ = ∂x
i + ∂y
j + ∂z
k
∂(x 3 +y 3 +z 3 +3xyz) ∂(x 3 +y 3 +z 3 +3xyz) ∂(x 3 +y 3 +z 3 +3xyz)
∇ϕ = ∂x
i + ∂y
j + ∂z
k
therefore ∇ϕ = (3x 2 + 3yz)i + (3y 2 + 3xz)j + (3z 2 + 3xy )k
[∇ϕ](1,2,−1) = −3i + 9j + 9k
The required unit normal vector
n̂ = (∇ϕ)
|∇ϕ|
=√(−3i+9j+9k)
= (−3i+9j+9k)
√
171
.
(9+81+81)

Department of Mathematics (AIT) Mathematics-II for EC stream ( 10/32) May 5, 2023 10 / 32

3. Find the unit vector normal to the surface ϕ = 2xz − y 2 at (1, 3, 2)
Solution: Let ϕ = 2xz − y 2 .
∇ϕ is a vector normal to the surface
∂ϕ ∂ϕ ∂ϕ
We have ∇ϕ = ∂x
i + ∂y
j + ∂z
k
∂(2xz−y 2 ) ∂(2xz−y 2 ) ∂(2xz−y 2 )
∇ϕ = ∂x
i + ∂y
j + ∂z
k
therefore ∇ϕ = 2zi − 2yj + 2xk
[∇ϕ](1,3,2) = 4i − 6j + 2k
(∇ϕ)
The required unit normal vector n̂ = |∇ϕ|
(4i−6j+2k) (4i−6j+2k)
n̂ = √ = √
56
.
(16+36+4)

Department of Mathematics (AIT) Mathematics-II for EC stream ( 11/32) May 5, 2023 11 / 32

4. Find the unit vector normal to the surface
x 2 y − 2xz + 2y 2 z 4 = 10 at (2, 1, −1)
Solution: Let ϕ = x 2 y − 2xz + 2y 2 z 4 .
∇ϕ is a vector normal to the surface
∂ϕ ∂ϕ ∂ϕ
We have ∇ϕ = ∂x
i + ∂y
j + ∂z
k
∂(x 2 y −2xz+2y 2 z 4 ) ∂(x 2 y −2xz+2y 2 z 4 ) ∂(x 2 y −2xz+2y 2 z 4 )
∇ϕ = ∂x
i + ∂y
j + ∂z
k
therefore ∇ϕ = (2xy − 2z)i + (x 2 + 4yz 4 )j + (−2x + 8y 2 z 3 )k
[∇ϕ](2,1,−1) = 6i + 8j − 12k = 2(3i + 4j − 6k)
(∇ϕ)
The required unit normal vector n̂ = |∇ϕ|

n̂ = √2(3i+4j−6k)
2
= (3i+4j−6k)
√
61
.
(2 (9+16+36)

Department of Mathematics (AIT) Mathematics-II for EC stream ( 12/32) May 5, 2023 12 / 32

5. Find the unit vector normal to the surface x 2 y + y 2 z + z 2 x = 5 at
(1, 2, −1) (July-2017)
Solution: Let ϕ = x 2 y + y 2 z + z 2 x.
∇ϕ is a vector normal to the surface
∂ϕ ∂ϕ ∂ϕ
We have ∇ϕ = ∂x
i + ∂y
j + ∂z
k
∂(x 2 y +y 2 z+z 2 x) ∂(x 2 y +y 2 z+z 2 x) ∂(x 2 y +y 2 z+z 2 x)
∇ϕ = ∂x
i + ∂y
j + ∂z
k
therefore ∇ϕ = (2xy + z 2 )i + (x 2 + 2yz)j + (y 2 + 2xz)k
[∇ϕ](1,2,−1) = 5i − 3j + 2k
(∇ϕ)
The required unit normal vector n̂ = |∇ϕ|
(5i−3j+2k)
n̂ = √5i−3j+2k = √
38
.
(25+9+4)

Department of Mathematics (AIT) Mathematics-II for EC stream ( 13/32) May 5, 2023 13 / 32

6. If ϕ = x 3 + y 3 + z 3 − 3xyz, find ∇ϕ, |∇ϕ| at (2,1,-2)
Solution: Let ϕ = x 3 + y 3 + z 3 − 3xyz.
∂ϕ ∂ϕ ∂ϕ
We have ∇ϕ = ∂x
i + ∂y
j + ∂z
k
∂(x 3 +y 3 +z 3 −3xyz) ∂(x 3 +y 3 +z 3 −3xyz) ∂(x 3 +y 3 +z 3 −3xyz)
∇ϕ = ∂x
i + ∂y
j + ∂z
k
∇ϕ = (3x 2 − 3yz)i + (3y 2 − 3xz)j + (3z 2 − 3xy )k
[∇ϕ](2,1,−2) = 18i + 15j + 6k = 3(6i + 5j + 2k)
p √
therefore |∇ϕ|(2,1,−2) = 32 (62 + 52 + 22 ) = 585.

Department of Mathematics (AIT) Mathematics-II for EC stream ( 14/32) May 5, 2023 14 / 32

7. If ϕ = xy + yz + zx and F⃗ = x 2 yi + y 2 zj + z 2 xk, find F⃗ · gradϕ
and F⃗ × gradϕ at the point (3, −1, 2)
Solution: Given ϕ = xy + yz + zx and F⃗ = x 2 yi + y 2 zj + z 2 xk
∂ϕ ∂ϕ ∂ϕ
We have gradϕ = ∂x
i + ∂y
j + ∂z
k
∂(xy +yz+zx) ∂(xy +yz+zx) ∂(xy +yz+zx)
gradϕ = ∂x
i + ∂y
j + ∂z
k
gradϕ = (y + z)i + (x + z)j + (x + y )k
therefore
F⃗ · gradϕ = (x 2 yi + y 2 zj + z 2 xk) · [(y + z)i + (x + z)j + (x + y )k]
F⃗ · gradϕ = x 2 y (y + z) + y 2 z(x + z) + z 2 x(x + y )
F⃗ · gradϕ = x 2 y 2 + x 2 yz + y 2 zx + y 2 z 2 + x 2 z 2 + z 2 xy
[F⃗ · gradϕ](3,−1,2) = 9 − 18 + 6 + 4 + 36 − 12 = 25.

Department of Mathematics (AIT) Mathematics-II for EC stream ( 15/32) May 5, 2023 15 / 32

 i j k
F⃗ × gradϕ = x 2y y 2z z 2x
y +z x +z x +y

F⃗ × gradϕ = i(xy 2 z + y 3 z − z 2 x 2 − z 3 x)
−j(x 3 y + x 2 y 2 − z 2 xy − z 3 x)
+k(x 3 y + x 2 yz − y 3 z − y 2 z 2 )

[F⃗ × gradϕ](3,−1,2) = −56i + 30j − 47k

Department of Mathematics (AIT) Mathematics-II for EC stream ( 16/32) May 5, 2023 16 / 32

8. In which direction the directional derivative of ϕ = xy 2 z 3 is
maximum at (−1, 1, 2). What is the magnitude of this maximum?
Solution: Given ϕ = xy 2 z 3
∂ϕ ∂ϕ ∂ϕ
We have ∇ϕ = ∂x
i + ∂y
j + ∂z
k
∂(xy 2 z 3 ) ∂(xy 2 z 3 ) ∂(xy 2 z 3 )
∇ϕ = ∂x
i + ∂y
j + ∂z
k
∇ϕ = y 2 z 3 i + 2xyz 3 j + 3xy 2 z 2 k
[∇ϕ](−1,1,2) = 8i − 16j − 12k is the direction in which it is maximum.
therefore the magnitude
√ of this is given
√ by
|∇ϕ|(−1,1,2) = 64 + 256 + 144 = 464

Department of Mathematics (AIT) Mathematics-II for EC stream ( 17/32) May 5, 2023 17 / 32

9. In which direction the directional derivative of ϕ = x 2 yz 3 is
maximum at (2, 1, −1). What is the magnitude of this maximum?
Solution: Given ϕ = x 2 yz 3
∂ϕ ∂ϕ ∂ϕ
We have ∇ϕ = ∂x
i + ∂y
j + ∂z
k
∂(x 2 yz 3 ) ∂(x 2 yz 3 ) ∂(x 2 yz 3 )
∇ϕ = ∂x
i + ∂y
j + ∂z
k
∇ϕ = 2xyz 3 i + x 2 z 3 j + 3x 2 yz 2 k
[∇ϕ](2,1,−1) = −4i − 4j + 12k is the direction in which it is maximum.
therefore the magnitude
√ of this is given
√ by
|∇ϕ|(2,1,−1) = 16 + 16 + 144 = 4 11

Department of Mathematics (AIT) Mathematics-II for EC stream ( 18/32) May 5, 2023 18 / 32

10. Find the direction that a person standing at the origin should
move to get warm as quickly as possible given that the temperature
field is xsinz − ycosz.
Solution: Given ϕ = xsinz − ycosz
∂ϕ ∂ϕ ∂ϕ
We have ∇ϕ = ∂x
i + ∂y
j + ∂z
k
∂(xsinz−ycosz) ∂(xsinz−ycosz) ∂(xsinz−ycosz)
∇ϕ = ∂x
i + ∂y
j + ∂z
k
∇ϕ = sinzi − coszj + (xcosz + ysinz)k
[∇ϕ](0,0,0) = −j is the required direction which means to stay that
the person should move down along the y-axis to get warm quickly.

Department of Mathematics (AIT) Mathematics-II for EC stream ( 19/32) May 5, 2023 19 / 32

11. If the directional derivative of ϕ = axy 2 + byz + cz 2 x 3 at the
point (−1, 1, 2) has a maximum magnitude of 32 units in the
direction parallel to y-axis find a,b,c.
Solution: Maximum directional derivative is along ∇ϕ and in the
direction parallel to y-axis the magnitude is given to be 32 units.
Therefore ∇ϕ.j = 32 at (−1, 1, 2)
∂ϕ ∂ϕ ∂ϕ
We have ∇ϕ = ∂x
i + ∂y
j + ∂z
k
∂(axy 2 +byz+cz 2 x 3 ) ∂(axy 2 +byz+cz 2 x 3 ) ∂(axy 2 +byz+cz 2 x 3 )
∇ϕ = ∂x
i + ∂y
j + ∂z
k
∇ϕ = (ay 2 + 3cx 2 z 2 )i + (2axy + bz)j + (by + 2cx 3 z)k
[∇ϕ](−1,1,2) = (a + 12c)i + (−2a + 2b)j + (b − 4c)k

Department of Mathematics (AIT) Mathematics-II for EC stream ( 20/32) May 5, 2023 20 / 32

Now ∇ϕ.j = −2a + 2b = 32 or −a + b = 16
Also since ∇ϕ is parallel to the y-axis we must have
a+12c=0 and b-4c=0
By solving the three equations we get
a=-12, b=4, c=1
12. If the directional derivative of ϕ = axy 2 + byz + cz 2 x 3 at the
point (1, 2, −1) has a maximum magnitude of 64 units in the
direction of z-axis find a,b,c. (Jan-2018)
Solution: Maximum directional derivative is along ∇ϕ and in the
direction of z-axis the magnitude is given to be 64 units.
Therefore ∇ϕ.k = 64 at (1, 2, −1)
∂ϕ ∂ϕ ∂ϕ
We have ∇ϕ = ∂x
i + ∂y
j + ∂z
k

Department of Mathematics (AIT) Mathematics-II for EC stream ( 21/32) May 5, 2023 21 / 32

 ∂(axy 2 +byz+cz 2 x 3 ) ∂(axy 2 +byz+cz 2 x 3 ) ∂(axy 2 +byz+cz 2 x 3 )
∇ϕ = ∂x
i + ∂y
j + ∂z
k
∇ϕ = (ay 2 + 3cx 2 z 2 )i + (2axy + bz)j + (by + 2cx 3 z)k
[∇ϕ](1,2,−1) = (4a + 3c)i + (4a − b)j + (2b − 2c)k
Now ∇ϕ.k = 2b − 2c = 64 or b − c = 32
Also since ∇ϕ is parallel to the z-axis we must have
4a+3c=0 and 4a-b=0
By solving the three equations we get
a=6, b=24, c=-8

Department of Mathematics (AIT) Mathematics-II for EC stream ( 22/32) May 5, 2023 22 / 32

13. Find the directional derivative of ϕ = 2xy + 5yz + zx at the point
(1, 2, 3) in the direction of 3i − 5j + 4k
Solution: Given ϕ = 2xy + 5yz + zx and direction d⃗ = 3i − 5j + 4k
∂ϕ ∂ϕ ∂ϕ
We have ∇ϕ = ∂x
i + ∂y
j + ∂z
k
∂(2xy +5yz+zx) ∂(2xy +5yz+zx) ∂(2xy +5yz+zx)
∇ϕ = ∂x
i + ∂y
j + ∂z
k
∇ϕ = (2y + z)i + (2x + 5z)j + (x + 5y )k
[∇ϕ](1,2,3) = 7i + 17j + 11k
The unit vector in the direction of d⃗ = 3i − 5j + 4k is
3i−5j+4k 3i−5j+4k
n̂ = √
9+25+16
= √
50
the required directional derivative is
∇ϕ.n̂ = (7i + 17j + 11k).( 3i−5j+4k
√
50
)= 21−85+44
√
50
= −20
√
50

Department of Mathematics (AIT) Mathematics-II for EC stream ( 23/32) May 5, 2023 23 / 32

14. Find the directional derivative of ϕ = x 2 yz + 4xz 2 at the point
(1, −2, −1) in the direction of 2i − j − 2k
Solution: Given ϕ = x 2 yz + 4xz 2 and direction d⃗ = 2i − j − 2k
∂ϕ ∂ϕ ∂ϕ
We have ∇ϕ = ∂x
i + ∂y
j + ∂z
k
∂(x 2 yz+4xz 2 ) ∂(x 2 yz+4xz 2 ) ∂(x 2 yz+4xz 2 )
∇ϕ = ∂x
i + ∂y
j + ∂z
k
∇ϕ = (2xyz + 4z 2 )i + (x 2 z)j + (x 2 y + 8xz)k
[∇ϕ](1,−2,−1) = 8i − j − 10k
The unit vector in the direction of d⃗ = 2i − j − 2k is
2i−j−2k 2i−j−2k
n̂ = √
4+1+4
= 3

the required directional derivative is

∇ϕ.n̂ = (8i − j − 10k).( 2i−j−2k
3
)= 16+1+20
3
= 37
3

Department of Mathematics (AIT) Mathematics-II for EC stream ( 24/32) May 5, 2023 24 / 32

15. Find the directional derivative of ϕ = 4xz 3 − 3x 2 y 2 z at the point
(2, −1, 2) in the direction of 2i − 3j + 6k (Jan-2016, Jan-2020)
Solution: Given ϕ = 4xz 3 − 3x 2 y 2 z and direction d⃗ = 2i − 3j + 6k
∂ϕ ∂ϕ ∂ϕ
We have ∇ϕ = ∂x
i + ∂y
j + ∂z
k
∂(4xz 3 −3x 2 y 2 z) ∂(4xz 3 −3x 2 y 2 z) ∂(4xz 3 −3x 2 y 2 z)
∇ϕ = ∂x
i + ∂y
j + ∂z
k
∇ϕ = (4z 3 − 6xy 2 z)i − (6x 2 yz)j + (12xz 2 − 3x 2 y 2 )k
[∇ϕ](2,−1,2) = 8i + 48j + 84k
The unit vector in the direction of d⃗ = 2i − 3j + 6k is
2i−3j+6k 2i−3j+6k
n̂ = √
4+9+36
= 7

the required directional derivative is

∇ϕ.n̂ = (8i + 48j + 84k).( 2i−3j+6k
7
)= 376
7

Department of Mathematics (AIT) Mathematics-II for EC stream ( 25/32) May 5, 2023 25 / 32

16. Find the directional derivative of ϕ = xy 3 + yz 3 at the point
(2, −1, 1) in the direction of i + 2j + 2k. (July-2017)
Solution: Given ϕ = xy 3 + yz 3 and direction d⃗ = i + 2j + 2k
∂ϕ ∂ϕ ∂ϕ
We have ∇ϕ = ∂x
i + ∂y
j + ∂z
k
∂(xy 3 +yz 3 ) ∂(xy 3 +yz 3 ) ∂(xy 3 +yz 3 )
∇ϕ = ∂x
i + ∂y
j + ∂z
k
∇ϕ = (y 3 )i + (3xy 2 + z 3 )j + (3yz 2 )k
[∇ϕ](2,−1,1) = −i + 7j − 3k
The unit vector in the direction of d⃗ = i + 2j + 2k is
i+2j+2k i+2j+2k
n̂ = √
1+4+4
= 3

the required directional derivative is

∇ϕ.n̂ = (−i + 7j − 3k).( i+2j+2k
3
)= −1+14+−6
3
= 7
3

Department of Mathematics (AIT) Mathematics-II for EC stream ( 26/32) May 5, 2023 26 / 32

17. Find the directional derivative of ϕ = x 2xz
+y 2
at the point
⃗
(1, −1, 1) in the direction of d = i − 2j + k
Solution: Given ϕ = xz
x 2 +y 2
and direction d⃗ = 2i − j − 2k
∂ϕ ∂ϕ ∂ϕ
We have ∇ϕ = ∂x
i + ∂y
j + ∂z
k
xz xz xz
∂( ) ∂( ) ∂( )
x 2 +y 2 x 2 +y 2 x 2 +y 2
∇ϕ = ∂x
i+ ∂y
j+ ∂z
k
2 +y 2 )1−x(2x)
∇ϕ = z[ (x (x 2 +y 2 )2
]i + ( (x−2xyz x
2 +y 2 )2 )j + ( x 2 +y 2 )k

−z ) 2 2
∇ϕ = ( z(y
(x 2 +y 2 )2
)i − ( (x 22xyz
+y 2 )2
x
)j + ( x 2 +y 2 )k

[∇ϕ](1,−1,1) = 21 j + 21 k = 12 (j + k)
The unit vector in the direction of d⃗ = i − 2j + k is
i−2j+k i−2j+k
n̂ = √
1+4+1
= √
6

Department of Mathematics (AIT) Mathematics-II for EC stream ( 27/32) May 5, 2023 27 / 32

the required directional derivative is
∇ϕ.n̂ = 12 (j + k).( i−2j+k
√
6
)= 0−2+1
√
2 6
= −1
√
2 6

18. f (x, y , z) = x 2 y 2 z 2 at the point (1, 1, −1) in the direction of the

tangent to the curve x = e t , y = 1 + 2sint, z = t − cost where
−1 ≤ t ≤ 1
Solution: Given f = x 2 y 2 z 2
∂ϕ ∂ϕ ∂ϕ
We have ∇f = ∂x
i + ∂y
j + ∂z
k
∇f = 2xy 2 z 2 i + 2x 2 yz 2 j + 2x 2 y 2 zk
[∇f ](1,1,−1) = 2i + 2j − 2k) = 2(i + j − k)
In order to find the direction of the tangent, let ⃗r = xi + yj + zk
⃗r = e t i + (1 + 2sint)j + (t − cost)k
d⃗
r
dt
= e t i + 2costj + (1 + sint)k is the tangent vector.
Department of Mathematics (AIT) Mathematics-II for EC stream ( 28/32) May 5, 2023 28 / 32
We have P = (x, y , z) = (1, 1, −1) by data implies
e t = 1, 1 + 2sint = 1, t − cost = −1
here e t = 1 ⇒ t = 0
therefore [ d⃗
r
]
dt t=0
= i + 2j + k
i+2j+k
The unit vector in this direction is n̂ = √
6
the required directional derivative is
∇f .n̂ = 2(i + j − k).( i+2j+k
√
6
)= √2 [1
6
+ 2 − 1] = √4
6

19. Find the directional derivative of ϕ = xy 2 + yz 3 at the

P(2, −1, 1) in the direction of the normal to the surface
xlogz − y 2 = −4 taken at the point Q(−1, 2, 1)

Department of Mathematics (AIT) Mathematics-II for EC stream ( 29/32) May 5, 2023 29 / 32

Solution:
Given ϕ = xy 2 + yz 3
∂ϕ ∂ϕ ∂ϕ
We have ∇ϕ = ∂x
i + ∂y
j + ∂z
k
∇ϕ = y 2 i + (2xy + z 3 )j + 3yz 2 k
[∇ϕ](2,−1,1) = i − 3j − 3k
The equation of the given surface S is Ψ(x, y , z) = −4 where
Ψ(x, y , z) = xlogz − y 2
Now we find ∇Ψ = logzi − 2yj + xz k normal to the surface
[∇Ψ](−1,2,1) = −4j − k
Ψ −4j−k −4j−k
n̂ = |Ψ|
= √
16+1
= √
17
the required directional derivative is
∇ϕ.n̂ = (i − 3j − 3k).( −4j−k
√
17
)= 0+12+3
√
17
= √15
17
Department of Mathematics (AIT) Mathematics-II for EC stream ( 30/32) May 5, 2023 30 / 32
20. Find the directional derivative of xyz along the direction of the
normal to the surface xy 2 + yz 2 + zx 2 = 3 at the P(1, 1, 1).
Solution: Let ϕ = xyz
∂ϕ ∂ϕ ∂ϕ
We have ∇ϕ = ∂x
i + ∂y
j + ∂z
k
∇ϕ = yzi + zxj + xyk implies [∇ϕ](1,1,1) = i + j + k
The equation of the given surface S is Ψ(x, y , z) = 3 where
Ψ = xy 2 + yz 2 + zx 2
Now we find ∇Ψ = (y 2 + 2xz)i + (2xy + z 2 )j + (2yz + x 2 )k normal
to the surface
[∇Ψ](1,1,1) = 3(i + j + k)
Ψ 3(i+j+k) (i+j+k)
n̂ = |Ψ|
= √
3 1+1+1
= √
3
the required directional derivative is
√
∇ϕ.n̂ = (i + j + k).( (i+j+k)
√
3
)= 1+1+1
√
3
= 3
Department of Mathematics (AIT) Mathematics-II for EC stream ( 31/32) May 5, 2023 31 / 32
21. Find the angle between the surfaces x 2 + y 2 + z 2 = 9,
z = x 2 + y 2 − 3 at the point (2, −1, 2). (July-2017,2018,2019)
Solution: The angle between the surfaces = Angle between the
normals.
let ϕ1 = x 2 + y 2 + z 2 and ϕ2 = x 2 + y 2 − z
∂ϕ ∂ϕ ∂ϕ
We have ∇ϕ = ∂x
i + ∂y
j + ∂z
k
∇ϕ1 = 2xi + 2yj + 2zk and ∇ϕ2 = 2xi + 2yj − k
[∇ϕ1 ](2,−1,2) = 4i − 2j + 4k and [∇ϕ2 ](2,−1,2) = 4i − 2j − k
If θ is the angle between these two normals then we have
∇ϕ1 .∇ϕ2 16+4−4 √8
cosθ = |∇ϕ1 ||∇ϕ2 |
= √ √
16+4+14 16+4+1
= 3 21

therefore θ = cos − 1( 3√821 )

Department of Mathematics (AIT) Mathematics-II for EC stream ( 32/32) May 5, 2023 32 / 32