This set of Basic Engineering Mathematics Questions and Answers focuses on “Partial Differentiation – 2”.

1. Differentiation of function f(x,y,z) = Sin(x)Sin(y)Sin(z)-Cos(x) Cos(y) Cos(z) w.r.t `y` is

a) f’(x,y,z) = Cos(x)Cos(y)Sin(z) + Sin(x)Sin(y)Cos(z)

b) f’(x,y,z) = Sin(x)Cos(y)Sin(z) + Cos(x)Sin(y)Cos(z)

c) f’(x,y,z) = Cos(x)Cos(y)Cos(z) + Sin(x)Sin(y)Sin(z)

d) f’(x,y,z) = Sin(x)Sin(y)Sin(z) + Cos(x)Cos(y)Cos(z)

View Answer

Explanation:

f(x,y,z) = Sin(x)Sin(y)Sin(z)-Cos(x) Cos(y) Cos(z)

Since the function has 3 independent variables hence during differentiation we have to consider x and z as constant and differentiate it w.r.t. y,

f’(x,y,z) = Sin(x)Cos(y)Sin(z) + Cos(x)Sin(y)Cos(z).

2. In euler theorem x ^{∂z}⁄_{∂x} + y ^{∂z}⁄_{∂y} = nz, here `n` indicates

a) order of z

b) degree of z

c) neither order nor degree

d) constant of z

View Answer

Explanation: Statement of euler theorem is “if z is an homogeneous function of x and y of order `n` then x

^{∂z}⁄

_{∂x}+ y

^{∂z}⁄

_{∂y}= nz ”.

3. If z = x^{n} f(^{y}⁄_{x}) then

a) y ^{∂z}⁄_{∂x} + x ^{∂z}⁄_{∂y} = nz

b) 1/y ^{∂z}⁄_{∂x} + 1/x ^{∂z}⁄_{∂y} = nz

c) x ^{∂z}⁄_{∂x} + y ^{∂z}⁄_{∂y} = nz

d) 1/x ^{∂z}⁄_{∂x} + 1/y ^{∂z}⁄_{∂y} = nz

View Answer

Explanation: Since the given function is homogeneous of order n , hence by euler’s theorem

x

^{∂z}⁄

_{∂x}+ y

^{∂z}⁄

_{∂y}= nz.

4. Necessary condition of euler’s theorem is

a) z should be homogeneous and of order n

b) z should not be homogeneous but of order n

c) z should be implicit

d) z should be the function of x and y only

View Answer

Explanation:

Answer `a` is correct as statement of euler’s theorem is “if z is an homogeneous function of x and y of order `n` then x

^{∂z}⁄

_{∂x}+ y

^{∂z}⁄

_{∂y}= nz”

Answer `b` is incorrect as z should be homogeneous.

Answer `c` is incorrect as z should not be implicit.

Answer `d` is incorrect as z should be the homogeneous function of x and y not non-homogeneous functions.

5. If f(x,y) = ^{x+y}⁄_{y} , x ^{∂z}⁄_{∂x} + y ^{∂z}⁄_{∂y} = ?

a) 0

b) 1

c) 2

d) 3

View Answer

6. Does function can be solved by euler’ s theorem

a) True

b) False

View Answer

Explanation:

No this function cannot be written in form of x

^{n}f(

^{y}⁄

_{x}) hence it does not satisfies euler’s theorem.

7. Value of is ,

a) -2.5 u

b) -1.5 u

c) 0

d) -0.5 u

View Answer

8. If u = x^{x} + y^{y} + z^{z} , find ^{du}⁄_{dx} + ^{du}⁄_{dy} + ^{du}⁄_{dz} at x = y = z = 1

a) 1

b) 0

c) 2u

d) u

View Answer

Explanation:

^{du}⁄

_{dx}= x

^{x}(1+log(x) )

^{du}⁄

_{dy}= y

^{y}(1+log(x))

^{du}⁄

_{dz}= z

^{z}(1+log(x))

At x = y = z = 1,

^{du}⁄

_{dx}+

^{du}⁄

_{dy}+

^{du}⁄

_{dz}= u.

10. If f(x,y)is a function satisfying euler’ s theorem then

View Answer

11. Find the approximate value of [0.98^{2} + 2.01^{2} + 1.94^{2} ]^{(1⁄2)}

a) 1.96

b) 2.96

c) 0.04

d)-0.04

View Answer

Explanation: Let f(x,y,z) = (x

^{2}+ y

^{2}+ z

^{2})

^{(1⁄2)}……………..(1)

Hence, x = 1, y = 2, z = 2 so that, dx = -0.02, dy = 0.01, dz = -0.06

From (1),

^{∂f}⁄

_{∂x}=

^{x}⁄

_{f}

^{∂f}⁄

_{∂y}=

^{y}⁄

_{f}

^{∂f}⁄

_{∂z}=

^{z}⁄

_{f}

12. The happiness(H) of a person depends upon the money he earned(m) and the time spend by him with his family(h) and is given by equation H=f(m,h)=400mh^{2} whereas the money earned by him is also depends upon the time spend by him with his family and is given by m(h)=√(1-h^{2} ). Find the time spend by him with his family so that the happiness of a person is maximum.

a) √(^{1}⁄_{3})

b) √(^{2}⁄_{3})

c) √(^{4}⁄_{3})

d) 0

View Answer

Explanation: Given,H=400mh

^{2}and m=√(1-h

^{2})

Let, ϑ=m

^{2}+ h

^{2}+ 1 = 0, hence

By lagrange’s method,

^{∂H}⁄

_{∂m}+ α

^{∂v}⁄

_{∂m}= 0

400h

^{2}+ 2m(α) = 0 ……………………………..(1)

Similarly,

^{∂H}⁄

_{∂h}+ α

^{∂h}⁄

_{∂v}=0

800mh + 2h(α) = 0…………………………………(2)

Multiply by m and h in eq’1’ and eq’2’ respectively and adding them

α=-600mh

^{2}

Now from eq. 1 and 2 we get putting value of α,

m =

^{1}⁄

_{√3}and h = √(

^{2}⁄

_{3}) .

Hence, total time spend by a person with his family is √(

^{2}⁄

_{3}).

**Sanfoundry Global Education & Learning Series – Engineering Mathematics.**

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