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)
Answer: b
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
Answer: a
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 = xn 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
Answer: c
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
Answer: a
Explanation: Answer `z should be homogeneous and of order n` 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 `z should not be homogeneous but of order n` is incorrect as z should be homogeneous.
Answer `z should be implicit` is incorrect as z should not be implicit.
Answer `z should be the function of x and y only` 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
Answer: a
Explanation: Given function f(x,y)=(frac{x+y}{y}) can be written as f(x,y) =(frac{[1+frac{y}{x}]}{frac{y}{x}} = x^0 f(frac{y}{x})),
Hence by euler’s theorem,
(x frac{∂z}{∂x}+y frac{∂z}{∂y})=0
6. Does function f(x,y) = (Sin^{-1} [frac{(sqrt x+sqrt y)}{sqrt{x+y}}]) can be solved by euler’ s theorem.
a) True
b) False
Answer: b
Explanation: No this function cannot be written in form of xn f(y⁄x) hence it does not satisfies euler’s theorem.
7. Value of (x frac{∂u}{∂x}+y frac{∂u}{∂y}) if (u=frac{Sin^{-1} (frac{y}{x})(sqrt x+sqrt y)}{x^3+y^3}) is?
a) -2.5 u
b) -1.5 u
c) 0
d) -0.5 u
Answer: a
Explanation: Since the function can be written as,
(u=x^{-frac{5}{2}} frac{Sin^{-1} (frac{y}{x})(1+sqrt{frac{y}{x}})}{1+(frac{y}{x})^3}=x^n f(frac{y}{x})), by equler’s theorem,
(x frac{∂u}{∂x}+y frac{∂u}{∂y}= -frac{5}{2} u)
8. If u = xx + yy + zz , find du⁄dx + du⁄dy + du⁄dz at x = y = z = 1.
a) 1
b) 0
c) 2u
d) u
Answer: d
Explanation: du⁄dx = xx (1+log(x))
du⁄dy = yy (1+log(x))
du⁄dz = zz (1+log(x))
At x = y = z = 1,
du⁄dx + du⁄dy + du⁄dz = u.
9. If (u=x^2 tan^{-1} (frac{y}{x})-y^2 tan^{-1} (frac{x}{y})) then (frac{∂^2 u}{∂x∂y}) is?
a) (frac{x^2+y^2}{x^2-y^2})
b) (frac{x^2-y^2}{x^2+y^2})
c) (frac{x^2}{x^2+y^2})
d) (frac{y^2}{x^2+y^2})
Answer: b
Explanation:
(frac{∂^2 u}{∂x∂y}=frac{∂}{∂x} frac{∂u}{∂y})
(frac{∂u}{∂y}=-2y ,Tan^{-1} (frac{x}{y})+frac{x^3}{x^2+y^2}+frac{xy^2}{x^2+y^2})
(frac{∂u}{∂y}=-2y ,Tan^{-1} (frac{x}{y})+x)
Now,
(frac{∂^2 u}{∂x∂y}=frac{∂}{∂x} frac{∂u}{∂y}= -frac{2y}{1+frac{x^2}{y^2}} (frac{1}{y})+1=frac{x^2-y^2}{x^2+y^2})
10. If f(x,y)is a function satisfying euler’ s theorem then?
a) (x^2frac{∂^2 f}{∂x^2}+2xy frac{∂^2 f}{∂x∂y}+y^2 frac{∂^2 f}{∂y^2}=n(n-1)f)
b) (frac{1}{x}^2frac{∂^2 f}{∂x^2}+frac{2}{xy} frac{∂^2 f}{∂x∂y}+frac{1}{y}^2 frac{∂^2 f}{∂y^2}=n(n-1)f)
c) (x^2frac{∂^2 f}{∂x^2}+2xy frac{∂^2 f}{∂x∂y}+y^2 frac{∂^2 f}{∂y^2}=nf)
d) (y^2frac{∂^2 f}{∂x^2}+2xy frac{∂^2 f}{∂x∂y}+x^2 frac{∂^2 f}{∂y^2}=n(n-1)f)
Answer: a
Explanation: Since f satisfies euler’s theorem,
Since f satisfies euler’s theorem,
(x frac{∂z}{∂x}+y frac{∂z}{∂y})=nz
Differentiating it w.r.t x and y respectively we get,
(xfrac{∂^2 u}{∂x^2}+frac{∂u}{∂x}+y frac{∂^2 u}{∂x∂y}=n frac{∂u}{∂x}),
and
(x frac{∂^2 u}{∂y∂x}+frac{∂u}{∂y}+y frac{∂^2 u}{∂y^2}=n frac{∂u}{∂y})
Multiplying with x and y respectively,
(x^2 frac{∂^2 u}{∂x^2}+x frac{∂u}{∂x}+xy frac{∂^2 u}{∂x∂y}=nx frac{∂u}{∂x}),
and
(xy frac{∂^2 u}{∂y∂x}+y frac{∂u}{∂y}+y^2 frac{∂^2 u}{∂y^2}=ny frac{∂u}{∂y})
Adding above equations we get
(x^2 frac{∂^2 u}{∂x^2}+y^2 frac{∂^2 u}{∂y}+2xy frac{∂^2 u}{∂x∂y}=n(n-1)u )
11. Find the approximate value of [0.982 + 2.012 + 1.942](1⁄2).
a) 1.96
b) 2.96
c) 0.04
d) -0.04
Answer: b
Explanation: Let f(x,y,z) = (x2 + y2 + z2)(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
And df=(frac{∂f}{∂x} dx+frac{∂f}{∂y} dy+frac{∂f}{∂z} dz=frac{(xdx+ydy+zdz)}{f}=frac{-0.02+0.02-0.12}{3})= -0.04
Hence,
([0.98^2+2.01^2+1.94^2]^{frac{1}{2}})=f(1,2,2)+df=3-0.04=2.96
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)=400mh2 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-h2). 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
Answer: b
Explanation: Given,H=400mh2 and m=√(1-h2)
Let, ϑ=m2 + h2 + 1 = 0, hence
By lagrange’s method,
∂H⁄∂m + α∂v⁄∂m = 0
400h2 + 2m(α) = 0 …….(1)
Similarly,
∂H⁄∂h + α ∂v⁄∂h=0
800mh + 2h(α) = 0 ………(2)
Multiply by m and h in eq’1’ and eq’2’ respectively and adding them
α=-600mh2
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).
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