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 = 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
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 `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
View Answer
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
View Answer
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
View Answer
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
View Answer
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) \(\frac{y^2}{x^2+y^2}\)
View Answer
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) \(x^2\frac{∂^2 f}{∂x^2}+2xy \frac{∂^2 f}{∂x∂y}+y^2 \frac{∂^2 f}{∂y^2}=n(n-1)f\)
b) \(\frac{1}{x}^2\frac{∂^2 f}{∂x^2}+\frac{2}{xy} \frac{∂^2 f}{∂x∂y}+\frac{1}{y}^2 \frac{∂^2 f}{∂y^2}=n(n-1)f\)
c) \(x^2\frac{∂^2 f}{∂x^2}+2xy \frac{∂^2 f}{∂x∂y}+y^2 \frac{∂^2 f}{∂y^2}=nf\)
d) \(y^2\frac{∂^2 f}{∂x^2}+2xy \frac{∂^2 f}{∂x∂y}+x^2 \frac{∂^2 f}{∂y^2}=n(n-1)f\)
View Answer
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,
\(x\frac{∂^2 u}{∂x^2}+\frac{∂u}{∂x}+y \frac{∂^2 u}{∂x∂y}=n \frac{∂u}{∂x}\),
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
View Answer
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
View Answer
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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