- Consider x, y, z to be right-handed Cartesian coordinates. A vector function defined in this coordinate system as v = 3xi + 3xyj − yz2k, where i, j and k are unit vectors along x, y and z axes, respectively. The curl of v is given by
- z2i − 3yk
- z2j + 3yk
- z2i + 3yj
- −z2i + 3yk
- Which of the following functions is periodic?
- ƒ(x) = x2
- ƒ(x) = log x
- ƒ(x) = ex
- ƒ(x) = const.
Answer:- ƒ(x) = const.
- The function ƒ(x1, x2, x3) = x12 + x22 + x32 − 2x1 − 4x2 − 6x3+ 14 has its minimum value at
- (1, 2, 3)
- (0, 0, 0)
- (3, 2, 1)
- (1, 1, 3)
Answer:- (1, 2, 3)
The critical points satisfy ƒx1 = ƒx2 = ƒx2 = 0
Therefore, ƒx1 = 2x1 − 2 = 0 ⇒ x1 = 1
ƒx2 = 2x2 − 4 = 0 ⇒ x2 = 2
ƒx3 = 2x3 − 6 = 0 ⇒ x3 = 3
So, (1, 2, 3) (denoting by p) is a critical point. Now, check whether it is maximum, minimum or saddle point.
Δ1 = ƒx1x1(p) = 2 > 0
Δ2 = = = 4 > 0
Δ3 = = = 8 > 0
As Δ1 > 0, Δ2 > 0 and Δ3 > 0, (1, 2, 3) is the local minimum of the given function. - Consider the function ƒ(x1, x2) = x12 + 2x22 + e− x1 − x2. The vector pointing in the direction of maximum increase of the function at the point (1, -1) is
- Two simultaneous equations given by y = π + x and y = x − π have
- a unique solution
- infinitely many solutions
- no solution
- a finite number of multiple solutions
Answer:- no solution
- In three-dimensional linear elastic solids, the number of non-trivial stress-strain relations, strain-displacement equations and equations of equilibrium are, respectively,
- 3, 3 and 3
- 6, 3 and 3
- 6, 6 and 3
- 6, 3 and 6
Answer:- 6, 6 and 3
- An Euler-Bernoulli beam in bending is assumed to satisfy
- both plane stress as well as plane strain conditions
- plane strain condition but not plane stress condition
- plane stress condition but not plane strain condition
- neither plane strain condition nor plane stress condition
Answer:- neither plane strain condition nor plane stress condition
- A statically indeterminate frame structure has
- same number of joint degrees of freedom as the number of equilibrium equations
- number of joint degrees of freedom greater than the number of equilibrium equations
- number of joint degrees of freedom less than the number of equilibrium equations
- unknown number of joint degrees of freedom, which cannot be solved using laws of mechanics
Answer:-
- Consider a single degree of freedom spring-mass-damper system with mass, damping and stiffness of m, c, and k, respectively. The logarithmic decrement of this system can be calculated using
Answer:-
logarithmic decrement =
ζ = , substituting in above equation we get
logarithmic decrement = - Consider a single degree of freedom spring-mass system of spring stiffness k1 and mass m which has a natural frequency of 10 rad/s. Consider another single degree of freedom spring-mass system of spring stiffness k2 and mass m which has a natural frequency of 20 rad/s. The spring stiffness k2 is equal to
- k1
- 2k1
- k1/4
- 4k1
Answer:- 4k1
anuradhapallati said:
may i know why u r complicated 3rd problem.trial&error is enough know
LikeLike
JIGAR SURA said:
I have given the method by which answer can be obtained. One should also know the reason of the answer.
LikeLike
anuradhapallati said:
where will i can found this method
LikeLike
JIGAR SURA said:
Any engineering mathematics book which has the chapter for function of 2 variables.
LikeLike