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Fundamentals of Stress and Vibration 1. Mathematics for Structural mechanics
[A Practical guide for aspiring Designers / Analysts]
‘‡ —•– ϐ‹”•– Ž‘ ƒ–‡ ȋšǡ›Ȍ „› ‡“—ƒ–‹‰ –Š‡ †‡”‹˜ƒ–‹˜‡ ‘ˆ ˆ— –‹‘ ™‹–Š ”‡•’‡ – –‘ Ǯšǯ ƒ† Ǯ›ǯ –‘ œ‡”‘Ǥ
—„•‡“—‡–Ž›ǡ –Š‡ ƒ–”‹š —•– „‡ ‘•–”— –‡† –‘ —‡”‹ ƒŽŽ› ‡˜ƒŽ—ƒ–‡ –Š‡ †‡–‡”‹ƒ–Ǥ
‹ˆˆ‡”‡–‹ƒ–‹‰ –Š‡ ˆ— –‹‘ ™‹–Š ”‡•’‡ – –‘ Ǯšǯ ™‡ ‰‡–ǣ

∂f x, y
2
= 2x + 2y = 0
∂x
Simplifying the above expression we get [x = −y ]
2
Differentiating the function with respect to ‘y’ we get:

∂f x, y
= 2y + 4xy = 0
∂y

1
Simplifying the above expression we get x = − and y = 0
2

Substituting the values [y = 0] in the expression [x = −y ], we get [x = 0]. Therefore we have:
2
1
x = 0 and x = −
2

In order to find the minimum value of the function let us choose the coordinates (x = 0) and (y = 0).

Let us begin by computing the remaining component of the Hessian matrix.

2
∂ f x, y
= 2
∂x 2
2
∂ f x, y
= 2 + 4x = 0 , since we have chosen the coordinates x = 0 and y = 0 , we get:
∂y 2
2
∂ f x, y
= 2 + 4 0 = 2
∂y 2
The cross derivatives are computed as follows:

∂ ∂f x, y
= 4y
∂x ∂y

∂ ∂f x, y
= 4y
∂x ∂x

Substituting the values (x = 0) and (y = 0) in the cross derivatives, we get:

2
2
∂ f x, y ∂ f x, y
= = 0
∂x ∂y ∂y ∂x
QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries, Page 69