By H.S. Gangwar, Dr. Prabhakar Gupta

ISBN-10: 8122424759

ISBN-13: 9788122424751

Written for the scholars of BTech I yr of UP Technical college, Lucknow and different states, this booklet discusses intimately the thoughts and strategies in Engineering arithmetic.

**Read Online or Download A Textbook of Engineering Mathematics-I, 2nd Edition PDF**

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**Additional info for A Textbook of Engineering Mathematics-I, 2nd Edition**

**Sample text**

T. x. Geometrical interpretation of Hence, and Z A P O ∂z = tan θ (slope of the curve APB at the point P) ∂x ∂z = tan φ (slope of the curve CPD at point P) ∂y Y f q B D X Fig. 2 22 A TEXTBOOK OF ENGINEERING MATHEMATICS—I Example 1. , 2007) 2 Sol. t. t. y. t. t. x. (B) 2 Thus, from (A) and (B), we find ∂ 2u ∂ 2u = . ∂y∂x ∂x∂y Hence proved. FG y IJ , verify that ∂ f H xK ∂y∂x F yI f = tan G J H xK 2 Example 2. If f = tan–1 Sol. We have = ∂2 f . (iv) 2 Differentiating (iii) partially with respect to x, we get F GH ∂2 f x ∂ = 2 ∂x∂y ∂x x + y 2 = Example 3.

Cos ∂ 2u ∂ 2u 2 Hence, x + 2xy +y ∂y2 = 2 cos 3u · cos u. ∂x∂y ∂x2 2 ∂ 2u Example 8. If z = xm f x2 Sol. (i) Now, u is homogeneous function of degree m. (iii) Adding (ii) and (iii), we get ∂ ∂2 ∂2 2 x 2 (u + v) + 2xy ∂x∂y (u + v) + y ∂y 2 (u + v) = m (m – 1) u + n (n – 1) v ∂x 2 ⇒ x2 ∂ 2z ∂2z ∂ 2z 2 + 2xy + y ∂y2 = m (m – 1) u + n (n – 1) v (As z = u + v). (v) ∂x m (m –1) u + n (n – 1) v = (m2 u + n2v) – (mu + nv) = m (m + n) u + n (m + n) v – mn (u + v) – (mu + nv) = (mu + nv) (m + n) − (mu + nv) − mnz = (mu + nv) (m + n – 1) – mnz ⇒ x Now, = (m + n –1) F x ∂z + y ∂z I GH ∂x ∂y JK – mnz, from (v) Putting this value in equation (iv), we get F x ∂z + y ∂z I GH ∂x ∂y JK – mnz F x ∂z + y ∂z I ∂ 2z ∂ z ∂ 2z ⇒ x 2 + 2xy ∂x∂y + y ∂y2 + mnz = (m + n –1) GH ∂x ∂y JK .

If u = sin–1 (x3 + y3)2/5, evaluate 2 ∂ 2u x 2∂ 2u 2 ∂ u . + + xy y 2 ∂x∂y ∂x 2 ∂y 2 u = sin–1 (x3 + y3)2/5 Sol. (i) 6 . r. to ’x‘, we get 6 ∂u ∂u ∂ 2u ∂ 2u sec2 u. +x 2 +y = 5 ∂x ∂x ∂x∂y ∂x Multiplying by x x ∂u ∂ 2u ∂ 2u 6 ∂u + x 2 2 + xy = sec2 u. r. to ‘y’, we get x 6 ∂ 2u ∂u ∂ 2u ∂u sec2 u + +y 2 = ∂y 5 ∂y∂x ∂y ∂y Multiplying by y y ∂u ∂ 2u ∂ 2u 6 ∂u + y 2 2 + xy sec2 u. (iv) Adding (iii) and (iv), we get F x ∂u + y ∂u I FG 6 sec u − 1IJ GH ∂x ∂y JK H 5 K ∂u ∂u 6 ∂ 2u ∂ 2u ∂ 2u 6 6 As x +y = tan u x 2 2 + 2xy + y2 2 = tan u FG sec2 u − 1IJ K H ∂ x ∂y 5 ∂x∂y 5 5 ∂x ∂y F xI F yI Example 14.

### A Textbook of Engineering Mathematics-I, 2nd Edition by H.S. Gangwar, Dr. Prabhakar Gupta

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