By H.S. Gangwar, Dr. Prabhakar Gupta
Written for the scholars of BTech I 12 months of UP Technical college, Lucknow and different states, this booklet discusses intimately the techniques and strategies in Engineering arithmetic.
Read Online or Download A Textbook of Engineering Mathematics-I, 2nd Edition PDF
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Extra info for A Textbook of Engineering Mathematics-I, 2nd Edition
X 2 ∂y 2 6. If z = tan (y – ax) + (y + ax)3/2, prove that 7. If u = b OP Q IK OP PQ ∂u ∂u = x xy y log x + y ; = x xy + 1 log x ∂x ∂y ∂ 2u ∂ 2u + = 0. ∂x 2 ∂y 2 2 ∂ 2z 2 ∂ z a – = 0. ∂x 2 ∂y 2 ∂u ∂u ∂u + ∂y + = 0. ∂x ∂z j FH x + −1 2 x2 − y2 −1 34 A TEXTBOOK OF ENGINEERING MATHEMATICS—I 8. If u = 2(ax + by)2 – (x2 + y2) and a2 + b2 = 1, find the value of ∂ 2u ∂ 2u . 2 + ∂x ∂y 2 LM MMAns. ex N ∂ 2u . 9. If u = log (x2 + y2 + z2), find the value of ∂y∂z e 10. If u = x 2 + y 2 + z 2 j 1 2 , then prove that ∂2u ∂x 2 11.
If f (x, y) = 0, φ (y, z) = 0, show that ∂f ∂φ dz ∂y ∂z dx = Sol. We have ∂f ∂φ . (ii) FG ∂f IJ H ∂x K = − F ∂f I GH ∂y JK F ∂φ I GH ∂y JK = − FG ∂φ IJ H ∂z K 51 DIFFERENTIAL CALCULUS-I Multiplying these two results, we get dz = dx ∂f ∂φ dz ∂y ∂z dx or Example 3. If u = u x2 F ∂φ I GH ∂y JK FG ∂φ IJ H ∂z K × ∂f ∂φ · ⋅ ∂x ∂y Hence proved. F y − x , z − x I , show that GH xy xz JK ∂u ∂u ∂u + y2 + z2 = 0. y ∂ ∂x ∂z Sol. , 2005) y−x 1 1 1 1 z− x = – and t = = – xy x y x z zx 1 = – , x2 ∂s ∂t 1 ∂t 1 ∂t 1 = 2 , = − 2, = 2, =0 ∂y ∂x y y ∂ ∂ z x z ∂s = 0 ∂z u = u(s, t) Since ∴ ∂u ∂x = ∂u ∂s ∂u ∂t ⋅ ⋅ + ∂s ∂x ∂t ∂x ⇒ ∂u ∂x = ∂u ∂s x2 or Next, ∂u ∂x ∂u ∂y ∂u ∂y or and = FG ∂f IJ H ∂x K F ∂f I GH ∂y JK FG – 1 IJ H xK 2 + ∂u ∂t FG − 1 IJ H xK 2 ∂u ∂u – ∂s ∂t ∂u ∂s ∂u ∂t + ∂s ∂y ∂t ∂y = – = ∂u ∂u ∂u 2 + 0 ⇒ y y ∂y = ∂s ∂s 1 ∂u ∂u ∂s ∂u ∂t + =0+ 2 z ∂s ∂z ∂t ∂z ∂t ∂u ∂t 1 = 2 ∂u = ∂z ∂u ⇒ z2 = ∂z Adding (i), (ii) and (iii), we get ∂u ∂u ∂u ∂u ∂u ∂u ∂u x2 + y2 ∂y + z2 = – – + + = 0.
We have u = tan –1 x 3 + y3 x 3 + y3 ⇒ tan u = x− y x− y x 3 + y3 Let f = tan u = x− y ∴ Since f (x, y) is a homogeneous function of degree n = 3–1=2 By Euler’s theorem, we have x ⇒ x ∂f ∂f ∂ ∂ +y = nf ⇒ x (tan u) + y (tan u) = 2 tan u ∂x ∂x ∂y ∂y ∂u ∂u · sec2 u + y · sec2 u = 2 tan u ∂y ∂x DIFFERENTIAL CALCULUS-I or x ∂u tan u ∂u +y = 2 = sin 2u. Proved. t. x, we get x or ∂u ∂ 2u ∂u ∂ 2u + + y = 2 cos 2u . t. (iii) Adding (ii) and (iii), we get x2 ∂ 2u ∂ 2u ∂ 2u 2 = ( 2cos 2u – 1) 2 + 2xy ∂x∂y + y ∂x ∂y2 F x ∂u + y ∂u I GH ∂x ∂y JK = (2cos 2u – 1) sin 2u, (from (i)) = (2sin 2u cos 2u – sin 2u) = sin 4u – sin 2u = 2 cos FG 4u + 2uIJ H 2 K FG 4u − 2uIJ .