Calculus ppt on "Partial Differentiation"#2
- 1. L.D COLLEGE OF ENGINEERING RUBBER TECHNOLOGY
- 2. ROLL NO. NAME 126001 ADHVARYU UTKARSH 126002 RAMOLIYA NAVDEEP 126003 BARAIYA GOPAl 126004 ITALIYA SHANKESH 126005 KANETIYA JAYDEEP 126006 LAKUM BHAVESH 126007 AMIPARA KEYUR 126008 SHAH ARTH 126010 SURANI PUNAL
- 3. Contents Function of two variables Limits and continuity Partial derivatives in first order Partial derivatives in higher order PARTIAL DIFFERENTIATION
- 5. WHAT IS PARTIAL DIFFERENTIATION ? Let z=f(x,y) be function of two individual variables x & y the derivative with respect to x keeping y constant Is called partial derivative of z with respect to x. It is denoted by ∂z , ∂f , fx . ∂x ∂x It is denoted as ∂z = lim f(x+∂x ,y) – f(x,y) . ∂x ∂x→0 ∂x
- 8. EXAMPLE : By considering differnent paths of approach, show that the function f(x,y) = x4 – y2 has no limit as (x,y)→(0,0). x4 + y2 Solution : lim lim x4 – y2 = lim x4 = lim 1 = 1 x→0 y→0 x4 + y2 x→0 x4 x→0 lim lim x4 – y2 = lim (–y2) = lim (-1) = -1 y→0 x→0 x4 + y2 y→0 y2 x→0 Since both the limits are different, f(x,y) has no limit as (x,y)→(0,0).
- 11. EXAMPLE : If a2x2+ b2y 2 = c2z2 , evaluate 1 ∂2z + 1 ∂2z . a2 ∂x2 b2 ∂y2 Solution : a2x2+ b2y2 = c2z2 Differentiating partially w.r.t. x, 2a2x = 2c2z ∂z ∂x ∂z = a2x ∂x c2z Differentiating ∂z partially w.r.t. x, ∂x ∂2z = a2 (1 - x ∂z) ∂x2 c2z z2 ∂x = a2 (1 – x a2x) c2z z c2z
- 15. EXAMPLE : If u = x3y3z3 then prove that x ∂u +y ∂u +z ∂u =6u x3+y3+z3 ∂x ∂y ∂z Solution : Replacing x by xt and y by yt, u = t6 x3y3z3 x3+y3+z3 Hence,u is a homogeneous function of degree 6. By euler’s theorem, x ∂u +y ∂u +z ∂u =6u ∂x ∂y ∂z
- 18. EXAMPLE : If u = y2-4ax,x = at2,y = 2at,find du . dt Solution : du = ∂u .dx + ∂u .dy dt ∂x dt ∂y dt =(-4a)2at+2(2at)(2a) du =-8a2t+2(2at)(2a) dt =-8a2t+8a2t =0
- 23. EXAMPLE : If x3+y3 = 3axy,find dy . dx Solution : Let f(x,y)= x3+y3-3axy dy = ∂u/ ∂x dx ∂f/ ∂y = 3x2-3ay 3y2-3ax = x2- ay y2-ax = ay-x2 y2-ax
- 24. YOU ROLL NO. 1 TO 10