If u = f (x − y , y − z , z − x) then prove that ∂u∂x+ ∂u∂y+ ∂u∂z= 0

so ∂z/∂u is  Answer.and ∂z/∂v is  Answer.and ∂z/∂w is

If v = 3x2yi + 2xyzj − 3x4y2k, find ∂v∂x , ∂v∂y , ∂v∂z . Further, find ∂2v∂x2 and ∂2v∂x∂z .Solution We find∂v∂x = 6xyi + 2yzj − 12x3y2k∂v∂y = 3x2i + 2xzj − 6x4yk∂v∂z = 2xyj∂2v∂x2 = 6yi − 36x2y2k∂2v∂x∂z = 2yjEXERCISES 26.21 Given v = 2xi + 3yzj + 5xz2k find(a) ∂v∂x (b) ∂v∂y (c) ∂v∂z2 If f = 2i − xyzj + 3x2zk find(a) ∂f∂x (b) ∂f∂y (c) ∂f∂z(d) ∂2f∂x2 (e) ∂2f∂y2 (f) ∂2f∂z23 Given E = (x2 + y)i + (1 − z)j + (x + 2z)k find(a) ∂E∂x (b) ∂E∂y (c) ∂E∂z(d) ∂2E∂x2 (e) ∂2E∂y2 (f) ∂2E∂z24 If v = 3xyzi + (x2 − y2 + z2 )j + (x + y2 )k find∂v∂x , ∂v∂y , ∂v∂z , ∂2v∂x2 , ∂2v∂y2 , and ∂2v∂z2 .5 If v = sin(xyz)i + z exyj − 2xyk find∂v∂x , ∂v∂y , ∂v∂z .6 If v = xi + x2yj − 3x3k, and φ = xyz, findφv, ∂∂x (φv), ∂φ∂x , ∂v∂x . Deduce that∂∂x (φv) = φ ∂v∂x + ∂φ∂x v7 If v = ln(xy)i + 2xy cos zj − x4yzk, find∂v∂x , ∂v∂y , ∂v∂z , ∂2v∂x2 , ∂2v∂y2 , and ∂2v∂z2 .Solutions1 (a) 2i + 5z2k (b) 3zj(c) 3yj + 10xzk2 (a) −yzj + 6xzk (b) −xzj(c) −xyj + 3x2k (d) 6zk(e) 0 (f) 0Downloaded From :www.EasyEngineering.netDownloaded From :www.EasyEngineering.netwww.EasyEngineering.net

The curl of F⃗ =(z−x−y)i⃗ +(x−y−z)j⃗ +(y−z−x)k⃗ 𝐹→=(𝑧−𝑥−𝑦)𝑖→+(𝑥−𝑦−𝑧)𝑗→+(𝑦−𝑧−𝑥)𝑘→ is

If ∂f∂x=0∂𝑓∂𝑥=0, the function f(x,y)𝑓(𝑥,𝑦) has no dependence on the variable x𝑥.Select one:TrueFalse

Suppose thatF (x, y) = g(u(x, y), v(x, y))with u(x, y) = x2 and v(x, y) = h(x, y) − x. Furthermore suppose thath(1, 1) = 2, hx(1, 1) = 3, hy(1, 1) = −2, g(1, 1) = 6, gu(1, 1) = −4 andgv(1, 1) = 5. Determine the partial derivatives Fx(1, 1) and Fy(1, 1).