Applied Calculus: UNIT II: Functions of Several Variables

Partial derivatives: Problems under type-4

Explanation, Formula, Equation, Example and Solved Problems - Partial derivatives: Problems under type-4

PROBLEMS UNDER TYPE 4

Let u= f(r,s,t). r, s and t are functions of x, y, z.Then


 

Example 34. If u = f(x‒y, yz, z‒x), then prove that ∂u/∂x + ∂u/∂y + ∂u/∂z =0.

Solution: Let r=x‒y, s=y‒z, tz‒x.

:. u = f(r,s,t)


 

Example 35. If u = f(2x‒3y, 3y‒4z, 4z‒2x), then prove that .

Solution: Let r = 2x‒3y, s = 3y‒4z, t = 4z ‒ 2x

u = f(r,s,t)


 

Example 36. If u = f(x/y, y/z, z/x), then prove that x.∂u/∂x + y.∂u/∂y + z.∂u/∂z = 0.

Solution:

Let r =x/y, s =y/z, t =z/x.

:. u = f(r,s,t)


 

Example 37. If u = u( y‒x / xy, z‒x / xz), then prove that .

Solution: 


 

Example 38. If u = f(x, y), where x = r cos 0, y = r sin 0, then P.T 

Solution.

Given x = r cos 0, y = r sin 0


 

Example 39. If Z = f (u,v), where u = lx + my, v = lymx then prove that .

Solution: Given u = lx + my, v = ly ‒ mx


 

Example 40. Prove that , u = excosy, v = exsiny and that f is a function of u and v and also of x and y.

Solution: Given u = excosy, v = exsin y


Hence the proof

 

Example 41. If f = f(u, v), where u = x2y2 and v = 2xy, then show that .

Solution: Given u = x2 ‒ y2; v = 2xy


 

Example 42. Transform the equation ∂2z/∂x2 + ∂2z/∂y2 = 0 into polar co ordinates.

Solution: Polar co‒ordinate is x = r cos 0, y = r sine

x2 + y2 = r2

r = √[x2 + y2]

y/x = tanθ

θ= tan‒1x/y


 

 

EXERCISE

 

46. Find du/dx when u = sin(x2 + y2), where x2 + 4y2 = 9.

Ans: 3xcos(x2+y2) / 2

 

47. Find du/dx when u = tan‒1(y/x), where x2 + y2 = a2.

Ans: ‒1/y

 

48. Find du/dt if u=x/y where x = et, y = log t.

Ans: dz/dt = et/logt [ 1‒ 1/tlogt ]

 

49. If z = x2 + y2, where x = t3, y = 1 + t2, find dz/dt.

Ans: = dz/dt  = 6t6 + 4t + 4t3

 

50. If ax2 + 2hxy + by2 = 1, then prove that d2y/dx2 = h2‒ab / (hx+by)3

 

51. Using partial Derivative find dy/dx for (sec x)y = (cot y)x.

Ans. [ log coty ‒ y tan x ] [log secx + x secy cosec y]

 

52. Using partial Derivative find dy/dx for (cos x)y = (sin x)y.

Ans. [ y tan x + log sin y ] / [ log cos x ‒ x coty ]

 

53. If ƒ (cx ‒ az, cy ‒ bz) = 0, where z is a function of x and y, prove that a.∂z/∂x + b. ∂z/∂y = c.

 

54. If ƒ is a function of u, v, w where u = √(yz), v = √(zx), w= √(xy) show that  

 

55. If u = u(x, y) and x = er cos θ, y = er sin θ, show that 

 

56. If z = f(u, v), where u = coshx cosy and v = sin hx sin y, then prove that 

 

57. Transform the equation , by changing the independent variables using u = 2x + y and v = 3x + y.

Ans:2z/∂uu = 0

 

58. Using partial differentiation, prove d2y/dx2 = b2‒ac / (ay+b)3 when ay2+2by + c = x2.

 

59. If u = f(r,s), r=x+at, s= y + bt, then show that ∂u/∂t = a [∂u/∂x] + b [∂u/∂x]

 

60. If z = f(x, y) where, x = Xcosa ‒ Ysina and y = Xsina + Ycosa show .

 

61. Transform the equation  = 0 by changing the independent variables using u = x ‒ y and v = x + y.

 

Applied Calculus: UNIT II: Functions of Several Variables : Tag: Applied Calculus : - Partial derivatives: Problems under type-4


Applied Calculus: UNIT II: Functions of Several Variables



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Applied Calculus

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