Question: 3]b) If u = $x^2+y^2+z^2where x= e^t , y= e^t sint , z= e^t$ cost Prove that $du/dt=4e^{2t}$
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Subject:- Applied Mathematics

Marks:- 3

Mumbai Unversity>FE>Sem1>Applied Maths1

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modified 9 weeks ago  • written 9 weeks ago by gravatar for Mayank Aggarwal Mayank Aggarwal0
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$du/dt = ∂u/∂x.dx/dt+∂u/∂y.dy/dt+ ∂u/∂z.dz/dt$ $∂u/∂x = 2x ,∂u/∂y = 2y , ∂u/∂z = 2z$ $dx/dt = e^t, dy/dt= e^t (sint+cost) ,dz/dt = e^t (-sint+cost)$ $du/dt = ∂u/∂x.dx/dt+∂u/∂y.dy/dt+ ∂u/∂z.dz/dt$ $= 2x(e^t) + 2y(e^t (sint+cost) ) + 2z(e^t (-sint+cost) )$ $ = 2x(x) + 2y(y+z) + 2z(z-y)$ $ = 2x^2+2y^2+2z^2+2xy-2xy$ $ = 2x^2+2y^2+2z^2$ $ = 2 (x^2+y^2+z^2)$ = 2u ………………………………………(1) $u = x^2+y^2+z^2 = (e^t )^2+(e^t sint)^2+(e^t cost)^2$ $ = e^{2t}+ e^{2t} (sin^{2⁡t}+cos^{2⁡t} )$ $ = e^{2t}+e^{2t}= 2e^{2t}$ Substituting value of u in equation (1) $du/dt = 2u = 2(2e^{2t}) = 4e^{2t}$ Hence proved $Du/dt=4e^{2t}$

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modified 9 weeks ago  • written 9 weeks ago by gravatar for Mayank Aggarwal Mayank Aggarwal0
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