Question

If ${\displaystyle u=x^2+y^2+z^2}$ where ${\displaystyle x=e^t,y=e^t\sin t,z=e^t\cos t}$

Prove that ${\displaystyle \frac{du}{dt}=4e^{2t}}$

Answer 1

${\displaystyle \frac{du}{dt}=\frac{\partial u}{\partial x}.\frac{dx}{dt}+\frac{\partial u}{\partial y}.\frac{dy}{dt}+\frac{\partial u}{\partial z}.\frac{dz}{dt}}$

${\displaystyle \frac{\partial u}{\partial x}=2x,\frac{\partial u}{\partial y}=2y,\frac{\partial u}{\partial z}=2z}$

${\displaystyle \frac{dx}{dt}=e^t,\frac{dy}{dt}=e^t(\sin t+\cos t),\frac{dz}{dt}=e^t(-\sin t+\cos t)}$

${\displaystyle \frac{du}{dt}=\frac{\partial u}{\partial x}.\frac{dx}{dt}+\frac{\partial u}{\partial y}.\frac{dy}{dt}+\frac{\partial u}{\partial z}.\frac{dz}{dt}}$

${\displaystyle =2x\left(e^t\right)+2y\left(e^t(\sin t+\cos t)\right)+2z\left(e^t(-\sin t+\cos t)\right)}$

= 2x(x)+2y(y+z)+2z(z-y)

${\displaystyle =2x^2+2y^2+2z^2+2xy-2xy}$

${\displaystyle =2x^2+2y^2+2z^2}$

${\displaystyle =2(x^2+y^2+z^2)}$

=2u ………………………………………(1)

${\displaystyle u=x^2+y^2+z^2={\left(e^t\right)}^2+{\left(e^t\sin t\right)}^2+{\left(e^t\cos t\right)}^2}$

${\displaystyle =e^{2t}+e^{2t}({\sin }^{2}t+{\cos }^{2}t)}$

${\displaystyle =e^{2t}+e^{2t}=2e^{2t}}$

Substituting value of u in equation (1)

${\displaystyle \frac{du}{dt}=2u=2\left(2e^{2t}\right)=4e^{2t}}$

Hence proved

${\displaystyle \frac{Du}{dt}=4e.}$