find $\frac{dt}{dt}$, $t(x,y,z)=y^{2}e^{xz}$, $x = cos t$, $y=sin t$, $z = 3 + t^{2}$

Explanation:

Step1: Apply the chain - rule formula

The chain - rule for $\frac{dT}{dt}=\frac{\partial T}{\partial x}\frac{dx}{dt}+\frac{\partial T}{\partial y}\frac{dy}{dt}+\frac{\partial T}{\partial z}\frac{dz}{dt}$. First, find the partial derivatives of $T(x,y,z)=y^{2}e^{xz}$ and the derivatives of $x = \cos t$, $y=\sin t$, $z = 3 + t^{2}$.

Step2: Calculate $\frac{\partial T}{\partial x}$

Using the product - rule and the chain - rule for partial derivatives, $\frac{\partial T}{\partial x}=y^{2}ze^{xz}$. Since $x = \cos t$, $y=\sin t$, $z = 3 + t^{2}$, then $\frac{\partial T}{\partial x}=\sin^{2}t(3 + t^{2})e^{\cos t(3 + t^{2})}$. And $\frac{dx}{dt}=-\sin t$.

Step3: Calculate $\frac{\partial T}{\partial y}$

$\frac{\partial T}{\partial y}=2ye^{xz}$. Substituting $y = \sin t$, $x=\cos t$, $z = 3 + t^{2}$, we get $\frac{\partial T}{\partial y}=2\sin t e^{\cos t(3 + t^{2})}$. And $\frac{dy}{dt}=\cos t$.

Step4: Calculate $\frac{\partial T}{\partial z}$

$\frac{\partial T}{\partial z}=xy^{2}e^{xz}$. Substituting $x=\cos t$, $y = \sin t$, $z = 3 + t^{2}$, we have $\frac{\partial T}{\partial z}=\cos t\sin^{2}t e^{\cos t(3 + t^{2})}$. And $\frac{dz}{dt}=2t$.

Step5: Calculate $\frac{dT}{dt}$

$\frac{dT}{dt}=\frac{\partial T}{\partial x}\frac{dx}{dt}+\frac{\partial T}{\partial y}\frac{dy}{dt}+\frac{\partial T}{\partial z}\frac{dz}{dt}$ [ \begin{align*} \frac{dT}{dt}&=\sin^{2}t(3 + t^{2})e^{\cos t(3 + t^{2})}\times(-\sin t)+2\sin t e^{\cos t(3 + t^{2})}\times\cos t+\cos t\sin^{2}t e^{\cos t(3 + t^{2})}\times2t\ &=e^{\cos t(3 + t^{2})}\left(-\sin^{3}t(3 + t^{2}) + 2\sin t\cos t+2t\cos t\sin^{2}t\right) \end{align*} ]

Answer:

$e^{\cos t(3 + t^{2})}\left(-\sin^{3}t(3 + t^{2})+2\sin t\cos t + 2t\cos t\sin^{2}t\right)$