Question

z=e^(x-2y) x=cost y=sint dz/dt=?

Answer 1

Given: $z=e^{x-2y},x=cost,y=sint$

Require to find $\frac{dz}{dt}$

Recollect the following Chain Rule:

If $z=f(x,y),x=g(t),y=h(t)$ , then $\frac{dz}{dt}=\frac{\partial z}{\partial x}\frac{dx}{dt}+\frac{\partial z}{\partial y}\frac{dy}{dt}$

Now $z=e^{x-2y}\Rightarrow \frac{\partial z}{\partial x}=e^{x-2y}(1-2(0))=e^{x-2y}$ , $\frac{\partial z}{\partial y}=e^{x-2y}(0-2(1))=-2e^{x-2y}$

And $x=cost,y=sint\Rightarrow \frac{dx}{dt}=-sint, \frac{dy}{dt}=cost$

Using the Chain Rule, we get

$\frac{dz}{dt}=\frac{\partial z}{\partial x}\frac{dx}{dt}+\frac{\partial z}{\partial y}\frac{dy}{dt}$

$\Rightarrow \frac{dz}{dt}=(e^{x-2y})(-sint)+(-2e^{x-2y})(cost)$

$\Rightarrow \frac{dz}{dt}=-sinte^{x-2y}-2coste^{x-2y}$

$\Rightarrow \frac{dz}{dt}=-e^{x-2y}[sint+2cost]$

Therefore $\Rightarrow \frac{dz}{dt}=-(sint+2cost)e^{x-2y}$