Answer
a) 1 b) 1
Work Step by Step
a) Using chain rule.
$\dfrac{dw}{dt}=\dfrac{\partial w}{\partial x}\dfrac{dx}{dt}+\dfrac{\partial w}{\partial y}\dfrac{dy}{dt}+\dfrac{\partial w}{\partial z}\dfrac{dz}{dt}$
or, $=\dfrac{-2}{z}\sin t cos t +\dfrac{2}{z}\sin t cos t+\dfrac{x+y}{z^2 t^2}$
or, $\dfrac{dw}{dt}=\dfrac{\cos^2 t+\sin^2 t}{(1/t^2)(t^2)}=1$
Using direct differentiation.
since, $w=\dfrac{x}{z}+\dfrac{y}{z}=\dfrac{\cos^2 t}{(1/t)}+\dfrac{\sin^2 t}{(1/t)}t$
and $\dfrac{dw}{dt}=1$
Now, b) $\dfrac{dw}{dt}(3)=1$
