Answer
$ \frac{dw}{dt}\Big|_{t=1}=7-2\pi$
Work Step by Step
\[
\frac{dw}{dt}=f_x(x,y,z)\frac{dx}{dt}+f_y(x,y,z)\frac{dy}{dt}+f_z(x,y,z)\frac{dz}{dt}.
\]
Compute the derivatives of the parameter functions:
\[
\frac{dx}{dt}=1,\qquad \frac{dy}{dt}=\pi\cos(\pi t),\qquad \frac{dz}{dt}=2t.
\]
At \(t=1\) we have \(x=1,\ y=\sin(\pi)=0,\ z=1^2+1=2\), so we may use the given partial derivatives:
\[
\frac{dw}{dt}\Big|_{t=1}
= f_x(1,0,2)\cdot 1 + f_y(1,0,2)\cdot \pi\cos(\pi) + f_z(1,0,2)\cdot 2(1).
\]
Substituting the values,
\[
\frac{dw}{dt}\Big|_{t=1} = 1\cdot 1 + 2\cdot \pi\cos(\pi) + 3\cdot 2
= 1 + 2\pi(-1) + 6 = 7 - 2\pi.
\]
\[
\boxed{\displaystyle \frac{dw}{dt}\Big|_{t=1}=7-2\pi}
\]
