The velocity vector is given by v=vxi^+vyj^+vzk^=−xi^+2yj^−zk^.
The components of acceleration are:
ax=vx∂x∂vx+vy∂y∂vx+vz∂z∂vx=(−x)(−1)+0+0=x
ay=vx∂x∂vy+vy∂y∂vy+vz∂z∂vy=0+(2y)(2)+0=4y
az=vx∂x∂vz+vy∂y∂vz+vz∂z∂vz=0+0+(−z)(−1)=z
The acceleration vector is a=xi^+4yj^+zk^.
At the point (1,2,4), substituting x=1, y=2, and z=4:
a=1i^+4(2)j^+4k^=i^+8j^+4k^
The magnitude of acceleration is:
∣a∣=12+82+42=1+64+16=81=9 m/s2
Answer: 9