## Calculus: Early Transcendentals 8th Edition

(a) Positive (b) $curlF$ is zero
(a) The vector field shown is a 2D vector field of the form $F=Pi+Qj$ We know that $divF=\frac{∂P}{∂x}+\frac{∂Q}{∂y}$ Here, $\frac{∂P}{∂x}$ is positive, since the $x$ components of the vectors are increasing in length, as we move along the positive $x-direction$. $\frac{∂Q}{∂y}$, is positive, since the $y$ components of the vectors are increasing in length, as we move along the positive $y-direction$. This implies that divergence is positive. That is, $divF=\frac{∂P}{∂x}+\frac{∂Q}{∂y}=0+(+ve)=+ve$ Hence, $divF$ is positive. (b) The vector field shown is a 2D vector field of the form $F=Pi+Qj$ We know that $curlF=(\frac{∂Q}{∂x}-\frac{∂P}{∂y})k$ Here $\frac{∂Q}{∂x}=0$, since the $y$ components of the vectors have same length, as we move along the positive x-direction. $\frac{∂P}{∂y}$ is $0$, since the $x$ components of the vectors have same length as we move along the positive y-direction. That is, $curlF=(\frac{∂Q}{∂x}-\frac{∂P}{∂y})k=(0-0)k=0$ Hence, $curlF$ is zero.