Find curl F for the vector field at the point (3, -9, 1). F(x, y, z) = x^2 z i - 2xz j + yz k

Question:

Find curl {eq}F {/eq} for the vector field at the point {eq}(3, -9, 1) {/eq}.

{eq}F(x, y, z) = x^2 z \ i - 2xz \ j + yz \ k \\ {/eq}

Curl F for the Vector Field:

You have to find the curl F of given vector field. You know that if vector F is the velocity field of a flowing fluid then curl F for the vector field {eq}{\rm{curl}}\vec F = \left| {\begin{array}{*{20}{c}} {\vec i}&{\vec j}&{\vec k}\\ {\frac{\partial }{{\partial x}}}&{\frac{\partial }{{\partial y}}}&{\frac{\partial }{{\partial z}}}\\ {{x^2}z}&{ - 2xz}&{yz} \end{array}} \right| {/eq}, substitute the value and find curl F for the vector field at the given point.

Answer and Explanation:

The given vector F will be {eq}F\left( {x,y,z} \right) = x^2zi - 2xzj + yzk. {/eq}

First find the curl of vector F and you have $$\begin{align*} F\left( {x,y,z} \right) &= {x^2}z\vec i - 2xz\vec j + yz\vec k\\ {\rm{curl}}\vec F &= \left| {\begin{array}{*{20}{c}} {\vec i}&{\vec j}&{\vec k}\\ {\frac{\partial }{{\partial x}}}&{\frac{\partial }{{\partial y}}}&{\frac{\partial }{{\partial z}}}\\ {{x^2}z}&{ - 2xz}&{yz} \end{array}} \right|\\ &= \left[ {\frac{\partial }{{\partial y}}\left\{ {yz} \right\} - \frac{\partial }{{\partial z}}\left( { - 2xz} \right)} \right]\vec i - \left[ {\frac{\partial }{{\partial x}}\left\{ {yz} \right\} - \frac{\partial }{{\partial z}}\left\{ {{x^2}z} \right\}} \right]\vec j + \left[ {\frac{\partial }{{\partial x}}\left( { - 2xz} \right) - \frac{\partial }{{\partial y}}\left\{ {{x^2}z} \right\}} \right]\vec j\\ {\rm{curl}}\vec F &= \left( {z + 2x} \right)\vec i - \left( {0 - {x^2}} \right)\vec j + \left[ { - 2z - 0} \right]\vec k\\ {\rm{curl}}\vec F &= \left( {z + 2x} \right)\vec i + {x^2}\vec j - 2z\vec k. \end{align*} $$

At point {eq}(3, -9, 1) {/eq}, the curl F for the vector field will be $$\begin{align*} {\rm{curl}}\vec F &= \left( {z + 2x} \right)\vec i + {x^2}\vec j - 2z\vec k\\ {\left. {{\rm{curl}}\vec F} \right|_{\left( {3, - 9,1} \right)}} &= \left( {1 + 2 \cdot 3} \right)\vec i + {\left( 3 \right)^2}\vec j - 2\left( { - 9} \right)\vec k\\ &= 7\vec i + 9\vec j + 18\vec k. \end{align*} $$


Learn more about this topic:

Vectors: Definition, Types & Examples

from Common Entrance Test (CET): Study Guide & Syllabus

Chapter 57 / Lesson 3
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