Chapter 15: Problem 19
Find div \(\mathbf{F}\) and curl \(\mathbf{F}\).\(\mathbf{F}(x, y, z)=7 y^{3} z^{2} \mathbf{i}-8 x^{2} z^{5} \mathbf{j}-3 x y^{4} \mathbf{k}\)
Short Answer
Expert verified
div \( \mathbf{F} = 0 \), curl \( \mathbf{F} = (40 x^2 z^4 - 12 x y^3) \mathbf{i} + (3 y^4 + 14 y^3 z) \mathbf{j} + (-16 x z^5 - 21 y^2 z^2) \mathbf{k} \).
Step by step solution
01
Understand the vector field
The given vector field is \( \mathbf{F}(x, y, z) = 7 y^{3} z^{2} \mathbf{i} - 8 x^{2} z^{5} \mathbf{j} - 3 x y^{4} \mathbf{k} \). It has components \( F_1 = 7 y^3 z^2 \), \( F_2 = -8 x^2 z^5 \), and \( F_3 = -3 x y^4 \). Our task is to find its divergence and curl.
02
Compute the divergence of \( \mathbf{F} \)
The divergence of a vector field \( \mathbf{F} = (F_1, F_2, F_3) \) is given by \( \text{div} \; \mathbf{F} = \frac{\partial F_1}{\partial x} + \frac{\partial F_2}{\partial y} + \frac{\partial F_3}{\partial z} \). Calculate each partial derivative: \( \frac{\partial F_1}{\partial x} = 0 \), \( \frac{\partial F_2}{\partial y} = 0 \), \( \frac{\partial F_3}{\partial z} = 0 \). Therefore, \( \text{div} \; \mathbf{F} = 0 + 0 + 0 = 0 \).
03
Compute partial derivatives for curl
Since curl involves cross partial derivatives, start by computing: \( \frac{\partial F_3}{\partial y} = -12 x y^3 \), \( \frac{\partial F_2}{\partial z} = -40 x^2 z^4 \), \( \frac{\partial F_1}{\partial z} = 14 y^3 z \), \( \frac{\partial F_3}{\partial x} = -3 y^4 \), \( \frac{\partial F_1}{\partial y} = 21 y^2 z^2 \), \( \frac{\partial F_2}{\partial x} = -16 x z^5 \).
04
Compute the curl of \( \mathbf{F} \)
The curl of a vector field \( \mathbf{F} = (F_1, F_2, F_3) \) is given by:\( \text{curl} \; \mathbf{F} = \left( \frac{\partial F_3}{\partial y} - \frac{\partial F_2}{\partial z} \right) \mathbf{i} - \left( \frac{\partial F_3}{\partial x} - \frac{\partial F_1}{\partial z} \right) \mathbf{j} + \left( \frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y} \right) \mathbf{k} \).Substitute the derivatives: \[ \left( -12 x y^3 + 40 x^2 z^4 \right) \mathbf{i} - \left( -3 y^4 - 14 y^3 z \right) \mathbf{j} + \left( -16 x z^5 - 21 y^2 z^2 \right) \mathbf{k} \].So, \( \text{curl} \; \mathbf{F} = (40 x^2 z^4 - 12 x y^3) \mathbf{i} + (3 y^4 + 14 y^3 z) \mathbf{j} + (-16 x z^5 - 21 y^2 z^2) \mathbf{k} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Divergence
Understanding divergence is key when analyzing vector fields like \( \mathbf{F} \). In simpler terms, divergence measures how much a vector field spreads out from a given point or if it's converging into that point. For the vector field \( \mathbf{F}(x, y, z) = 7 y^{3} z^{2} \mathbf{i} - 8 x^{2} z^{5} \mathbf{j} - 3 x y^{4} \mathbf{k} \), we use the formula:
- \( \text{div} \, \mathbf{F} = \frac{\partial F_1}{\partial x} + \frac{\partial F_2}{\partial y} + \frac{\partial F_3}{\partial z} \)
Curl
The curl of a vector field like \( \mathbf{F} \) provides information about the rotation or swirling of the vector field around a point. If you imagine a small paddle wheel placed in the field, the curl tells you how the wheel would spin. For our vector field, \( \mathbf{F} \), the curl is calculated using the cross partial derivatives and the formula:
- \( \text{curl} \, \mathbf{F} = \left( \frac{\partial F_3}{\partial y} - \frac{\partial F_2}{\partial z} \right) \mathbf{i} - \left( \frac{\partial F_3}{\partial x} - \frac{\partial F_1}{\partial z} \right) \mathbf{j} + \left( \frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y} \right) \mathbf{k} \)
Vector Field Analysis
Vector field analysis involves exploring the characteristics of vector functions, like \( \mathbf{F}(x, y, z) \), particularly through divergence and curl. It helps in visualizing complex fields in physics and engineering.
- Divergence: Assesses how a vector field behaves via net flow out of or into a region, offering insight into source/sink dynamics.
- Curl: Captures rotational features, analogous to identifying tiny circular motions within the field.