Math Problem Statement

We define a cosmic energy field M(x, y, z) = (3x^2 + y^3/2 - z^4/2) * ln(x + sqrt(y)). Determine the partial derivatives of M with respect to x, y, and z, and construct the gradient ∇M. What type of field does this represent?

Solution

This task involves finding the partial derivatives of the scalar field M(x,y,z)M(x, y, z) with respect to xx, yy, and zz, and then constructing the gradient vector field M\nabla M. Let's proceed step by step:

Given function:

M(x,y,z)=(3x2+y32z42)ln(x+y).M(x, y, z) = \left( 3x^2 + \frac{y^3}{2} - \frac{z^4}{2} \right) \cdot \ln(x + \sqrt{y}).

Steps:

  1. Partial derivative with respect to xx: Use the product rule x[fg]=fg+fg\frac{\partial}{\partial x} [f \cdot g] = f' \cdot g + f \cdot g'.

  2. Partial derivative with respect to yy: Take into account the term ln(x+y)\ln(x + \sqrt{y}) and its dependence on yy.

  3. Partial derivative with respect to zz: Focus on the term z42-\frac{z^4}{2}, which only involves zz directly.

  4. Construct M\nabla M: Combine the partial derivatives into a vector field: M=(Mx,My,Mz).\nabla M = \left( \frac{\partial M}{\partial x}, \frac{\partial M}{\partial y}, \frac{\partial M}{\partial z} \right).

I will now compute these derivatives explicitly. sympy import symbols, diff, sqrt, ln

Define variables

x, y, z = symbols('x y z')

Define the scalar field M

M = (3*x2 + y3/2 - z**4/2) * ln(x + sqrt(y))

Compute partial derivatives

partial_x = diff(M, x) partial_y = diff(M, y) partial_z = diff(M, z)

Gradient vector field (∇M)

gradient_M = (partial_x, partial_y, partial_z) partial_x, partial_y, partial_z, gradient_M### Results: The partial derivatives of M(x,y,z)M(x, y, z) are as follows:

  1. Partial derivative with respect to xx: Mx=6xln(x+y)+3x2+y32z42x+y.\frac{\partial M}{\partial x} = 6x \ln(x + \sqrt{y}) + \frac{3x^2 + \frac{y^3}{2} - \frac{z^4}{2}}{x + \sqrt{y}}.

  2. Partial derivative with respect to yy: My=3y22ln(x+y)+3x2+y32z422y(x+y).\frac{\partial M}{\partial y} = \frac{3y^2}{2} \ln(x + \sqrt{y}) + \frac{3x^2 + \frac{y^3}{2} - \frac{z^4}{2}}{2\sqrt{y}(x + \sqrt{y})}.

  3. Partial derivative with respect to zz: Mz=2z3ln(x+y).\frac{\partial M}{\partial z} = -2z^3 \ln(x + \sqrt{y}).

Gradient vector field (M\nabla M):

= \left( 6x \ln(x + \sqrt{y}) + \frac{3x^2 + \frac{y^3}{2} - \frac{z^4}{2}}{x + \sqrt{y}}, \ \frac{3y^2}{2} \ln(x + \sqrt{y}) + \frac{3x^2 + \frac{y^3}{2} - \frac{z^4}{2}}{2\sqrt{y}(x + \sqrt{y})}, \ -2z^3 \ln(x + \sqrt{y}) \right).$$ This gradient forms a **vector field**, which represents the rate and direction of change of the scalar field $$M(x, y, z)$$ in space. --- Would you like a detailed explanation of any specific derivative? Or help visualizing this gradient field? ### Related Questions: 1. How do we compute directional derivatives using the gradient field? 2. What does the scalar field $$M(x, y, z)$$ physically represent in this context? 3. How can we interpret the components of $$\nabla M$$ geometrically? 4. How would the gradient behave if $$y \to 0$$? 5. Can the vector field $$\nabla M$$ be classified as conservative? ### Tip: The gradient vector always points in the direction of the steepest ascent of the scalar field $$M(x, y, z)$$.

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Math Problem Analysis

Mathematical Concepts

Partial derivatives
Gradient vector
Scalar fields

Formulas

Product rule: d(fg)/dx = f'g + fg'
Gradient: ∇M = (∂M/∂x, ∂M/∂y, ∂M/∂z)

Theorems

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Suitable Grade Level

University-level Mathematics or Advanced High School Calculus