Question

G find the gradient field of the function g(x, y, z) = 8 x y + 5yz + 5xz

Answer 1

If the original volumes of liquid is represented by x, we have x=x for A and B. Since 120mL is poured from A to B, A loses 120 mL (x-120) and B gains 120 mL (x+120). Since we know that B contains 4 times as much liquid as A is now (x-120), we have 4*(x-120)=(x+120)=B. Using the distributive property, we get 4x-480=x+120. Adding 480 to both sides, we get 4x=x+600. After that, we can subtract both sides by x to get 3x=600. Lastly, we can divide both sides by 3 to get x=200=the original amount of liquid. Since A=x-120, we get A=200-120=80
 

Answer 2

Answer:

$\displaystyle \nabla g(x, y, z) = \Big( 8y + 5z \Big) \hat{\i} + \Big( 8x + 5z \Big) \hat{\j} + \Big( 5y + 5x \Big) \hat{\text{k}}$

General Formulas and Concepts:

Calculus

Differentiation

• Derivatives
• Derivative Notation

Derivative Rule [Basic Power Rule]:

1. f(x) = cxⁿ
2. f’(x) = c·nxⁿ⁻¹

Multivariable Calculus

Differentiation

• Partial Derivatives
• Derivative Notation

Gradient:                                                                                                             $\displaystyle \nabla f(x, y, z) = \frac{\partial f}{\partial x} \hat{\i} + \frac{\partial f}{\partial y} \hat{\j} + \frac{\partial f}{\partial z} \hat{\text{k}}$

Gradient Property [Addition/Subtraction]:                                                          $\displaystyle \nabla \big[ f(x) + g(x) \big] = \nabla f(x) + \nabla g(x)$

Gradient Property [Multiplied Constant]:                                                            $\displaystyle \nabla \big[ \alpha f(x) \big] = \alpha \nabla f(x)$

Gradient Rule [Product Rule]:                                                                              $\displaystyle \nabla \big[ f(x)g(x) \big] = f(x) \nabla g(x) + \nabla f(x) g(x)$

Step-by-step explanation:

Step 1: Define

Identify.

$\displaystyle g(x, y, z) = 8xy + 5yz + 5xz$

Step 2: Find Gradient

1. [Function] Differentiate [Gradient]:                                                              $\displaystyle \nabla g(x, y, z) = \frac{\partial}{\partial x} \Big( 8xy + 5yz + 5xz \Big) \hat{\i} + \frac{\partial}{\partial y} \Big( 8xy + 5yz + 5zx \Big) \hat{\j} + \frac{\partial}{\partial z} \Big( 8xy + 5yz + 5zx \Big) \hat{\text{k}}$
2. [Gradient] Rewrite [Gradient Property - Addition/Subtraction]:                $\displaystyle \nabla g(x, y, z) = \Big[ \frac{\partial}{\partial x}(8xy) + \frac{\partial}{\partial x}(5yz) + \frac{\partial}{\partial x}(5xz) \Big] \hat{\i} + \Big[ \frac{\partial}{\partial y}(8xy) + \frac{\partial}{\partial y}(5yz) + \frac{\partial}{\partial y}(5zx) \Big] \hat{\j} + \Big[ \frac{\partial}{\partial z}(8xy) + \frac{\partial}{\partial z}(5yz) + \frac{\partial}{\partial z}(5zx) \Big] \hat{\text{k}}$
3. [Gradient] Rewrite [Gradient Property - Multiplied Constants]:                $\displaystyle \nabla g(x, y, z) = \Big[ 8 \frac{\partial}{\partial x}(xy) + 5 \frac{\partial}{\partial x}(yz) + 5 \frac{\partial}{\partial x}(xz) \Big] \hat{\i} + \Big[ 8 \frac{\partial}{\partial y}(xy) + 5 \frac{\partial}{\partial y}(yz) + 5 \frac{\partial}{\partial y}(zx) \Big] \hat{\j} + \Big[ 8 \frac{\partial}{\partial z}(xy) + 5 \frac{\partial}{\partial z}(yz) + 5 \frac{\partial}{\partial z}(zx) \Big] \hat{\text{k}}$
4. [Gradient] Differentiate [Derivative Rule - Basic Power Rule]:                  $\displaystyle \nabla g(x, y, z) = \Big( 8y + 5z \Big) \hat{\i} + \Big( 8x + 5z \Big) \hat{\j} + \Big( 5y + 5x \Big) \hat{\text{k}}$

∴ the gradient of the function is <8y + 5z, 8x + 5z, 5y + 5x>.

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Learn more about gradients: https://brainly.com/question/6158243
Learn more about multivariable calculus: https://brainly.com/question/17433118

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Topic: Multivariable Calculus

Unit: Directional Derivatives