The expression in the image is:

$\phi(x, y, z) = 3x^2y - y^2z$

We are asked to find the gradient of $\phi$ (denoted as $\nabla \phi$) at the point $(1, 2, 3)$.

Step 1: Calculate the Partial Derivatives

The gradient $\nabla \phi$ is a vector of partial derivatives of $\phi$ with respect to $x$, $y$, and $z$:

$\nabla \phi = \left( \frac{\partial \phi}{\partial x}, \frac{\partial \phi}{\partial y}, \frac{\partial \phi}{\partial z} \right)$

Partial Derivative with respect to $x$:

$\frac{\partial \phi}{\partial x} = \frac{\partial}{\partial x}(3x^2y - y^2z) = 6xy$

Partial Derivative with respect to $y$:

$\frac{\partial \phi}{\partial y} = \frac{\partial}{\partial y}(3x^2y - y^2z) = 3x^2 - 2yz$

Partial Derivative with respect to $z$:

$\frac{\partial \phi}{\partial z} = \frac{\partial}{\partial z}(3x^2y - y^2z) = -y^2$

Step 2: Evaluate the Gradient at the Point $(1, 2, 3)$

Substitute $x = 1$, $y = 2$, and $z = 3$ into the partial derivatives:

• $\frac{\partial \phi}{\partial x} = 6xy = 6(1)(2) = 12$
• $\frac{\partial \phi}{\partial y} = 3x^2 - 2yz = 3(1)^2 - 2(2)(3) = 3 - 12 = -9$
• $\frac{\partial \phi}{\partial z} = -y^2 = -(2)^2 = -4$

Final Answer: The Gradient at $(1, 2, 3)$

$\nabla \phi(1, 2, 3) = \langle 12, -9, -4 \rangle$

Would you like more details or have any questions?

Related Questions:

1. How is the gradient of a function related to the direction of steepest ascent?
2. Can you explain why the gradient is a vector and not a scalar?
3. How would you find the gradient of a function with more than three variables?
4. What is the physical interpretation of the gradient in a 3D space?
5. How would the gradient change if the function $\phi(x, y, z)$ included trigonometric terms?

Tip:

When computing gradients, remember that each partial derivative represents the rate of change of the function with respect to one variable, holding the others constant.