Question

Find the directional derivative of \(f\) at the point \(\mathbf{p}\) in the direction of \(\mathbf{a}\). \(f(x, y, z)=x^{3} y-y^{2} z^{2} ; \mathbf{p}=(-2,1,3) ; \mathbf{a}=\mathbf{i}-2 \mathbf{j}+2 \mathbf{k}\)

Short answer

The directional derivative is \( \frac{52}{3} \).

Step-by-step solution

Calculate the Gradient of f

The gradient of a function \( f(x, y, z) \) is given by \( abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right) \). Compute each partial derivative:- \( \frac{\partial f}{\partial x} = 3x^2y \)- \( \frac{\partial f}{\partial y} = x^3 - 2yz^2 \)- \( \frac{\partial f}{\partial z} = -2y^2z \) Thus, the gradient is \( abla f = (3x^2y, x^3 - 2yz^2, -2y^2z) \).

Evaluate the Gradient at Point p

Plug the coordinates of the point \( \mathbf{p} = (-2, 1, 3) \) into the gradient to find \( abla f(-2, 1, 3) \):- \( \frac{\partial f}{\partial x} = 3(-2)^2 \cdot 1 = 12 \)- \( \frac{\partial f}{\partial y} = (-2)^3 - 2 \cdot 1 \cdot 3^2 = -8 - 18 = -26 \)- \( \frac{\partial f}{\partial z} = -2 \cdot 1^2 \cdot 3 = -6 \) Thus, \( abla f(-2, 1, 3) = (12, -26, -6) \).

Compute the Unit Vector in Direction a

First, find the magnitude of \( \mathbf{a} = \mathbf{i} - 2\mathbf{j} + 2\mathbf{k} \):\[ ||\mathbf{a}|| = \sqrt{1^2 + (-2)^2 + 2^2} = \sqrt{1 + 4 + 4} = 3 \]The unit vector in the direction of \( \mathbf{a} \) is given by:\[ \hat{\mathbf{a}} = \frac{1}{3}(1, -2, 2) = \left( \frac{1}{3}, -\frac{2}{3}, \frac{2}{3} \right) \]

Calculate the Directional Derivative

The directional derivative of \( f \) at \( \mathbf{p} \) in the direction of \( \mathbf{a} \) is given by:\[ D_\mathbf{a} f = abla f(-2, 1, 3) \cdot \hat{\mathbf{a}} \]Calculate the dot product:\[ = (12, -26, -6) \cdot \left( \frac{1}{3}, -\frac{2}{3}, \frac{2}{3} \right) \]\[ = 12 \cdot \frac{1}{3} + (-26) \cdot \left(-\frac{2}{3}\right) + (-6) \cdot \frac{2}{3} \]\[ = 4 + \frac{52}{3} - 4 \]\[ = \frac{52}{3} \]Thus, the directional derivative is \( \frac{52}{3} \).

Key concepts

Gradient

The gradient of a function is a way of capturing all the partial derivative information of a function in one vector.
Think of this as a multi-dimensional derivative.
If you are dealing with a multivariable function, the gradient points in the direction of the greatest increase of the function. The gradient is noted as \( abla f \).
For a function \( f(x, y, z) \), its gradient is represented as \( abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right) \).
This means you find how the function changes as each variable (\( x \), \( y \), and \( z \)) changes while keeping the others constant.

• \( \frac{\partial f}{\partial x} \) gives change in the \( x \) direction.
• \( \frac{\partial f}{\partial y} \) gives change in the \( y \) direction.
• \( \frac{\partial f}{\partial z} \) gives change in the \( z \) direction.

When you compute the gradient at a specific point, such as \( \mathbf{p} = (-2, 1, 3) \), you are evaluating how the function changes from that point.
The resulting vector is used to understand the slope or steepness of the function in relation to the point.

Partial Derivatives

Partial derivatives are foundational in understanding how functions change when you tweak one variable at a time, while holding the other variables constant. This is like measuring how the temperature varies as you walk in one direction and fixing your position in all other directions.
For our function \( f(x, y, z) = x^3y - y^2z^2 \), we find each partial derivative by taking the derivative with respect to one variable and treating the others as constants:

• \( \frac{\partial f}{\partial x} \) measures sensitivity to changes in \( x \), calculated here as \( 3x^2y \).
• \( \frac{\partial f}{\partial y} \) finds how \( f \) changes with \( y \), given by \( x^3 - 2yz^2 \).
• \( \frac{\partial f}{\partial z} \) shows the rate of change with \( z \), found as \( -2y^2z \).

These calculations form the components of the gradient and each tells a small part of the full story of how the function behaves at every point in space.
At a specific point like \( \mathbf{p} = (-2, 1, 3) \), evaluating these partial derivatives gives us precise directional information about the surface behavior of \( f \).

Unit Vector

When discussing directional derivatives, unit vectors become essential. They transform a direction into a simplified, magnitude-free form.
A unit vector maintains the same direction but has a length of one, making it extremely useful for direction-based measurements without altering the size of change.

To find a unit vector from any vector, like \( \mathbf{a} = \mathbf{i} - 2\mathbf{j} + 2\mathbf{k} \), one follows this formula: divide each component by the vector's magnitude.
Calculating the magnitude involves squaring each component, adding them, and taking the square root:

• The magnitude \( ||\mathbf{a}|| \) then calculates as \( \sqrt{1^2 + (-2)^2 + 2^2} = \sqrt{9} = 3 \).

The unit vector \( \hat{\mathbf{a}} \) is then:

• \( \hat{\mathbf{a}} = \frac{1}{3}(1, -2, 2) \)
• This simplifies to \( \left( \frac{1}{3}, -\frac{2}{3}, \frac{2}{3} \right) \).

With the unit vector, any change is purely directional, which is perfect for finding how steep our function slope gets in that direction at a given point.