Problem 25 - APEX Calculus

Question

In Exercises \(25-28,\) a function \(w=F(x, y, z),\) a vector \(\vec{v}\) and a point \(P\) are given. (a) Find \(\nabla F(x, y, z)\). (b) Find \(D_{\vec{u}} F\) at \(P,\) where \(\vec{u}\) is the unit vector in the direction of \(\vec{v}\). $$ F(x, y, z)=3 x^{2} z^{3}+4 x y-3 z^{2}, \vec{v}=\langle 1,1,1\rangle, P=(3,2,1) $$

Step-by-Step Solution

Answer

The gradient is \( \nabla F(3, 2, 1) = \langle 26, 12, 75 \rangle \) and \( D_{\vec{u}} F \) is \( \frac{113 \sqrt{3}}{3} \).

Step 1: Compute Partial Derivatives

To find the gradient \(abla F(x, y, z)\), we compute the partial derivatives of \(F\) with respect to \(x\), \(y\), and \(z\). - The partial derivative \(\frac{\partial F}{\partial x}\) is computed by treating \(y\) and \(z\) as constants: \[ \frac{\partial F}{\partial x} = 6x z^3 + 4y \]- The partial derivative \(\frac{\partial F}{\partial y}\) is computed by treating \(x\) and \(z\) as constants: \[ \frac{\partial F}{\partial y} = 4x \]- The partial derivative \(\frac{\partial F}{\partial z}\) is computed by treating \(x\) and \(y\) as constants: \[ \frac{\partial F}{\partial z} = 9x^2 z^2 - 6z \].

Step 2: Construct the Gradient Vector

Using the partial derivatives, construct the gradient vector:\[ abla F(x, y, z) = \left\langle 6x z^3 + 4y, 4x, 9x^2 z^2 - 6z \right\rangle \]

Step 3: Evaluate the Gradient at the Point P

Substitute the given point \(P = (3, 2, 1)\) into the gradient vector to evaluate it at \(P\):- Substitute into \(\frac{\partial F}{\partial x}:\) \[ 6(3)(1)^3 + 4(2) = 18 + 8 = 26 \]- Substitute into \(\frac{\partial F}{\partial y}:\) \[ 4(3) = 12 \]- Substitute into \(\frac{\partial F}{\partial z}:\) \[ 9(3)^2 (1)^2 - 6(1) = 81 - 6 = 75 \]Thus, \[ abla F(3, 2, 1) = \langle 26, 12, 75 \rangle \].

Step 4: Compute the Unit Vector \(\vec{u}\)

First, find the magnitude of \(\vec{v} = \langle 1, 1, 1 \rangle\):\[ \| \vec{v} \| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3} \].Then, the unit vector \(\vec{u}\) in the direction of \(\vec{v}\) is:\[ \vec{u} = \left\langle \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \right\rangle \].

Step 5: Evaluate the Directional Derivative

The directional derivative \(D_{\vec{u}} F\) at point \(P\) is given by the dot product of the gradient vector \(abla F\) evaluated at \(P\) and the unit vector \(\vec{u}\):\[ D_{\vec{u}} F = abla F(3, 2, 1) \cdot \vec{u} = \langle 26, 12, 75 \rangle \cdot \left\langle \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \right\rangle \].- Compute the dot product: \[ \frac{1}{\sqrt{3}}(26) + \frac{1}{\sqrt{3}}(12) + \frac{1}{\sqrt{3}}(75) = \frac{26}{\sqrt{3}} + \frac{12}{\sqrt{3}} + \frac{75}{\sqrt{3}} \]- Simplify: \[ = \frac{113}{\sqrt{3}} \],which can also be rationalized to: \[ \frac{113 \sqrt{3}}{3} \].

Key Concepts

Partial Derivatives

Partial derivatives help in understanding how a multivariable function changes as we vary each variable independently. When calculating a partial derivative, we treat all other variables as constants. For example in the context of the function \( F(x, y, z)=3 x^{2} z^{3}+4 x y-3 z^{2} \):

• The partial derivative with respect to \(x\) is calculated by holding \(y\) and \(z\) constant, which results in \( \frac{\partial F}{\partial x} = 6x z^3 + 4y \).
• The partial derivative with respect to \(y\) simply treats \(x\) and \(z\) as constants, giving us \( \frac{\partial F}{\partial y} = 4x \).
• Lastly, the partial derivative with respect to \(z\) yields \( \frac{\partial F}{\partial z} = 9x^2 z^2 - 6z \), treating \(x\) and \(y\) as constants.

Understanding these derivations clarifies the reliance of \(F\) upon its variables.

Gradient Vector

The gradient vector is a powerful tool in multivariable calculus that all pointing in the direction of most rapid increase of the function. It is formed by combining the partial derivatives into a vector. For our function, this looks like:

• \( abla F(x, y, z) = \left\langle 6x z^3 + 4y, 4x, 9x^2 z^2 - 6z \right\rangle \)

To evaluate the gradient at a specific point, such as \(P = (3, 2, 1)\), we substitute the values of \(x, y,\) and \(z\) into each component:

• \(26 = 6(3)(1)^3 + 4(2) \)
• \(12 = 4(3) \)
• \(75 = 9(3)^2(1)^2 - 6(1) \)

Thus, \( abla F(3, 2, 1) = \langle 26, 12, 75 \rangle \), indicating the direction of maximum increase from the point \(P\).

Unit Vector

A unit vector is a vector with a magnitude of 1, used to indicate direction. It’s essential in the context of directional derivatives. To find a unit vector in the direction of the given vector \( \vec{v} = \langle 1, 1, 1 \rangle \), we first compute its magnitude:

• \( \| \vec{v} \| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3} \).

Then, each component of \(\vec{v}\) is divided by this magnitude:

• \( \vec{u} = \left\langle \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \right\rangle \).

This unit vector \( \vec{u} \) maintains the direction of \( \vec{v} \) but rescales it to a unit length.

Directional Derivative

The directional derivative gives us the rate of change of a function as we move in the direction of a specific vector. It is computed as the dot product of the gradient vector and a unit vector representing the direction. For our exercise, the directional derivative at point \( P \) in the direction of \( \vec{u} \) is found by:

• Taking the dot product \( abla F(3, 2, 1) \cdot \vec{u} = \langle 26, 12, 75 \rangle \cdot \left\langle \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \right\rangle \)
• Calculating \( \frac{26}{\sqrt{3}} + \frac{12}{\sqrt{3}} + \frac{75}{\sqrt{3}} \)
• Simplifying to \( \frac{113}{\sqrt{3}} \), or \( \frac{113 \sqrt{3}}{3} \) after rationalization.

Understanding the directional derivative allows us to predict behavior at any point in a given direction.