1) Find The Directional Derivatives Of $f(x, Y, Z) = E^{2x} \cos(yz$\] At $(0, 0, 0$\] In The Direction Of The Tangent To The Curve:$x = A \sin \theta, \quad Y = A \cos \theta, \quad Z = A \theta$ At $\theta = \pi / 4$.

Introduction

In the realm of multivariable calculus, directional derivatives play a crucial role in understanding the behavior of functions in various directions. Given a function $f(x, y, z)$, the directional derivative at a point $(x_0, y_0, z_0)$ in the direction of a unit vector $\mathbf{u}$ is denoted by $D_{\mathbf{u}}f(x_0, y_0, z_0)$. In this article, we will delve into the concept of directional derivatives and explore how to find them in the direction of the tangent to a curve.

Directional Derivatives

The directional derivative of a function $f(x, y, z)$ at a point $(x_0, y_0, z_0)$ in the direction of a unit vector $\mathbf{u}$ is given by:

$$D_{\mathbf{u}}f(x_0, y_0, z_0) = \nabla f(x_0, y_0, z_0) \cdot \mathbf{u}$$

where $\nabla f(x_0, y_0, z_0)$ is the gradient of $f$ at $(x_0, y_0, z_0)$.

Gradient of a Function

The gradient of a function $f(x, y, z)$ is a vector field that points in the direction of the maximum rate of increase of the function at a given point. It is denoted by $\nabla f$ and is given by:

$$\nabla f = \frac{\partial f}{\partial x} \mathbf{i} + \frac{\partial f}{\partial y} \mathbf{j} + \frac{\partial f}{\partial z} \mathbf{k}$$

where $\mathbf{i}$, $\mathbf{j}$, and $\mathbf{k}$ are the unit vectors in the $x$, $y$, and $z$ directions, respectively.

Tangent Curve

The tangent curve to a surface at a given point is the curve that lies in the tangent plane to the surface at that point. In this article, we will consider the curve given by:

$$x = a \sin \theta, \quad y = a \cos \theta, \quad z = a \theta$$

at $\theta = \pi / 4$.

Finding the Directional Derivative

To find the directional derivative of $f(x, y, z) = e^{2x} \cos(yz)$ at $(0, 0, 0)$ in the direction of the tangent to the curve at $\theta = \pi / 4$, we need to follow these steps:

1. Find the gradient of $f$: We need to find the partial derivatives of $f$ with respect to $x$, $y$, and $z$.
2. Find the tangent vector: We need to find the tangent vector to the curve at $\theta = \pi / 4$.
3. Find the unit vector: We need to find the unit vector in the direction of the tangent vector.
4. Find the directional derivative: We need to find the directional derivative of $f$ at $(0, 0, 0)$ in the direction of the unit vector.

Step 1: Find the Gradient of $f$

To find the gradient of $f$, we need to find the partial derivatives of $f$ with respect to $x$, $y$, and $z$.

$$\frac{\partial f}{\partial x} = 2e^{2x} \cos(yz)$$

$$\frac{\partial f}{\partial y} = -ze^{2x} \sin(yz)$$

$$\frac{\partial f}{\partial z} = -ye^{2x} \sin(yz)$$

Step 2: Find the Tangent Vector

To find the tangent vector to the curve at $\theta = \pi / 4$, we need to find the partial derivatives of the curve with respect to $\theta$.

$$\frac{\partial x}{\partial \theta} = a \cos \theta$$

$$\frac{\partial y}{\partial \theta} = -a \sin \theta$$

$$\frac{\partial z}{\partial \theta} = a$$

Evaluating these partial derivatives at $\theta = \pi / 4$, we get:

$$\frac{\partial x}{\partial \theta} = \frac{a}{\sqrt{2}}$$

$$\frac{\partial y}{\partial \theta} = -\frac{a}{\sqrt{2}}$$

$$\frac{\partial z}{\partial \theta} = a$$

The tangent vector to the curve at $\theta = \pi / 4$ is given by:

$$\mathbf{T} = \frac{\partial x}{\partial \theta} \mathbf{i} + \frac{\partial y}{\partial \theta} \mathbf{j} + \frac{\partial z}{\partial \theta} \mathbf{k}$$

$$\mathbf{T} = \frac{a}{\sqrt{2}} \mathbf{i} - \frac{a}{\sqrt{2}} \mathbf{j} + a \mathbf{k}$$

Step 3: Find the Unit Vector

To find the unit vector in the direction of the tangent vector, we need to divide the tangent vector by its magnitude.

$$\|\mathbf{T}\| = \sqrt{\left(\frac{a}{\sqrt{2}}\right)^2 + \left(-\frac{a}{\sqrt{2}}\right)^2 + a^2}$$

$$\|\mathbf{T}\| = \sqrt{\frac{a^2}{2} + \frac{a^2}{2} + a^2}$$

$$\|\mathbf{T}\| = \sqrt{2a^2}$$

$$\|\mathbf{T}\| = \sqrt{2}a$$

The unit vector in the direction of the tangent vector is given by:

$$\mathbf{u} = \frac{\mathbf{T}}{\|\mathbf{T}\|}$$

$$\mathbf{u} = \frac{1}{\sqrt{2}} \mathbf{i} - \frac{1}{\sqrt{2}} \mathbf{j} + \frac{1}{\sqrt{2}} \mathbf{k}$$

Step 4: Find the Directional Derivative

To find the directional derivative of $f$ at $(0, 0, 0)$ in the direction of the unit vector, we need to evaluate the gradient of $f$ at $(0, 0, 0)$ and take the dot product with the unit vector.

$$\nabla f(0, 0, 0) = \left(2e^{2(0)} \cos(0 \cdot 0), -0e^{2(0)} \sin(0 \cdot 0), -0e^{2(0)} \sin(0 \cdot 0)\right)$$

$$\nabla f(0, 0, 0) = (2, 0, 0)$$

The directional derivative of $f$ at $(0, 0, 0)$ in the direction of the unit vector is given by:

$$D_{\mathbf{u}}f(0, 0, 0) = \nabla f(0, 0, 0) \cdot \mathbf{u}$$

$$D_{\mathbf{u}}f(0, 0, 0) = (2, 0, 0) \cdot \left(\frac{1}{\sqrt{2}} \mathbf{i} - \frac{1}{\sqrt{2}} \mathbf{j} + \frac{1}{\sqrt{2}} \mathbf{k}\right)$$

$$D_{\mathbf{u}}f(0, 0, 0) = \frac{2}{\sqrt{2}}$$

$$D_{\mathbf{u}}f(0, 0, 0) = \sqrt{2}$$

Therefore, the directional derivative of $f(x, y, z) = e^{2x} \cos(yz)$ at $(0, 0, 0)$ in the direction of the tangent to the curve at $\theta = \pi / 4$ is $\sqrt{2}$.

Conclusion

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Introduction

In our previous article, we explored the concept of directional derivatives and how to find them in the direction of the tangent to a curve. We considered the function $f(x, y, z) = e^{2x} \cos(yz)$ and the curve given by $x = a \sin \theta, \quad y = a \cos \theta, \quad z = a \theta$ at $\theta = \pi / 4$. In this article, we will provide a Q&A guide to help you better understand the concept of directional derivatives and how to apply it to various problems.

Q: What is a directional derivative?

A: A directional derivative is a measure of the rate of change of a function in a specific direction. It is denoted by $D_{\mathbf{u}}f(x_0, y_0, z_0)$ and is given by the dot product of the gradient of the function at the point $(x_0, y_0, z_0)$ and the unit vector in the direction of the tangent to the curve.

Q: How do I find the directional derivative of a function?

A: To find the directional derivative of a function, you need to follow these steps:

1. Find the gradient of the function.
2. Find the tangent vector to the curve.
3. Find the unit vector in the direction of the tangent vector.
4. Take the dot product of the gradient of the function and the unit vector.

Q: What is the gradient of a function?

A: The gradient of a function is a vector field that points in the direction of the maximum rate of increase of the function at a given point. It is denoted by $\nabla f$ and is given by:

$$\nabla f = \frac{\partial f}{\partial x} \mathbf{i} + \frac{\partial f}{\partial y} \mathbf{j} + \frac{\partial f}{\partial z} \mathbf{k}$$

Q: How do I find the tangent vector to a curve?

A: To find the tangent vector to a curve, you need to find the partial derivatives of the curve with respect to the parameter.

Q: What is the unit vector in the direction of the tangent vector?

A: The unit vector in the direction of the tangent vector is given by:

$$\mathbf{u} = \frac{\mathbf{T}}{\|\mathbf{T}\|}$$

where $\mathbf{T}$ is the tangent vector and $\|\mathbf{T}\|$ is the magnitude of the tangent vector.

Q: How do I find the directional derivative of a function at a point?

A: To find the directional derivative of a function at a point, you need to evaluate the gradient of the function at the point and take the dot product with the unit vector in the direction of the tangent to the curve.

Q: What are some common applications of directional derivatives?

A: Directional derivatives have many applications in various fields, including:

• Physics: to find the rate of change of a physical quantity in a specific direction.
• Engineering: to design and optimize systems that involve directional derivatives.
• Computer Science: to develop algorithms that involve directional derivatives.

Q: Can I use directional derivatives to find the maximum or minimum of a function?

A: Yes, you can use directional derivatives to find the maximum or minimum of a function. The directional derivative of a function at a point is a measure of the rate of change of the function in a specific direction. By finding the directional derivative of the function at a point, you can determine whether the function is increasing or decreasing in that direction.

Conclusion

In this article, we provided a Q&A guide to help you better understand the concept of directional derivatives and how to apply it to various problems. We hope that this guide has been helpful in clarifying any doubts you may have had about directional derivatives. If you have any further questions or need additional clarification, please don't hesitate to ask.

Additional Resources

For further reading on directional derivatives, we recommend the following resources:

• "Calculus" by Michael Spivak
• "Multivariable Calculus" by James Stewart
• "Vector Calculus" by Peter Baxandall

We hope that this guide has been helpful in your understanding of directional derivatives. Happy learning!