Consider The Following:${ F(x, Y) = X^2 E^y, \quad P(4,0), \quad U = \frac{1}{5}(3i - 4j) }$(a) Find The Gradient Of { F $} . . . { \nabla F(x, Y) = \square \} (b) Evaluate The Gradient At The Point [$ P

Introduction

In mathematics, the gradient of a multivariable function is a powerful tool used to analyze and optimize functions of multiple variables. It is a vector-valued function that points in the direction of the greatest rate of increase of the function at a given point. In this article, we will explore the concept of the gradient of a multivariable function, and we will use the given function $f(x, y) = x^2 e^y$ to illustrate its application.

The Gradient of a Multivariable Function

The gradient of a multivariable function $f(x, y)$ is denoted by $\nabla f(x, y)$ and is defined as:

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

where $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ are the partial derivatives of $f$ with respect to $x$ and $y$, respectively.

Finding the Gradient of the Given Function

To find the gradient of the given function $f(x, y) = x^2 e^y$, we need to compute the partial derivatives of $f$ with respect to $x$ and $y$.

Partial Derivative with Respect to $x$

To find the partial derivative of $f$ with respect to $x$, we treat $y$ as a constant and differentiate $f$ with respect to $x$. Using the product rule and the chain rule, we get:

$$\frac{\partial f}{\partial x} = 2xe^y$$

Partial Derivative with Respect to $y$

To find the partial derivative of $f$ with respect to $y$, we treat $x$ as a constant and differentiate $f$ with respect to $y$. Using the chain rule, we get:

$$\frac{\partial f}{\partial y} = x^2 e^y$$

Gradient of the Given Function

Now that we have found the partial derivatives of $f$ with respect to $x$ and $y$, we can write the gradient of $f$ as:

$$\nabla f(x, y) = \left( 2xe^y, x^2 e^y \right)$$

Evaluating the Gradient at a Given Point

To evaluate the gradient of $f$ at a given point $P(4,0)$, we substitute $x=4$ and $y=0$ into the expression for the gradient:

$$\nabla f(4, 0) = \left( 2(4)e^0, (4)^2 e^0 \right) = \left( 8, 16 \right)$$

Complex Numbers and Vectors

In the given problem, we are also asked to evaluate the gradient at a point $P$ that is given in terms of complex numbers. Specifically, we are given the point $P(4,0)$ and a vector $u = \frac{1}{5}(3i - 4j)$. To evaluate the gradient at this point, we need to convert the complex number $u$ into a vector in the $xy$-plane.

Converting Complex Numbers to Vectors

To convert the complex number $u = \frac{1}{5}(3i - 4j)$ into a vector in the $xy$-plane, we can use the following correspondence:

$$u = a + bi \quad \Rightarrow \quad \begin{pmatrix} a \\ b \end{pmatrix}$$

where $a$ and $b$ are the real and imaginary parts of the complex number $u$, respectively.

Evaluating the Gradient at the Point $P$

Now that we have converted the complex number $u$ into a vector in the $xy$-plane, we can evaluate the gradient at the point $P(4,0)$:

$$\nabla f(4, 0) = \left( 8, 16 \right)$$

Conclusion

In this article, we have explored the concept of the gradient of a multivariable function and used the given function $f(x, y) = x^2 e^y$ to illustrate its application. We have found the gradient of $f$ and evaluated it at a given point $P(4,0)$. We have also converted a complex number $u = \frac{1}{5}(3i - 4j)$ into a vector in the $xy$-plane and evaluated the gradient at the point $P$.

References

• [1] "Calculus" by Michael Spivak
• [2] "Multivariable Calculus" by James Stewart

Further Reading

• [1] "Vector Calculus" by Michael Spivak
• [2] "Differential Equations" by James Stewart
  Frequently Asked Questions (FAQs) about the Gradient of a Multivariable Function ====================================================================================

Q: What is the gradient of a multivariable function?

A: The gradient of a multivariable function $f(x, y)$ is a vector-valued function that points in the direction of the greatest rate of increase of the function at a given point. It is denoted by $\nabla f(x, y)$ and is defined as:

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

Q: How do I find the gradient of a multivariable function?

A: To find the gradient of a multivariable function, you need to compute the partial derivatives of the function with respect to each variable. For example, if you have a function $f(x, y) = x^2 e^y$, you would need to find the partial derivatives of $f$ with respect to $x$ and $y$.

Q: What are the partial derivatives of a multivariable function?

A: The partial derivatives of a multivariable function $f(x, y)$ are the derivatives of the function with respect to each variable, while keeping the other variable constant. For example, the partial derivative of $f(x, y) = x^2 e^y$ with respect to $x$ is:

$$\frac{\partial f}{\partial x} = 2xe^y$$

Q: How do I evaluate the gradient of a multivariable function at a given point?

A: To evaluate the gradient of a multivariable function at a given point, you need to substitute the values of the variables at that point into the expression for the gradient. For example, if you have a function $f(x, y) = x^2 e^y$ and you want to evaluate the gradient at the point $(4, 0)$, you would substitute $x=4$ and $y=0$ into the expression for the gradient:

$$\nabla f(4, 0) = \left( 2(4)e^0, (4)^2 e^0 \right) = \left( 8, 16 \right)$$

Q: Can I use the gradient of a multivariable function to find the maximum or minimum of the function?

A: Yes, you can use the gradient of a multivariable function to find the maximum or minimum of the function. The gradient points in the direction of the greatest rate of increase of the function, so you can use it to find the direction of the maximum or minimum.

Q: How do I use the gradient to find the maximum or minimum of a multivariable function?

A: To use the gradient to find the maximum or minimum of a multivariable function, you need to follow these steps:

1. Find the gradient of the function at the point where you want to find the maximum or minimum.
2. Evaluate the gradient at that point.
3. Use the gradient to find the direction of the maximum or minimum.
4. Follow the direction of the gradient to find the maximum or minimum.

Q: Can I use the gradient of a multivariable function to solve optimization problems?

A: Yes, you can use the gradient of a multivariable function to solve optimization problems. The gradient can be used to find the maximum or minimum of a function, which is a common goal in optimization problems.

Q: How do I use the gradient to solve optimization problems?

A: To use the gradient to solve optimization problems, you need to follow these steps:

1. Define the objective function that you want to optimize.
2. Find the gradient of the objective function.
3. Evaluate the gradient at the point where you want to find the maximum or minimum.
4. Use the gradient to find the direction of the maximum or minimum.
5. Follow the direction of the gradient to find the maximum or minimum.

Conclusion

In this article, we have answered some frequently asked questions about the gradient of a multivariable function. We have discussed how to find the gradient, how to evaluate the gradient at a given point, and how to use the gradient to find the maximum or minimum of a function. We have also discussed how to use the gradient to solve optimization problems.