Find The Maximum Rate Of Change Of F F F At The Given Point And The Direction In Which It Occurs. F ( X , Y ) = 8 X Y 2 , ( 3 , − 5 F(x, Y)=8xy^2, \quad(3,-5 F ( X , Y ) = 8 X Y 2 , ( 3 , − 5 ]- Maximum Rate Of Change: □ \square □ - Direction Vector: □ \square □

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Introduction

In calculus, the rate of change of a function is a measure of how fast the output of the function changes when the input changes. This is a fundamental concept in mathematics and has numerous applications in various fields, including physics, engineering, and economics. In this article, we will discuss how to find the maximum rate of change of a function at a given point and the direction in which it occurs.

The Problem

We are given a function f(x,y)=8xy2f(x, y) = 8xy^2 and a point (3,−5)(3, -5). Our goal is to find the maximum rate of change of ff at the given point and the direction in which it occurs.

The Concept of Partial Derivatives

To find the rate of change of a function, we need to use partial derivatives. The partial derivative of a function f(x,y)f(x, y) with respect to xx is denoted by ∂f∂x\frac{\partial f}{\partial x}, and the partial derivative with respect to yy is denoted by ∂f∂y\frac{\partial f}{\partial y}. The partial derivatives represent the rate of change of the function with respect to each variable separately.

Calculating Partial Derivatives

To find the partial derivatives of the function f(x,y)=8xy2f(x, y) = 8xy^2, we will use the power rule of differentiation. The power rule states that if f(x)=xnf(x) = x^n, then f′(x)=nxn−1f'(x) = nx^{n-1}. We will apply this rule to each term in the function.

Partial Derivative with Respect to xx

To find the partial derivative with respect to xx, we will differentiate the function with respect to xx while treating yy as a constant.

∂f∂x=∂∂x(8xy2)\frac{\partial f}{\partial x} = \frac{\partial}{\partial x} (8xy^2)

Using the power rule, we get:

∂f∂x=8y2\frac{\partial f}{\partial x} = 8y^2

Partial Derivative with Respect to yy

To find the partial derivative with respect to yy, we will differentiate the function with respect to yy while treating xx as a constant.

∂f∂y=∂∂y(8xy2)\frac{\partial f}{\partial y} = \frac{\partial}{\partial y} (8xy^2)

Using the power rule, we get:

∂f∂y=16xy\frac{\partial f}{\partial y} = 16xy

Finding the Maximum Rate of Change

The maximum rate of change of a function at a given point is the magnitude of the gradient vector at that point. The gradient vector is a vector that points in the direction of the maximum rate of change and has a magnitude equal to the maximum rate of change.

To find the gradient vector, we need to find the partial derivatives of the function at the given point. We will substitute the values of xx and yy into the partial derivatives we found earlier.

Partial Derivative with Respect to xx at (3,−5)(3, -5)

∂f∂x(3,−5)=8(−5)2=200\frac{\partial f}{\partial x} (3, -5) = 8(-5)^2 = 200

Partial Derivative with Respect to yy at (3,−5)(3, -5)

∂f∂y(3,−5)=16(3)(−5)=−240\frac{\partial f}{\partial y} (3, -5) = 16(3)(-5) = -240

Finding the Gradient Vector

The gradient vector is a vector that points in the direction of the maximum rate of change and has a magnitude equal to the maximum rate of change. The gradient vector is given by:

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

Substituting the values of the partial derivatives, we get:

∇f(3,−5)=(200,−240)\nabla f (3, -5) = (200, -240)

Finding the Magnitude of the Gradient Vector

The magnitude of the gradient vector is the maximum rate of change of the function at the given point. It is given by:

∥∇f(x,y)∥=(∂f∂x)2+(∂f∂y)2\|\nabla f (x, y)\| = \sqrt{\left( \frac{\partial f}{\partial x} \right)^2 + \left( \frac{\partial f}{\partial y} \right)^2}

Substituting the values of the partial derivatives, we get:

∥∇f(3,−5)∥=(200)2+(−240)2=40000+57600=97600=312.5\|\nabla f (3, -5)\| = \sqrt{(200)^2 + (-240)^2} = \sqrt{40000 + 57600} = \sqrt{97600} = 312.5

Finding the Direction Vector

The direction vector is a vector that points in the direction of the maximum rate of change. It is given by:

u=∇f(x,y)∥∇f(x,y)∥\mathbf{u} = \frac{\nabla f (x, y)}{\|\nabla f (x, y)\|}

Substituting the values of the gradient vector and its magnitude, we get:

u=(200,−240)312.5=(200312.5,−240312.5)=(0.64,−0.77)\mathbf{u} = \frac{(200, -240)}{312.5} = \left( \frac{200}{312.5}, \frac{-240}{312.5} \right) = \left( 0.64, -0.77 \right)

Conclusion

In this article, we discussed how to find the maximum rate of change of a function at a given point and the direction in which it occurs. We used the concept of partial derivatives and the gradient vector to find the maximum rate of change and the direction vector. We applied these concepts to the function f(x,y)=8xy2f(x, y) = 8xy^2 at the point (3,−5)(3, -5) and found the maximum rate of change to be 312.5 and the direction vector to be (0.64,−0.77)\left( 0.64, -0.77 \right).

Discussion

The maximum rate of change of a function at a given point is an important concept in mathematics and has numerous applications in various fields. It is used to find the rate of change of a function with respect to each variable separately and to determine the direction in which the function changes most rapidly. The gradient vector is a powerful tool for finding the maximum rate of change and the direction vector, and it is widely used in various fields, including physics, engineering, and economics.

References

  • [1] Stewart, J. (2016). Calculus: Early Transcendentals. Cengage Learning.
  • [2] Anton, H. (2017). Calculus: A New Horizon. John Wiley & Sons.
  • [3] Edwards, C. H. (2019). Calculus: Early Transcendentals. Pearson Education.

Additional Resources

Introduction

In our previous article, we discussed how to find the maximum rate of change of a function at a given point and the direction in which it occurs. We used the concept of partial derivatives and the gradient vector to find the maximum rate of change and the direction vector. In this article, we will answer some frequently asked questions related to the maximum rate of change of a function.

Q1: What is the maximum rate of change of a function?

A1: The maximum rate of change of a function is the magnitude of the gradient vector at a given point. It represents the rate of change of the function with respect to each variable separately.

Q2: How do I find the maximum rate of change of a function?

A2: To find the maximum rate of change of a function, you need to find the partial derivatives of the function with respect to each variable, evaluate them at the given point, and then find the magnitude of the gradient vector.

Q3: What is the gradient vector?

A3: The gradient vector is a vector that points in the direction of the maximum rate of change and has a magnitude equal to the maximum rate of change. It is given by:

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

Q4: How do I find the direction vector?

A4: To find the direction vector, you need to divide the gradient vector by its magnitude. The direction vector is given by:

u=∇f(x,y)∥∇f(x,y)∥\mathbf{u} = \frac{\nabla f (x, y)}{\|\nabla f (x, y)\|}

Q5: What is the significance of the maximum rate of change of a function?

A5: The maximum rate of change of a function is an important concept in mathematics and has numerous applications in various fields, including physics, engineering, and economics. It is used to find the rate of change of a function with respect to each variable separately and to determine the direction in which the function changes most rapidly.

Q6: Can I use the maximum rate of change of a function to find the minimum rate of change?

A6: No, the maximum rate of change of a function is not directly related to the minimum rate of change. However, you can use the concept of the gradient vector to find the direction of the minimum rate of change.

Q7: How do I find the minimum rate of change of a function?

A7: To find the minimum rate of change of a function, you need to find the direction of the minimum rate of change, which is opposite to the direction of the maximum rate of change. You can use the concept of the gradient vector to find the direction of the minimum rate of change.

Q8: Can I use the maximum rate of change of a function to find the rate of change of a function at a point?

A8: Yes, you can use the maximum rate of change of a function to find the rate of change of a function at a point. The maximum rate of change of a function at a point is equal to the magnitude of the gradient vector at that point.

Q9: How do I find the rate of change of a function at a point?

A9: To find the rate of change of a function at a point, you need to find the partial derivatives of the function with respect to each variable, evaluate them at the given point, and then find the magnitude of the gradient vector.

Q10: What is the relationship between the maximum rate of change of a function and the minimum rate of change?

A10: The maximum rate of change of a function and the minimum rate of change are related in the sense that the direction of the minimum rate of change is opposite to the direction of the maximum rate of change.

Conclusion

In this article, we answered some frequently asked questions related to the maximum rate of change of a function. We discussed the concept of the maximum rate of change, the gradient vector, and the direction vector, and provided examples and explanations to help clarify the concepts.

Discussion

The maximum rate of change of a function is an important concept in mathematics and has numerous applications in various fields. It is used to find the rate of change of a function with respect to each variable separately and to determine the direction in which the function changes most rapidly. The gradient vector is a powerful tool for finding the maximum rate of change and the direction vector, and it is widely used in various fields, including physics, engineering, and economics.

References

  • [1] Stewart, J. (2016). Calculus: Early Transcendentals. Cengage Learning.
  • [2] Anton, H. (2017). Calculus: A New Horizon. John Wiley & Sons.
  • [3] Edwards, C. H. (2019). Calculus: Early Transcendentals. Pearson Education.

Additional Resources