Find The Directional Derivative Of The Function AtPin The Direction Ofv.f(x,y)=x3y3,P(8,5),v=22(i+j)Find
Understanding how to compute the directional derivative of a function provides valuable insight into how the function changes at a particular point in a specific direction. In this article, we will explore the step-by-step process to find the directional derivative of the function \(f(x, y) = x^3 y^3\) at the point \(P(8, 5)\), in the direction of the vector \(\mathbf{v} = 2 2 (\mathbf{i} + \mathbf{j})\). We'll break down the concepts involved, perform necessary calculations, and clarify each step to ensure a comprehensive understanding.
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Understanding the Concept of the Directional Derivative
What Is a Directional Derivative?
The directional derivative measures the rate at which a function \(f(x, y)\) changes at a point \((x0, y0)\) in the direction of a given vector \(\mathbf{v}\). Unlike partial derivatives which measure change along coordinate axes, the directional derivative assesses change in an arbitrary direction.Mathematically, the directional derivative of \(f\) at \((x0, y0)\) in the direction of a unit vector \(\mathbf{u} = (ux, uy)\) is defined as:
\[
D{\mathbf{u}}f(x0, y0) = \lim{h \to 0} \frac{f(x0 + h ux, y0 + h uy) - f(x0, y0)}{h}
\]
Alternatively, it can be computed using the gradient \(\nabla f\):
\[
D{\mathbf{u}}f(x0, y0) = \nabla f(x0, y_0) \cdot \mathbf{u}
\]
where \(\nabla f(x, y) = \left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}\right)\).
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Step-by-Step Approach to Find the Directional Derivative
1. Find the Gradient \(\nabla f(x, y)\)
The gradient vector contains the partial derivatives of \(f\): \[ \nabla f(x, y) = \left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}\right) \] Given \(f(x, y) = x^3 y^3\), compute: \[ \frac{\partial f}{\partial x} = 3x^2 y^3 \] \[ \frac{\partial f}{\partial y} = 3x^3 y^2 \]2. Evaluate the Gradient at the Given Point \(P(8, 5)\)
Substitute \(x=8\), \(y=5\): \[ \frac{\partial f}{\partial x} \bigg|_{(8, 5)} = 3 \times 8^2 \times 5^3 \] \[ \frac{\partial f}{\partial y} \bigg|_{(8, 5)} = 3 \times 8^3 \times 5^2 \]Calculate each:
- \(8^2 = 64\)
- \(5^3 = 125\)
- \(8^3 = 512\)
- \(5^2 = 25\)
Thus:
\[
\frac{\partial f}{\partial x} \bigg|_{(8, 5)} = 3 \times 64 \times 125 = 3 \times 8000 = 24,000
\]
\[
\frac{\partial f}{\partial y} \bigg|_{(8, 5)} = 3 \times 512 \times 25 = 3 \times 12,800 = 38,400
\]
So, the gradient vector at \(P(8, 5)\) is:
\[
\nabla f(8, 5) = (24,000, 38,400)
\]
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3. Normalize the Direction Vector \(\mathbf{v}\)
Given \(\mathbf{v} = 2 2 (\mathbf{i} + \mathbf{j})\). First, interpret this:- The vector \(\mathbf{v}\) is scaled by 2, and in the direction of \(\mathbf{i} + \mathbf{j}\).
Next, normalize \(\mathbf{v}\):
\[
|\mathbf{v}| = \sqrt{2^2 + 2^2} = \sqrt{4 + 4} = \sqrt{8} = 2 \sqrt{2}
\]
The unit vector \(\mathbf{u}\) in the direction of \(\mathbf{v}\) is:
\[
\mathbf{u} = \frac{\mathbf{v}}{|\mathbf{v}|} = \left(\frac{2}{2 \sqrt{2}}, \frac{2}{2 \sqrt{2}}\right) = \left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)
\]
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4. Compute the Directional Derivative
Using the gradient and the unit vector: \[ D_{\mathbf{u}}f(8, 5) = \nabla f(8, 5) \cdot \mathbf{u} \] \[ = (24,000, 38,400) \cdot \left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right) \] \[ = 24,000 \times \frac{1}{\sqrt{2}} + 38,400 \times \frac{1}{\sqrt{2}} = \frac{24,000 + 38,400}{\sqrt{2}} \] \[ = \frac{62,400}{\sqrt{2}} = 62,400 \times \frac{\sqrt{2}}{2} = 62,400 \times \frac{\sqrt{2}}{2} \] \[ = 31,200 \sqrt{2} \]---
Final Result and Interpretation
The directional derivative of the function \(f(x, y) = x^3 y^3\) at the point \(P(8, 5)\) in the direction of the vector \(\mathbf{v} = 2 \mathbf{i} + 2 \mathbf{j}\) is:
\[
\boxed{
D_{\mathbf{u}}f(8, 5) = 31,200 \sqrt{2}
}
\]
This value indicates the rate at which the function increases (or decreases) at the point \(P(8, 5)\) when moving in the specified direction. A positive value signifies an increase, while a negative value (not in this case) would indicate a decrease.
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Additional Insights and Applications
Why Is the Directional Derivative Important?
The directional derivative helps in understanding the behavior of multivariable functions, especially in optimization problems, gradient-based algorithms, and in fields like physics and engineering where the rate of change in specific directions is critical.Applications Include:
- Finding the maximum rate of increase of a function at a point (the magnitude of the gradient).
- Determining the direction in which a function increases most rapidly.
- Designing gradients for optimization algorithms like gradient descent.
- Modeling physical phenomena such as heat transfer or fluid flow where directional rates matter.
Summary
In this article, we've outlined the process to compute the directional derivative of a function at a specific point in a given direction. From calculating the gradient, normalizing the direction vector, to applying the dot product, each step is crucial. Mastering these steps enhances your understanding of multivariable calculus and its applications across various scientific disciplines.---
In conclusion, the key steps to find the directional derivative are:
- Compute the gradient \(\nabla f\).
- Evaluate the gradient at the point of interest.
- Normalize the direction vector to obtain a unit vector.
- Take the dot product of the gradient and the unit vector.
By applying these principles, you can analyze how functions change in any specified direction, empowering you to solve complex problems involving multivariable functions efficiently.
