Problem 12
Question
Find the directional derivative of \(f\) at the point \(P\) in the direction of a. $$ f(x, y, z)=e^{x^{2}+y^{2}+z^{2}} ; P=(0,0,0) ; \mathbf{a}=-\mathbf{i}+\mathbf{j}-\mathbf{k} $$
Step-by-Step Solution
Verified Answer
The directional derivative is 0.
1Step 1: Find the Gradient Vector of f
Start by calculating the gradient vector \( abla f \) of the function \( f(x, y, z) = e^{x^2 + y^2 + z^2} \). The gradient is found by taking the partial derivatives with respect to each variable. Evaluate these at point \( P = (0, 0, 0) \). We get: \[ abla f = \left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}\right) = (2xe^{x^2 + y^2 + z^2}, 2ye^{x^2 + y^2 + z^2}, 2ze^{x^2 + y^2 + z^2}) \] Evaluating at \( P = (0, 0, 0) \), \( abla f(0,0,0) = (0, 0, 0) \).
2Step 2: Normalize the Direction Vector
Normalize the vector \( \mathbf{a} = -\mathbf{i} + \mathbf{j} - \mathbf{k} \). The magnitude of \( \mathbf{a} \) is \( \sqrt{(-1)^2 + 1^2 + (-1)^2} = \sqrt{3} \). The normalized vector \( \mathbf{\hat{a}} \) is \( \left(-\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}} \right) \).
3Step 3: Calculate the Directional Derivative
The directional derivative of a function \( f \) at a point \( P \) in the direction of \( \mathbf{\hat{a}} \) is given by \( abla f \cdot \mathbf{\hat{a}} \). Since \( abla f(0,0,0) = (0, 0, 0) \), the dot product with any vector, including \( \mathbf{\hat{a}} \), is zero: \[ abla f \cdot \mathbf{\hat{a}} = (0, 0, 0) \cdot \left(-\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}} \right) = 0 \] Therefore, the directional derivative is 0.
Key Concepts
Gradient VectorPartial DerivativesNormalize Vector
Gradient Vector
When you need to find how a function changes at a given point, the gradient vector becomes your best friend. The gradient, represented by \( abla f \), consists of partial derivatives and points in the direction of greatest increase. Think of it as a multi-dimensional extension of a derivative.
For the function \( f(x, y, z) = e^{x^2 + y^2 + z^2} \), calculating the gradient involves taking the partial derivatives with respect to each variable \(x\), \(y\), and \(z\). These partial derivatives are \( \frac{\partial f}{\partial x} = 2xe^{x^2 + y^2 + z^2} \), \( \frac{\partial f}{\partial y} = 2ye^{x^2 + y^2 + z^2} \), and \( \frac{\partial f}{\partial z} = 2ze^{x^2 + y^2 + z^2} \).
To find the gradient vector at a specific point, plug in the coordinates. In this case, evaluating at \( P = (0, 0, 0) \) simplifies the gradient to \( abla f(0,0,0) = (0, 0, 0) \). This outcome tells us there's no change at that point in any direction, hence the directional derivative will be zero as well.
For the function \( f(x, y, z) = e^{x^2 + y^2 + z^2} \), calculating the gradient involves taking the partial derivatives with respect to each variable \(x\), \(y\), and \(z\). These partial derivatives are \( \frac{\partial f}{\partial x} = 2xe^{x^2 + y^2 + z^2} \), \( \frac{\partial f}{\partial y} = 2ye^{x^2 + y^2 + z^2} \), and \( \frac{\partial f}{\partial z} = 2ze^{x^2 + y^2 + z^2} \).
To find the gradient vector at a specific point, plug in the coordinates. In this case, evaluating at \( P = (0, 0, 0) \) simplifies the gradient to \( abla f(0,0,0) = (0, 0, 0) \). This outcome tells us there's no change at that point in any direction, hence the directional derivative will be zero as well.
Partial Derivatives
Partial derivatives allow you to focus on one variable at a time in a multi-variable function, treating all others as constants. They are the building blocks of a gradient vector.
Let's break it down further using our function: \( f(x, y, z) = e^{x^2 + y^2 + z^2} \). To compute the partial derivative \( \frac{\partial f}{\partial x} \), use the chain rule which involves taking the derivative of the exponent \( x^2 \) which is \( 2x \). The result is \( 2x e^{x^2 + y^2 + z^2} \).
Each partial derivative tells you the rate of change of the function along one axis only, while the others remain fixed. Hence, \( \frac{\partial f}{\partial y} = 2y e^{x^2 + y^2 + z^2} \) and \( \frac{\partial f}{\partial z} = 2z e^{x^2 + y^2 + z^2} \) represent changes in the \( y \) and \( z \) directions, respectively. Evaluating these at point \( (0,0,0) \) gives zero for each, showing no change at this particular point.
Let's break it down further using our function: \( f(x, y, z) = e^{x^2 + y^2 + z^2} \). To compute the partial derivative \( \frac{\partial f}{\partial x} \), use the chain rule which involves taking the derivative of the exponent \( x^2 \) which is \( 2x \). The result is \( 2x e^{x^2 + y^2 + z^2} \).
Each partial derivative tells you the rate of change of the function along one axis only, while the others remain fixed. Hence, \( \frac{\partial f}{\partial y} = 2y e^{x^2 + y^2 + z^2} \) and \( \frac{\partial f}{\partial z} = 2z e^{x^2 + y^2 + z^2} \) represent changes in the \( y \) and \( z \) directions, respectively. Evaluating these at point \( (0,0,0) \) gives zero for each, showing no change at this particular point.
Normalize Vector
Normalizing a vector is a method that makes it "unit-length," so it has a magnitude of 1. This is crucial for finding the directional derivative, as it ensures you're measuring the function's rate of change in a pure direction, unscaled by the vector's length.
Let's normalize vector \( \mathbf{a} = -\mathbf{i} + \mathbf{j} - \mathbf{k} \). The magnitude, or length, can be calculated using Pythagoras' theorem: \( \sqrt{(-1)^2 + 1^2 + (-1)^2} = \sqrt{3} \).
To get the unit vector \( \mathbf{\hat{a}} \), divide each component of \( \mathbf{a} \) by its magnitude. This results in \( \left(-\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}} \right) \). With this unit vector, you correctly capture the function's rate of change purely along that direction, crucial for the calculation of a directional derivative.
Let's normalize vector \( \mathbf{a} = -\mathbf{i} + \mathbf{j} - \mathbf{k} \). The magnitude, or length, can be calculated using Pythagoras' theorem: \( \sqrt{(-1)^2 + 1^2 + (-1)^2} = \sqrt{3} \).
To get the unit vector \( \mathbf{\hat{a}} \), divide each component of \( \mathbf{a} \) by its magnitude. This results in \( \left(-\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}} \right) \). With this unit vector, you correctly capture the function's rate of change purely along that direction, crucial for the calculation of a directional derivative.
Other exercises in this chapter
Problem 12
Compute \(\partial z / \partial r\) and \(\partial z / \partial s\). $$ z=\ln u+\ln v ; u=4^{r s}, v=4^{r / s} $$
View solution Problem 12
Find the gradient of the function at the given point. $$ f(x, y, z)=z-\sqrt{x^{2}+y^{2}} ;(3,-4,7) $$
View solution Problem 12
Find the first partial derivatives of the function. $$ z=x^{y} $$
View solution Problem 12
Evaluate the limit. $$ \lim _{(x, y) \rightarrow(0,0)} \frac{y\left(e^{x}-1\right)}{\sqrt{x^{2}+y^{2}}} $$
View solution