Problem 28
Question
A function \(w=F(x, y, z),\) a vector \(\vec{v}\) and a point \(P\) are given. (a) Find \(\nabla F(x, y, z)\). (b) Find \(D_{\vec{u}} F\) at \(P,\) where \(\vec{u}\) is the unit vector in the direction of \(\vec{v}\). $$ F(x, y, z)=\frac{2}{x^{2}+y^{2}+z^{2}}, \vec{v}=\langle 1,1,-2\rangle, P=(1,1,1) $$
Step-by-Step Solution
Verified Answer
The gradient at \( P \) is \( \langle -\frac{4}{9}, -\frac{4}{9}, -\frac{4}{9} \rangle \); the directional derivative is 0.
1Step 1: Calculate Partial Derivatives
To find the gradient \( abla F(x, y, z) \), we need the partial derivatives of \( F \) with respect to \( x, y, \) and \( z \). The function is \( F(x, y, z) = \frac{2}{x^2 + y^2 + z^2} \). Start by finding the partial derivative with respect to \( x \): \[ \frac{\partial F}{\partial x} = -\frac{4x}{(x^2+y^2+z^2)^2} \] Do similarly for \( y \) and \( z \): \[ \frac{\partial F}{\partial y} = -\frac{4y}{(x^2+y^2+z^2)^2} \] \[ \frac{\partial F}{\partial z} = -\frac{4z}{(x^2+y^2+z^2)^2} \]
2Step 2: Find the Gradient Vector
Combine the partial derivatives to form the gradient vector: \[ abla F(x, y, z) = \left\langle -\frac{4x}{(x^2+y^2+z^2)^2}, -\frac{4y}{(x^2+y^2+z^2)^2}, -\frac{4z}{(x^2+y^2+z^2)^2} \right\rangle \]
3Step 3: Evaluate Gradient at Point P
Substitute \( P = (1, 1, 1) \) into the gradient vector: Calculate the denominator at \( P \): \[ (1^2 + 1^2 + 1^2)^2 = 9 \] Evaluate each component:\[ \frac{\partial F}{\partial x} \bigg|_P = -\frac{4(1)}{9} = -\frac{4}{9} \] \[ \frac{\partial F}{\partial y} \bigg|_P = -\frac{4(1)}{9} = -\frac{4}{9} \] \[ \frac{\partial F}{\partial z} \bigg|_P = -\frac{4(1)}{9} = -\frac{4}{9} \] Hence, \( abla F(1, 1, 1) = \langle -\frac{4}{9}, -\frac{4}{9}, -\frac{4}{9} \rangle \).
4Step 4: Find Unit Vector in the Direction of v
Given \( \vec{v} = \langle 1, 1, -2 \rangle \), find the magnitude: \[ |\vec{v}| = \sqrt{1^2 + 1^2 + (-2)^2} = \sqrt{6} \] The unit vector \( \vec{u} \) is: \[ \vec{u} = \left\langle \frac{1}{\sqrt{6}}, \frac{1}{\sqrt{6}}, -\frac{2}{\sqrt{6}} \right\rangle \]
5Step 5: Compute Directional Derivative
The directional derivative \( D_{\vec{u}} F \) is given by the dot product \( abla F \cdot \vec{u} \): \[ D_{\vec{u}} F = \left\langle -\frac{4}{9}, -\frac{4}{9}, -\frac{4}{9} \right\rangle \cdot \left\langle \frac{1}{\sqrt{6}}, \frac{1}{\sqrt{6}}, -\frac{2}{\sqrt{6}} \right\rangle \] Calculate the dot product: \[ D_{\vec{u}} F = \left(-\frac{4}{9} \cdot \frac{1}{\sqrt{6}}\right) + \left(-\frac{4}{9} \cdot \frac{1}{\sqrt{6}}\right) + \left(-\frac{4}{9} \cdot -\frac{2}{\sqrt{6}}\right) \] Simplify the expression:\[ D_{\vec{u}} F = -\frac{4}{9\sqrt{6}} - \frac{4}{9\sqrt{6}} + \frac{8}{9\sqrt{6}} = 0 \]
6Step 6: Conclusion
The gradient \( abla F(1, 1, 1) = \langle -\frac{4}{9}, -\frac{4}{9}, -\frac{4}{9} \rangle \) and the directional derivative \( D_{\vec{u}} F \) at \( P \) is 0.
Key Concepts
GradientPartial DerivativesUnit Vector
Gradient
The gradient is one of the most critical concepts when dealing with directional derivatives. It provides a vector representation of a function's steepest ascent rate at any point in its domain. For a multivariable function like \( F(x, y, z) \), the gradient \( abla F \) is a vector that consists of its partial derivatives: \( abla F = \left\langle \frac{\partial F}{\partial x}, \frac{\partial F}{\partial y}, \frac{\partial F}{\partial z} \right\rangle \).
This vector points in the direction where the function increases most rapidly and its magnitude gives the rate of increase.
To find the gradient vector, simply calculate each partial derivative separately, then combine them into a vector. For the function in our exercise, \( F(x, y, z) = \frac{2}{x^2 + y^2 + z^2} \), we computed:
This vector points in the direction where the function increases most rapidly and its magnitude gives the rate of increase.
To find the gradient vector, simply calculate each partial derivative separately, then combine them into a vector. For the function in our exercise, \( F(x, y, z) = \frac{2}{x^2 + y^2 + z^2} \), we computed:
- \( \frac{\partial F}{\partial x} = -\frac{4x}{(x^2+y^2+z^2)^2} \)
- \( \frac{\partial F}{\partial y} = -\frac{4y}{(x^2+y^2+z^2)^2} \)
- \( \frac{\partial F}{\partial z} = -\frac{4z}{(x^2+y^2+z^2)^2} \)
Partial Derivatives
Partial derivatives are a crucial aspect of multivariable calculus, as they allow us to examine how a function changes with respect to one variable while holding the others constant. This is like slicing the function's domain along one axis to see its behavior in just that direction.
For a function \( F(x, y, z) \), the partial derivative with respect to \( x \) is denoted as \( \frac{\partial F}{\partial x} \).
It shows the rate of change of the function as \( x \) changes, with \( y \) and \( z \) fixed. Similarly, you can find partial derivatives with respect to \( y \) and \( z \), which provide the function's surface slope in those directions.
In our exercise, for each variable, the partial derivative was computed to form the gradient.
This fundamental understanding of partial derivatives allows us to compile changes from multiple variables and apply them to practical problems, like finding the gradient or computing a directional derivative.
For a function \( F(x, y, z) \), the partial derivative with respect to \( x \) is denoted as \( \frac{\partial F}{\partial x} \).
It shows the rate of change of the function as \( x \) changes, with \( y \) and \( z \) fixed. Similarly, you can find partial derivatives with respect to \( y \) and \( z \), which provide the function's surface slope in those directions.
In our exercise, for each variable, the partial derivative was computed to form the gradient.
This fundamental understanding of partial derivatives allows us to compile changes from multiple variables and apply them to practical problems, like finding the gradient or computing a directional derivative.
Unit Vector
A unit vector is essential when exploring directional derivatives because it standardizes direction without affecting magnitude. It's a vector with a length of one, ensuring that only the true direction contributes to calculations. To calculate a unit vector \( \vec{u} \) from a given vector \( \vec{v} = \langle a, b, c \rangle \), divide each component by the vector's magnitude.
The magnitude \( |\vec{v}| \) is calculated as \( \sqrt{a^2 + b^2 + c^2} \).
Then, the unit vector becomes \( \vec{u} = \left\langle \frac{a}{|\vec{v}|}, \frac{b}{|\vec{v}|}, \frac{c}{|\vec{v}|} \right\rangle \).
In our case, \( \vec{v} = \langle 1, 1, -2 \rangle \) has a magnitude of \( \sqrt{6} \), resulting in the unit vector:
\( \vec{u} = \left\langle \frac{1}{\sqrt{6}}, \frac{1}{\sqrt{6}}, -\frac{2}{\sqrt{6}} \right\rangle \).
This ensures that when calculating the directional derivative, the vector influences the function solely through direction, giving us a true measure of change in that path.
The magnitude \( |\vec{v}| \) is calculated as \( \sqrt{a^2 + b^2 + c^2} \).
Then, the unit vector becomes \( \vec{u} = \left\langle \frac{a}{|\vec{v}|}, \frac{b}{|\vec{v}|}, \frac{c}{|\vec{v}|} \right\rangle \).
In our case, \( \vec{v} = \langle 1, 1, -2 \rangle \) has a magnitude of \( \sqrt{6} \), resulting in the unit vector:
\( \vec{u} = \left\langle \frac{1}{\sqrt{6}}, \frac{1}{\sqrt{6}}, -\frac{2}{\sqrt{6}} \right\rangle \).
This ensures that when calculating the directional derivative, the vector influences the function solely through direction, giving us a true measure of change in that path.
Other exercises in this chapter
Problem 27
Form a function \(z=f(x, y)\) such that \(f_{x}\) and \(f_{y}\) match those given. $$ f_{x}=\sin y+1, \quad f_{y}=x \cos y $$
View solution Problem 27
In Exercises \(27-30,\) describe the level surfaces of the given functions of three variables. $$ f(x, y, z)=x^{2}+y^{2}+z^{2} $$
View solution Problem 28
Form a function \(z=f(x, y)\) such that \(f_{x}\) and \(f_{y}\) match those given. $$ f_{x}=x+y, \quad f_{y}=x+y $$
View solution Problem 28
Describe the level surfaces of the given functions of three variables. $$ f(x, y, z)=z-x^{2}+y^{2} $$
View solution