Problem 10
Question
In Problems 9-12, find a unit vector in the direction in which \(f\) increases most rapidly at p. What is the rate of change in this direction? \(f(x, y)=e^{y} \sin x ; \mathbf{p}=(5 \pi / 6,0)\)
Step-by-Step Solution
Verified Answer
Unit vector: \((-\frac{\sqrt{3}}{2}, \frac{1}{2})\); Rate of change: 1.
1Step 1: Calculate the Gradient of f
The gradient of a function \(f(x, y) = e^{y} \sin x\) is given by \(abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right)\). First, calculate the partial derivatives: \(\frac{\partial f}{\partial x} = e^{y} \cos x\) and \(\frac{\partial f}{\partial y} = e^{y} \sin x\). Thus, \(abla f = (e^{y} \cos x, e^{y} \sin x)\).
2Step 2: Evaluate the Gradient at Point p
We evaluate the gradient \(abla f\) at the point \(\mathbf{p} = \left( \frac{5\pi}{6}, 0 \right)\). Substitute \(x = \frac{5\pi}{6}\) and \(y = 0\) into the gradient: \(e^{0} \cos \left( \frac{5\pi}{6} \right) = -\frac{\sqrt{3}}{2}\) and \(e^{0} \sin \left( \frac{5\pi}{6} \right) = \frac{1}{2}\). Therefore, \(abla f (\mathbf{p}) = \left( -\frac{\sqrt{3}}{2}, \frac{1}{2} \right)\).
3Step 3: Find the Unit Vector in the Gradient Direction
The unit vector in the direction of \(abla f (\mathbf{p})\) is found by dividing the gradient by its magnitude. The magnitude is \(\sqrt{\left( -\frac{\sqrt{3}}{2} \right)^2 + \left( \frac{1}{2} \right)^2} = \sqrt{1} = 1\). Therefore, the unit vector is the same as the evaluated gradient: \(\left( -\frac{\sqrt{3}}{2}, \frac{1}{2} \right)\).
4Step 4: Determine the Rate of Change
The rate of change of \(f\) in the direction of greatest increase at point \(\mathbf{p}\) is the magnitude of the gradient, \(\|abla f (\mathbf{p})\|\), which is \(1\). Therefore, the rate of change is \(1\).
Key Concepts
Partial DerivativesRate of ChangeUnit Vector
Partial Derivatives
Partial derivatives help us understand how a function changes as each variable changes, while keeping other variables fixed.
For example, in a function of two variables such as \( f(x, y) \), partial derivatives \( \frac{\partial f}{\partial x} \) and \( \frac{\partial f}{\partial y} \) indicate how \( f \) changes with \( x \) and \( y \) respectively, with the other variable held constant. These are calculated just like regular derivatives, but treat all other variables as constants.
In the given function \( f(x, y) = e^{y} \sin x \), its partial derivatives are:
For example, in a function of two variables such as \( f(x, y) \), partial derivatives \( \frac{\partial f}{\partial x} \) and \( \frac{\partial f}{\partial y} \) indicate how \( f \) changes with \( x \) and \( y \) respectively, with the other variable held constant. These are calculated just like regular derivatives, but treat all other variables as constants.
In the given function \( f(x, y) = e^{y} \sin x \), its partial derivatives are:
- \( \frac{\partial f}{\partial x} = e^{y} \cos x \)
- \( \frac{\partial f}{\partial y} = e^{y} \sin x \)
Rate of Change
The rate of change measures how a function's output changes as the input changes in a specific direction.
This is particularly useful to understand how fast something increases or decreases.
In our context, the rate of change in the direction of maximum increase is represented by the gradient vector's magnitude.
It quantifies how steeply \( f \) increases at a given point.
The direction of this change is specified by \( abla f \), and at point \( \mathbf{p} = \left( \frac{5\pi}{6}, 0 \right) \), the gradient \( abla f \) equals \(-\frac{\sqrt{3}}{2}, \frac{1}{2}\), resulting in a rate of change equal to \(1\), indicating how quickly \( f \) rises.
This is particularly useful to understand how fast something increases or decreases.
In our context, the rate of change in the direction of maximum increase is represented by the gradient vector's magnitude.
It quantifies how steeply \( f \) increases at a given point.
The direction of this change is specified by \( abla f \), and at point \( \mathbf{p} = \left( \frac{5\pi}{6}, 0 \right) \), the gradient \( abla f \) equals \(-\frac{\sqrt{3}}{2}, \frac{1}{2}\), resulting in a rate of change equal to \(1\), indicating how quickly \( f \) rises.
Unit Vector
A unit vector is a vector with a magnitude of 1.
It specifies a direction without considering length, acting as a "pure direction" indicator.
To find a unit vector in a certain direction, you divide the vector by its magnitude.
In our problem, after calculating the gradient vector, we found the magnitude to be 1.
This meant the gradient was already a unit vector: \( \left( -\frac{\sqrt{3}}{2}, \frac{1}{2} \right) \).
Using unit vectors is crucial in physics and engineering, ensuring that calculations focus on direction rather than magnitude, which simplifies complex problems where only direction matters.
It specifies a direction without considering length, acting as a "pure direction" indicator.
To find a unit vector in a certain direction, you divide the vector by its magnitude.
In our problem, after calculating the gradient vector, we found the magnitude to be 1.
This meant the gradient was already a unit vector: \( \left( -\frac{\sqrt{3}}{2}, \frac{1}{2} \right) \).
Using unit vectors is crucial in physics and engineering, ensuring that calculations focus on direction rather than magnitude, which simplifies complex problems where only direction matters.
Other exercises in this chapter
Problem 9
$$ \text { In Problems 1-10, find the gradient } \nabla f \text {. } $$ $$ f(x, y, z)=x^{2} y e^{x-z} $$
View solution Problem 10
Use the total differential dz to approximate the change in \(z\) as \((x, y)\) moves from \(P\) to \(Q\). Then use a calculator to find the corresponding exact
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In Problems 7-12, find \(\partial w / \partial t\) by using the Chain Rule. Express your final answer in terms of \(s\) and \(t\). $$ w=\ln (x+y)-\ln (x-y) ; x=
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\(\lim _{(x, y) \rightarrow(0,0)} \frac{x^{4}-y^{4}}{x^{2}+y^{2}}\)
View solution