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Magazine Chapter 9: Problem 15
Find the directional derivative of the function at the given point in the direction of the vector \(v\) . \(V(u, t)=e^{-u t}, \quad(0,3), \quad \mathbf{v}=[2,-1]\)
Short Answer
Expert verified
The directional derivative is \(-\frac{6}{\sqrt{5}}\).
Step by step solution
01
Understand the Problem
We need to find the directional derivative of the function \( V(u, t) = e^{-u t} \) at the point \((0, 3)\) in the direction of the vector \( \mathbf{v} = [2, -1] \). This involves calculating the gradient of the function and then projecting it in the direction of \( \mathbf{v} \).
02
Calculate the Gradient of the Function
The gradient \( abla V \) of the function \( V(u, t) = e^{-u t} \) is the vector of its partial derivatives. Compute the partial derivative with respect to \( u \), \( \frac{\partial V}{\partial u} = -te^{-u t} \), and with respect to \( t \), \( \frac{\partial V}{\partial t} = -ue^{-u t} \). Thus, the gradient is \( abla V = [-te^{-u t}, -ue^{-u t}] \).
03
Evaluate the Gradient at the Point
Substitute \( u = 0 \) and \( t = 3 \) into the gradient. This gives \( abla V(0, 3) = [-3e^{0}, -0e^{0}] = [-3, 0] \).
04
Normalize the Directional Vector
To find the unit vector in the direction of \( \mathbf{v} = [2, -1] \), compute its magnitude: \( \| \mathbf{v} \| = \sqrt{2^2 + (-1)^2} = \sqrt{5} \). The unit vector \( \mathbf{u}_v \) is then \( \frac{1}{\sqrt{5}} [2, -1] = [\frac{2}{\sqrt{5}}, \frac{-1}{\sqrt{5}}] \).
05
Compute the Directional Derivative
The directional derivative \( D_{\mathbf{u}_v} V \) is given by \( abla V \cdot \mathbf{u}_v \). Thus, it is \( [-3, 0] \cdot [\frac{2}{\sqrt{5}}, \frac{-1}{\sqrt{5}}] = -3 \times \frac{2}{\sqrt{5}} + 0 \times \frac{-1}{\sqrt{5}} = -\frac{6}{\sqrt{5}} \).
06
Finalize the Directional Derivative
Conclude that the directional derivative of \( V(u, t) = e^{-u t} \) at the point \((0, 3)\) in the direction of \( \mathbf{v} = [2, -1] \) is \( -\frac{6}{\sqrt{5}} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Gradient of a function
The gradient of a function is a fundamental concept in calculus, particularly when dealing with functions of several variables. It is represented as a vector that points in the direction of the greatest rate of increase of the function. For a given function, the gradient acts as a collection of all its partial derivatives.Here are some key points about gradients:
- The gradient is denoted by \( abla \) (nabla) and is expressed as \( abla V \) for a function \( V \).
- It consists of partial derivatives, each representing the rate of change along one of the function's input variables.
- For a function \( V(u, t) \), the gradient would be \( abla V = \left( \frac{\partial V}{\partial u}, \frac{\partial V}{\partial t} \right) \).
Partial derivatives
Partial derivatives are the building blocks for gradients and crucial for understanding how functions behave when variables are changed independently of each other.Essential points about partial derivatives:
- Partial derivatives \( \frac{\partial}{\partial x} \) of a function are concerned with how the function changes when one variable changes, keeping others constant.
- They provide insight into the function's sensitivity to each variable individually.
- In our example, the partial derivatives of \( V(u, t) = e^{-ut} \) are \( \frac{\partial V}{\partial u} = -te^{-ut} \) and \( \frac{\partial V}{\partial t} = -ue^{-ut} \).
Unit vector
Unit vectors are important in directional derivatives because they provide a standardized direction for measuring change.Here's what you need to know:
- A unit vector has a magnitude of 1, meaning it only indicates direction without affecting how large or small the measurement of direction is.
- To convert any vector to a unit vector, divide each of its components by the vector's magnitude. The magnitude of a vector \( \mathbf{v} = [a, b] \) is \( \| \mathbf{v} \| = \sqrt{a^2 + b^2} \).
- For our vector \( \mathbf{v} = [2, -1] \), the magnitude is \( \sqrt{5} \), and the unit vector becomes \( \left[ \frac{2}{\sqrt{5}}, \frac{-1}{\sqrt{5}} \right] \).
