Explanations 
Textbooks 
Flashcards 
91影视 AI 
Notes 
Study Plans 
Study Sets 
Find a degree 
Magazine Chapter 10: Problem 12
Find all directions in which the directional derivative of \(f(x, y)=y e^{-x y}\) is 1 at the point (0,2) .
Short Answer
Expert verified
The two directions in which the directional derivative of the function \(f(x, y) = ye^{-xy}\) is 1 at the point (0, 2) are given by the vectors:
\(\mathbf{u}_1 = \left\langle \frac{1 + \sqrt{6}}{4}, \frac{-1 + \sqrt{6}}{5} \right\rangle\)
and
\(\mathbf{u}_2 = \left\langle \frac{1 - \sqrt{6}}{4}, \frac{-1 - \sqrt{6}}{5} \right\rangle\)
Step by step solution
01
Calculate the gradient vector of the function
To find the gradient of the given function, we need to compute the partial derivatives with respect to x and y. The function is \(f(x, y) = ye^{-xy}\).
Now, let's find the partial derivatives:
The partial derivative of f with respect to x is:
\(\frac{\partial f}{\partial x} = -y^2 e^{-xy}\)
The partial derivative of f with respect to y is:
\(\frac{\partial f}{\partial y} = e^{-xy}(1-xy)\)
Thus, the gradient vector is:
\(\nabla f(x, y) = \left\langle -y^2 e^{-xy}, e^{-xy}(1-xy) \right\rangle\)
02
Find the directional derivative formula
The directional derivative of a function along a unit vector \(\mathbf{u} = \langle a, b \rangle\) is given by the formula:
\(D_{\mathbf{u}}f(x, y) = \nabla f(x, y) \cdot \mathbf{u}\)
Using the gradient vector we found earlier, and evaluating it at the point (0, 2), we have:
\(\nabla f(0, 2) = \left\langle -4, 3 \right\rangle\)
Now, let's compute the inner product of the gradient vector and the unit vector \(\mathbf{u}\):
\(\nabla f(0, 2) \cdot \mathbf{u} = -4a + 3b\)
03
Set the directional derivative equal to 1 and find the direction vectors
We want the directional derivative to be 1, so we now have the following equation:
\(-4a + 3b = 1\)
Additionally, since \(\mathbf{u}\) is a unit vector, it must satisfy the equation:
\(a^2 + b^2 = 1\)
We now have a system of two equations and two unknowns:
\[
\begin{cases}
-4a + 3b = 1 \\
a^2 + b^2 = 1
\end{cases}
\]
To solve for a, we can rewrite the first equation as:
\(a = \frac{1 + 3b}{4}\)
Substituting this into the second equation and solving for b:
\[\left(\frac{1 + 3b}{4}\right)^2 + b^2 = 1\]
Expanding and simplifying, we get:
\(\frac{1}{16} + \frac{3}{8}b + \frac{9}{16}b^2 + b^2 = 1\)
Multiplying by 16 and combining like terms, we have:
\(15b^2 + 6b - 15 = 0\)
Dividing by 3, we have:
\(5b^2 + 2b - 5 = 0\)
Using the quadratic formula, we find two possible values for b:
\(b_1 = \frac{-1 + \sqrt{6}}{5}\) and \(b_2 = \frac{-1 - \sqrt{6}}{5}\)
Plugging these values back into the equation \(a = \frac{1 + 3b}{4}\), we find the corresponding values for a:
\(a_1 = \frac{1 + \sqrt{6}}{4}\) and \(a_2 = \frac{1 - \sqrt{6}}{4}\)
Therefore, the two directions in which the directional derivative of the function is 1 at the point (0, 2) are given by the vectors:
\(\mathbf{u}_1 = \left\langle \frac{1 + \sqrt{6}}{4}, \frac{-1 + \sqrt{6}}{5} \right\rangle\)
and
\(\mathbf{u}_2 = \left\langle \frac{1 - \sqrt{6}}{4}, \frac{-1 - \sqrt{6}}{5} \right\rangle\)
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91影视!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Gradient Vector
Understanding the gradient vector is crucial when working with multivariable functions like the one given in our problem, which is represented as \(f(x, y) = ye^{-xy}\). The gradient vector is denoted as \(abla f\) (or simply 'grad f'), and it holds all the partial derivatives of the function with respect to each variable. For a two-variable function, such as the one we are considering, the gradient vector has two components, each representing how the function changes in the direction of each variable.
The calculation of the gradient involves finding the partial derivatives of \(f\) with respect to \(x\) and \(y\), which are the rates at which \(f\) changes in the directions of \(x\) and \(y\) respectively. After computing the partial derivatives, we plug them into a vector, getting the gradient vector \(abla f(x, y)\). In our exercise, we found that \(abla f(x, y) = \left\langle -y^2 e^{-xy}, e^{-xy}(1-xy) \right\rangle\), which points in the direction of the greatest rate of increase of the function and its magnitude gives the rate of that increase.
The calculation of the gradient involves finding the partial derivatives of \(f\) with respect to \(x\) and \(y\), which are the rates at which \(f\) changes in the directions of \(x\) and \(y\) respectively. After computing the partial derivatives, we plug them into a vector, getting the gradient vector \(abla f(x, y)\). In our exercise, we found that \(abla f(x, y) = \left\langle -y^2 e^{-xy}, e^{-xy}(1-xy) \right\rangle\), which points in the direction of the greatest rate of increase of the function and its magnitude gives the rate of that increase.
Partial Derivatives
Partial derivatives of a function like \( f(x, y) \) are derivatives that consider the change in the function with respect to one variable while keeping the others constant. This is akin to looking at a multi-dimensional landscape and focusing on the slope in one direction鈥攕ay, northward or eastward鈥攊gnoring the others.
To calculate a partial derivative, we apply the differentiation rules as if it's a single-variable function, 鈥榝reezing鈥� the other variables. For \(f(x, y) = ye^{-xy}\), we calculated two partial derivatives. The partial derivative with respect to \(x\) is \(\frac{\partial f}{\partial x} = -y^2 e^{-xy}\), which shows how \(f\) changes as \(x\) varies and y is kept constant. Similarly, the partial derivative with respect to \(y\) is \(\frac{\partial f}{\partial y} = e^{-xy}(1-xy)\), representing the change in \(f\) in the \(y\)-direction. These components are crucial to constructing the gradient vector that's used in finding directional derivatives.
To calculate a partial derivative, we apply the differentiation rules as if it's a single-variable function, 鈥榝reezing鈥� the other variables. For \(f(x, y) = ye^{-xy}\), we calculated two partial derivatives. The partial derivative with respect to \(x\) is \(\frac{\partial f}{\partial x} = -y^2 e^{-xy}\), which shows how \(f\) changes as \(x\) varies and y is kept constant. Similarly, the partial derivative with respect to \(y\) is \(\frac{\partial f}{\partial y} = e^{-xy}(1-xy)\), representing the change in \(f\) in the \(y\)-direction. These components are crucial to constructing the gradient vector that's used in finding directional derivatives.
Unit Vector
A unit vector, in the context of our problem, serves as the direction in which we want to measure the rate of change of the function. A vector is said to be a unit vector if it has a length (or magnitude) of 1. This property is essential because it allows us to measure the directional derivative without scaling, giving us the truest sense of the function's slope in that direction.
The unit vector can be expressed in the form \( \mathbf{u} = \langle a, b \rangle \) where the values of \(a\) and \(b\) must satisfy the equation \(a^2 + b^2 = 1\). This relationship ensures that the vector's magnitude is 1, hence qualifying as a unit vector. In our exercise, we sought a unit vector such that the directional derivative of our function along that vector would be exactly 1 at the point (0,2). Using these conditions, we establish two equations to solve for the components of possible direction vectors that constitute our unit vector.
The unit vector can be expressed in the form \( \mathbf{u} = \langle a, b \rangle \) where the values of \(a\) and \(b\) must satisfy the equation \(a^2 + b^2 = 1\). This relationship ensures that the vector's magnitude is 1, hence qualifying as a unit vector. In our exercise, we sought a unit vector such that the directional derivative of our function along that vector would be exactly 1 at the point (0,2). Using these conditions, we establish two equations to solve for the components of possible direction vectors that constitute our unit vector.
Quadratic Formula
The quadratic formula is an algebraic solution to find the roots of a quadratic equation, which is an equation of the form \( ax^2 + bx + c = 0 \). The formula given by \( x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}} \) is used to calculate the values of \(x\) that satisfy the equation. It's notably useful when factorization is not straightforward or applicable.
Our problem eventually led us to a quadratic equation after substituting and rearranging terms from the gradient and unit vector restrictions. By applying the quadratic formula to the equation \(5b^2 + 2b - 5 = 0\), we could solve for the values of \(b\), and subsequently find the corresponding values of \(a\) for our unit vector. This demonstrates how the formula plays a significant role in the solution when faced with quadratic relationships in calculus problems.
Our problem eventually led us to a quadratic equation after substituting and rearranging terms from the gradient and unit vector restrictions. By applying the quadratic formula to the equation \(5b^2 + 2b - 5 = 0\), we could solve for the values of \(b\), and subsequently find the corresponding values of \(a\) for our unit vector. This demonstrates how the formula plays a significant role in the solution when faced with quadratic relationships in calculus problems.
