Problem 15
Question
Find the direction in which \(f\) increases most rapidly at the given point, and find the maximal directional derivative at that point. $$ f(x, y)=e^{x}(\cos y+\sin y) ;(0,0) $$
Step-by-Step Solution
Verified Answer
The function increases most rapidly in the direction \((1, 1)\) with maximal directional derivative \(\sqrt{2}\).
1Step 1: Find the Gradient of the Function
To find the direction in which the function increases most rapidly, we need to compute its gradient. The gradient of a function \( f(x, y) \) is given by:\[ abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \]First, find the partial derivative with respect to \( x \):\[ \frac{\partial f}{\partial x} = e^x(\cos y + \sin y) \]Now, find the partial derivative with respect to \( y \):\[ \frac{\partial f}{\partial y} = e^x(-\sin y + \cos y) \]Thus, the gradient of the function \( f(x, y) \) is:\[ abla f = \left( e^x(\cos y + \sin y), e^x(-\sin y + \cos y) \right) \].
2Step 2: Evaluate the Gradient at the Given Point
Now that we have the gradient, evaluate it at the given point \((0,0)\).Substituting \(x = 0\) and \(y = 0\) into the gradient we get:\[ abla f(0,0) = \left( e^0(\cos 0 + \sin 0), e^0(-\sin 0 + \cos 0) \right) \] Calculating the trigonometric functions gives:\[ abla f(0,0) = (1(1 + 0), 1(0 + 1)) = (1, 1) \].
3Step 3: Determine the Direction of Maximum Increase
The direction in which \( f \) increases most rapidly corresponds to the direction of the gradient vector at the point of interest. Therefore, at point \((0,0)\), the direction of maximum increase is given by the vector \((1, 1)\).
4Step 4: Calculate the Maximal Directional Derivative
The maximal directional derivative at a point is the magnitude of the gradient vector at that point. For the vector \((1, 1)\), the magnitude is:\[ \|abla f(0,0)\| = \sqrt{1^2 + 1^2} = \sqrt{2} \].
Key Concepts
Directional DerivativePartial DerivativeGradient Vector
Directional Derivative
The directional derivative helps us understand how a function changes as we move in a specific direction. It's a fundamental concept when analyzing functions over multiple variables. The directional derivative calculates the rate of change of the function as you move from a point in a specified direction.
- To find this derivative, we need two main ingredients: the gradient of the function and a unit vector indicating direction.
- The formula is the dot product between the gradient vector and the unit vector.
Partial Derivative
Partial derivatives let us examine how a function changes with respect to one variable at a time, while all other variables are held constant. This is crucial when dealing with multivariable functions.
- The notation \( \frac{\partial f}{\partial x} \) indicates a partial derivative with respect to \( x \), meaning all other variables remain unchanged.
- Each partial derivative provides a slice of the "rate of change" landscape of the function.
Gradient Vector
The gradient vector is like a compass pointing in the direction of the steepest ascent of a function. It combines the partial derivatives of a function into a vector form.
- In a function \( f(x, y) \), its gradient vector \( abla f \) would be \( \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \).
- This vector not only shows the direction to move for the greatest increase but also gives the rate at which the function increases in this direction.
Other exercises in this chapter
Problem 15
Compute \(d w / d t\) $$ \mathrm{w}=\frac{x}{y}-\frac{z}{x} ; x=\sin t, y=\cos t, z=\tan t $$
View solution Problem 15
Determine \(d f\). $$ f(x, y)=x^{2}+y^{2} $$
View solution Problem 15
Let \(f(x, y)=\sinh (y \ln x)\). Find the directional derivative of \(f\) at \((2,-1)\) in the direction away from the origin.
View solution Problem 15
Find the first partial derivatives of the function. $$ f(x, y, z)=\frac{x+y+z}{x y+y z+z x} $$
View solution