Problem 295
Question
For the following exercises, find the derivative of the function. \(f(x, y)=e^{x y}\) at point \((6,7)\) in the direction the function increases most rapidly
Step-by-Step Solution
Verified Answer
The derivative in the direction of steepest ascent at point (6, 7) is \( e^{42} \sqrt{85} \).
1Step 1: Find the Gradient of the Function
The gradient of a function, denoted by \( abla f \), represents the direction of the steepest ascent of the function at any given point. For the function \( f(x, y) = e^{xy} \), compute the partial derivatives with respect to \( x \) and \( y \). \( \frac{\partial f}{\partial x} = y e^{xy} \) and \( \frac{\partial f}{\partial y} = x e^{xy} \). Therefore, the gradient is \( abla f = (y e^{xy}, x e^{xy}) \).
2Step 2: Evaluate the Gradient at the Given Point
Substitute the point \((6, 7)\) into the gradient. \( \frac{\partial f}{\partial x} = 7 e^{42} \) and \( \frac{\partial f}{\partial y} = 6 e^{42} \). Thus, the gradient at the point \((6, 7)\) is \( abla f = (7 e^{42}, 6 e^{42}) \).
3Step 3: Determine the Direction of Maximum Increase
The direction of maximum increase of the function is given by the vector pointing in the direction of the gradient, \( abla f \). Thus, the direction of steepest ascent at point \((6, 7)\) is along the vector \((7 e^{42}, 6 e^{42})\).
4Step 4: Find the Magnitude of the Gradient Vector
The magnitude of the gradient \( abla f \) is found using the formula for Euclidean norm: \( ||abla f|| = \sqrt{(7 e^{42})^2 + (6 e^{42})^2} = e^{42} \sqrt{49 + 36} = e^{42} \sqrt{85} \).
5Step 5: Write the Directional Derivative
The directional derivative along the gradient at point \((6, 7)\) in the direction of steepest ascent is equal to the magnitude of \( abla f \), which is \( e^{42} \sqrt{85} \). This value represents the rate of change of the function at the point in the direction where it increases most rapidly.
Key Concepts
Partial DerivativesDirectional DerivativeSteepest Ascent
Partial Derivatives
Partial derivatives are a fundamental concept in calculus, used to understand how a function changes as each variable is varied independently. Unlike ordinary derivatives, which measure change along a single dimension, partial derivatives look at change along multiple dimensions. For a function like \( f(x, y) = e^{xy} \), we want to see how the function behaves when only \( x \) or \( y \) changes while the other stays constant.
- The partial derivative with respect to \( x \), \( \frac{\partial f}{\partial x} = y e^{xy} \), tells us how much \( f \) will change per unit change in \( x \), holding \( y \) constant.
- Similarly, \( \frac{\partial f}{\partial y} = x e^{xy} \) measures the change in \( f \) when \( y \) changes, with \( x \) held constant.
Directional Derivative
The directional derivative is a crucial idea when dealing with functions of several variables. It extends the concept of the derivative to directions other than parallel to the coordinate axes. In simpler terms, it tells us the rate of change of a function in any specified direction.
To find the directional derivative at a point, we rely on the gradient. The formula is:
To find the directional derivative at a point, we rely on the gradient. The formula is:
- The directional derivative of a function \( f \) at a point \( (x_0, y_0) \) in the direction of a unit vector \( \mathbf{u} = (u_1, u_2) \) is given by: \( D_\mathbf{u}f(x_0, y_0) = abla f(x_0, y_0) \cdot \mathbf{u} \).
- This dot product results in a scalar value, representing how much \( f \) changes in that specific direction.
Steepest Ascent
Steepest ascent is the direction where a function increases the quickest. When we explore a function's behavior, we often look for paths of steepest ascent to describe how a function optimizes or climbs. This is inherently connected to the gradient of the function, \( abla f \).
- The vector of the gradient \( abla f = (y e^{xy}, x e^{xy}) \) points directly in the direction of the steepest ascent.
- At any point, its magnitude, given as \( e^{42} \sqrt{85} \) in the solution, tells us the rate of change in that direction.
Other exercises in this chapter
Problem 292
For the following exercises, find the gradient vector at the indicated point. $$ f(x, y, z)=x y-\ln (z), P(2,-2,2) $$
View solution Problem 293
For the following exercises, find the gradient vector at the indicated point. $$ f(x, y, z)=x \sqrt{y^{2}+z^{2}}, P(-2,-1,-1) $$
View solution Problem 296
For the following exercises, find the derivative of the function. \(f(x, y)=\arctan \left(\frac{y}{x}\right)\) at point \((-9,9) \quad\) in the direction the fu
View solution Problem 297
For the following exercises, find the derivative of the function. \(f(x, y, z)=\ln (x y+y z+z x) \) at point \((-9,-18,-27)\) in the direction the function incr
View solution