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Chapter 13: Problem 62

Find a unit vector in the direction in which \(f\) decreases most rapidly at \(P\), and find the rate of change of \(f\) at \(P\) in that direction. $$ f(x, y)=e^{x y} ; P(2,3) $$

Short Answer

Expert verified
The unit vector is \( (-\frac{3}{\sqrt{13}}, -\frac{2}{\sqrt{13}}) \) and the rate of change is \( -e^{6} \sqrt{13} \).

Step by step solution

01

Calculate the Gradient of f

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, we compute the partial derivatives:- \( \frac{\partial f}{\partial x} = y e^{xy} \), since the derivative of \( x \) with respect to \( x \) is 1.- \( \frac{\partial f}{\partial y} = x e^{xy} \), since the derivative of \( y \) with respect to \( y \) is 1.Thus, \( abla f(x, y) = ( y e^{xy}, x e^{xy} ) \).
02

Evaluate the Gradient at Point P

Substitute \( P(2, 3) \) into the gradient:\[ abla f(2, 3) = ( 3 e^{6}, 2 e^{6} ) \].
03

Determine the Direction of Maximum Decrease

The direction in which \( f \) decreases most rapidly is the opposite direction of the gradient. The unit vector in this direction is given by:\[ \mathbf{u} = -\frac{abla f(2,3)}{\|abla f(2,3)\|} \]First, calculate the magnitude of the gradient:\[ \|abla f(2, 3)\| = \sqrt{(3 e^{6})^2 + (2 e^{6})^2} = e^{6} \sqrt{9 + 4} = e^{6} \sqrt{13} \].Now, find the unit vector:\[ \mathbf{u} = \left( -\frac{3 e^{6}}{e^{6} \sqrt{13}}, -\frac{2 e^{6}}{e^{6} \sqrt{13}} \right) = \left( -\frac{3}{\sqrt{13}}, -\frac{2}{\sqrt{13}} \right) \].
04

Find Rate of Maximum Decrease

The rate of change of \( f \) in the direction of the unit vector is the negative magnitude of the gradient:\[ -\| abla f(2, 3) \| = -e^{6} \sqrt{13} \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Partial Derivatives
When we work with functions of multiple variables, like \( f(x, y) = e^{xy} \), we need to understand how the function changes with respect to each variable individually. This is where partial derivatives come in, and they help us find the rate at which the function \( f \) changes with respect to one variable while keeping the other fixed.
  • The partial derivative with respect to \( x \), denoted \( \frac{\partial f}{\partial x} \), tells us how much \( f \) changes if we only change \( x \) and keep \( y \) constant. In this exercise, we computed it as \( ye^{xy} \).
  • The partial derivative with respect to \( y \), denoted \( \frac{\partial f}{\partial y} \), indicates the rate of change as we vary \( y \) while keeping \( x \) the same. Here, it's calculated as \( xe^{xy} \).

Understanding partial derivatives is crucial because they are foundational to finding the gradient vector, which reveals the direction of most rapid change.
Gradient Vector
The gradient vector, represented as \( abla f \), encapsulates all the partial derivatives of a function. It points in the direction of steepest ascent or most rapid increase of the function. For our function \( f(x, y) = e^{xy} \), the gradient is given by \( \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) = (ye^{xy}, xe^{xy}) \).
  • Evaluating this at a point, like \( P(2,3) \), gives us a specific direction in which the function increases the fastest when you compute \( abla f(2, 3) = (3e^6, 2e^6) \).
  • The gradient is not just about direction, but also about the magnitude which indicates how steep the ascent is at that particular point.
  • The unit vector in the direction of the gradient indicates the normalized direction of maximum increase.
Understanding the gradient is key for optimization problems and is extensively used in techniques like gradient descent for finding minima.
Rate of Change
The rate of change in a specific direction gives us an idea of how fast the function's value is increasing or decreasing there. At point \( P \), the fastest decrease of the function occurs in the direction opposite the gradient.
  • To find this, we calculate the unit vector in the opposite direction of the gradient. In this case, at point \( P(2,3) \), the direction of the steepest decrease is \( \left( -\frac{3}{\sqrt{13}}, -\frac{2}{\sqrt{13}} \right) \).
  • The rate of change of \( f \) in this direction is captured by the negative of the magnitude of the gradient vector, reflecting how swiftly the function descends. Here, it's \( -e^6\sqrt{13} \).

The rate of change is a vital concept in calculus, helping to determine how a function behaves near a point, which is critical in various fields such as physics and economics.

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Find an equation for the tangent plane and parametric equations for the normal line to the surface at the point \(P\). $$ z=e^{3 y} \sin 3 x ; \quad P(\pi / 6,0,1) $$

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