Chapter 9: Problem 12
Find the directional derivative of the given function at the given point in the indicated direction. $$ f(x, y)=4 x+x y^{2}-5 y ;(3,-1), \quad \theta=\pi / 4 $$
Short Answer
Expert verified
The directional derivative is \(-3\sqrt{2}\).
Step by step solution
01
Compute the gradient of the function
The gradient \( abla f(x, y) \) is a vector of the partial derivatives \( \left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}\right) \). Calculate these partial derivatives:- \( \frac{\partial f}{\partial x} = \frac{\partial}{\partial x} (4x + xy^2 - 5y) = 4 + y^2 \).- \( \frac{\partial f}{\partial y} = \frac{\partial}{\partial y} (4x + xy^2 - 5y) = 2xy - 5 \).Thus, \( abla f(x, y) = \left(4 + y^2, 2xy - 5 \right) \).
02
Evaluate the gradient at the given point
Substitute the point \((3, -1)\) into the gradient:- \( \frac{\partial f}{\partial x} \bigg|_{(3, -1)} = 4 + (-1)^2 = 5 \).- \( \frac{\partial f}{\partial y} \bigg|_{(3, -1)} = 2(3)(-1) - 5 = -11 \).Hence, the gradient at \((3, -1)\) is \( abla f(3, -1) = (5, -11) \).
03
Compute the unit vector in the given direction
The direction is given by an angle \( \theta = \frac{\pi}{4} \). The corresponding unit vector is \( \mathbf{u} = (\cos \theta, \sin \theta) \).Compute the components:- \( \cos \frac{\pi}{4} = \frac{\sqrt{2}}{2} \).- \( \sin \frac{\pi}{4} = \frac{\sqrt{2}}{2} \).Thus, the unit vector is \( \mathbf{u} = \left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right) \).
04
Calculate the directional derivative
The directional derivative is the dot product of the gradient and the unit vector:\[ D_{\mathbf{u}} f = abla f(3, -1) \cdot \mathbf{u} = (5, -11) \cdot \left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right) \].Compute this dot product:- \( 5 \cdot \frac{\sqrt{2}}{2} + (-11) \cdot \frac{\sqrt{2}}{2} = \frac{5\sqrt{2}}{2} - \frac{11\sqrt{2}}{2} \).- Simplify to get \( \frac{-6\sqrt{2}}{2} = -3\sqrt{2} \).
05
Interpret and conclude
The directional derivative \( -3\sqrt{2} \) indicates the rate of change of the function \( f(x,y) \) at the point \((3, -1)\) in the direction of \( \theta = \frac{\pi}{4} \). A negative value implies the function decreases in this direction.
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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 one of the most crucial concepts when dealing with multivariable functions. It plays a similar role as the derivative for single-variable functions. The gradient of a function, often denoted as \( abla f(x, y) \), is a vector that consists of all the partial derivatives of the function. It points in the direction of the greatest rate of increase of the function.
For a function \( f(x, y) \), the gradient is given by:
For a function \( f(x, y) \), the gradient is given by:
- \( abla f(x, y) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \)
Partial Derivatives
Partial derivatives are like the regular derivatives you learned in basic calculus but are applied to functions with more than one variable. They help determine how a function changes as one particular variable changes, holding all other variables constant.
When you compute the partial derivative of a function with respect to \( x \), you differentiate as if \( x \) is the only variable, treating all other variables as constants:
When you compute the partial derivative of a function with respect to \( x \), you differentiate as if \( x \) is the only variable, treating all other variables as constants:
- \( \frac{\partial f}{\partial x} \)
- \( \frac{\partial f}{\partial y} \)
Unit Vector
A unit vector is a vector with a magnitude of 1. It's used to indicate direction without altering length. Unit vectors are particularly useful in directional derivatives because they provide a standardized way to express a direction, ensuring the derivative measures the rate of change purely in terms of direction and not magnitude.
To find a unit vector in a given direction described by an angle \( \theta \):
To find a unit vector in a given direction described by an angle \( \theta \):
- \( \mathbf{u} = (\cos \theta, \sin \theta) \)
Dot Product
The dot product is a fundamental operation when working with vectors. It's a method of multiplying two vectors that results in a scalar. In the context of directional derivatives, the dot product measures how much of one vector 'goes in the direction' of another.
When calculating a directional derivative, you compute the dot product between the gradient and the unit vector in the desired direction:
When calculating a directional derivative, you compute the dot product between the gradient and the unit vector in the desired direction:
- \( D_{\mathbf{u}}f = abla f \cdot \mathbf{u} \)
- \( 5 \cdot \frac{\sqrt{2}}{2} + (-11) \cdot \frac{\sqrt{2}}{2} \)