Problem 273
Question
For the following exercises, find the directional derivative of the function at point \(P\) in the direction of \(\mathbf{v}\). $$f(x, y)=y^{2}+x z, P(1,2,2), \quad \mathbf{v}=\langle 2,-1,2\rangle$$
Step-by-Step Solution
Verified Answer
Directional derivative is \(2\).
1Step 1: Apply differentiation rules
Use the appropriate differentiation rules (product, quotient, chain) to find the derivative.
2Step 2: Simplify the result
The derivative is Directional derivative is \(2\)..
Key Concepts
GradientPartial DerivativesMultivariable Calculus
Gradient
The gradient is a vector that points in the direction of the greatest rate of increase of a function. It is crucial in multivariable calculus, specifically when dealing with directional derivatives. The gradient of a function, denoted as \( abla f \), combines the partial derivatives of each variable into a single vector. This vector encapsulates how a function changes as each variable simultaneously changes.
For instance, if our function is \( f(x, y, z) = y^2 + xz \), the gradient \( abla f \) is calculated from the partial derivatives \( \frac{\partial f}{\partial x} = z \), \( \frac{\partial f}{\partial y} = 2y \), and \( \frac{\partial f}{\partial z} = x \). Thus, the gradient becomes a vector \((z, 2y, x)\).
For instance, if our function is \( f(x, y, z) = y^2 + xz \), the gradient \( abla f \) is calculated from the partial derivatives \( \frac{\partial f}{\partial x} = z \), \( \frac{\partial f}{\partial y} = 2y \), and \( \frac{\partial f}{\partial z} = x \). Thus, the gradient becomes a vector \((z, 2y, x)\).
- The direction of the gradient vector shows the quickest direction of increase of the function.
- The magnitude (or length) of the gradient vector indicates the rate of increase in that direction.
- To find the gradient at a point, substitute the point's coordinates into the gradient vector.
Partial Derivatives
Partial derivatives are the cornerstone of understanding how multivariable functions change when each variable is varied individually. When dealing with functions of more than one variable, partial derivatives allow us to focus on one variable at a time, holding the other variables constant.
For the function \( f(x, y, z) = y^2 + xz \), the partial derivatives are computed as follows:
For the function \( f(x, y, z) = y^2 + xz \), the partial derivatives are computed as follows:
- \( \frac{\partial f}{\partial x} = z \) - This considers \( x \) as the variable while treating \( y \) and \( z \) as constants.
- \( \frac{\partial f}{\partial y} = 2y \) - Here, \( y \) is the variable, with \( x \) and \( z \) held constant.
- \( \frac{\partial f}{\partial z} = x \) - In this case, \( z \) is varied, keeping \( x \) and \( y \) constant.
Multivariable Calculus
Multivariable calculus extends the concepts of calculus to functions of several variables. It plays a vital role in fields such as physics, engineering, and economics, where systems often depend on multiple changing factors.
Some vital concepts in multivariable calculus include:
Some vital concepts in multivariable calculus include:
- Gradients: These are crucial for optimization and understanding how functions change in multidimensional spaces.
- Directional Derivatives: These help in finding the rate of change of a function in any given direction, not just along an axis.
- Partial Derivatives: These allow for the breakdown and analysis of multi-variable functions step-by-step by focusing on one variable at a time.
Other exercises in this chapter
Problem 271
For the following exercises, find the directional derivative of the function at point \(P\) in the direction of \(\mathbf{v}\). $$f(x, y)=\ln \left(x^{2}+y^{2}\
View solution Problem 272
For the following exercises, find the directional derivative of the function at point \(P\) in the direction of \(\mathbf{v}\). $$f(x, y)=x^{2} y, P(-5,5), \qua
View solution Problem 274
For the following exercises, find the directional derivative of the function in the direction of the unit vector \(\mathbf{u}=\cos \theta \mathbf{i}+\sin \theta
View solution Problem 275
For the following exercises, find the directional derivative of the function in the direction of the unit vector \(\mathbf{u}=\cos \theta \mathbf{i}+\sin \theta
View solution