Problem 21
Question
In what direction does \(f(x, y)=\sqrt{x^{2}-y^{2}}\) increase most rapidly at \((5,3)\) ?
Step-by-Step Solution
Verified Answer
The function increases most rapidly in the direction of the vector \( \left( \frac{5}{4}, \frac{-3}{4} \right) \).
1Step 1: Calculate the Gradient Vector
To find the direction in which the function increases most rapidly, we need to calculate the gradient of the function, \( abla f(x, y) \). The gradient is calculated by finding the partial derivatives with respect to each variable.\\[ \frac{\partial}{\partial x} \left( \sqrt{x^2 - y^2} \right) = \frac{x}{\sqrt{x^2 - y^2}} \] \\[ \frac{\partial}{\partial y} \left( \sqrt{x^2 - y^2} \right) = \frac{-y}{\sqrt{x^2 - y^2}} \] \Thus, \( abla f(x, y) = \left( \frac{x}{\sqrt{x^2 - y^2}}, \frac{-y}{\sqrt{x^2 - y^2}} \right) \).
2Step 2: Evaluate the Gradient at the Given Point
Now we evaluate the gradient at the point \((5,3)\). \\[ abla f(5, 3) = \left( \frac{5}{\sqrt{5^2 - 3^2}}, \frac{-3}{\sqrt{5^2 - 3^2}} \right) = \left( \frac{5}{4}, \frac{-3}{4} \right) \].
3Step 3: Determine the Direction of Steepest Ascent
The gradient vector \( abla f(5, 3) = \left( \frac{5}{4}, \frac{-3}{4} \right) \) points in the direction of the steepest ascent of the function. To describe the direction as a vector, it can be written as \( \left( \frac{5}{4}, \frac{-3}{4} \right) \).
Key Concepts
Partial DerivativeSteepest AscentDirectional Derivative
Partial Derivative
The concept of a partial derivative is pivotal in understanding how functions behave in multiple dimensions. Imagine a function with two variables, like a landscape where the height is dictated by variables like x and y. You would want to know the rate of change of the height with respect to one of these variables. That's where partial derivatives come in. They allow us to see how the function changes as we vary one variable, while keeping the other constant.
To calculate a partial derivative with respect to x (i.e., \( \frac{\partial}{\partial x} \)), we treat y as a constant and differentiate the function accordingly. Similarly, for the partial derivative with respect to y, we hold x constant (i.e., \( \frac{\partial}{\partial y} \)).
To calculate a partial derivative with respect to x (i.e., \( \frac{\partial}{\partial x} \)), we treat y as a constant and differentiate the function accordingly. Similarly, for the partial derivative with respect to y, we hold x constant (i.e., \( \frac{\partial}{\partial y} \)).
- Partial derivatives help create the gradient vector.
- The gradient vector points towards the direction of the steepest ascent in the function's landscape.
- They also help in constructing tangent planes to the surface of the function.
Steepest Ascent
Steepest ascent refers to the direction in which a function increases at its fastest rate. Imagine you're standing on a hill, and you want to climb up as quickly as possible. The steepest ascent is the path that will take you to the top most efficiently.
This concept is crucial in optimization problems where we want to maximize a function. The gradient vector directly points in this direction. When we say the gradient \( abla f(x, y) \) dictates the direction of steepest ascent, we mean that this vector shows the path of the greatest increase.
This concept is crucial in optimization problems where we want to maximize a function. The gradient vector directly points in this direction. When we say the gradient \( abla f(x, y) \) dictates the direction of steepest ascent, we mean that this vector shows the path of the greatest increase.
- The length and direction of the gradient tell us the slope and direction to move.
- It's like a compass that navigates us through the function's increases.
- In practical applications, like machine learning, it helps in tuning parameters to optimize models.
Directional Derivative
The directional derivative gives us the rate at which a function changes as we move in any specified direction. Think of it as checking what happens when you take a step in a particular direction on our hill, rather than just going straight up or straight down.
The beauty of the directional derivative lies in its flexibility. By choosing different directions, you can explore how the function behaves from all angles. Mathematically, we combine the gradient vector with a unit vector in our chosen direction to find the directional derivative:
\[D_u f(x, y) = abla f(x, y) \cdot u\]
The beauty of the directional derivative lies in its flexibility. By choosing different directions, you can explore how the function behaves from all angles. Mathematically, we combine the gradient vector with a unit vector in our chosen direction to find the directional derivative:
\[D_u f(x, y) = abla f(x, y) \cdot u\]
- \( u \) is the unit vector in the direction of interest.
- The dot product tells us how aligned our direction is with the steepest ascent.
- A high directional derivative indicates alignment with steepest ascent.
- While a low (or negative) one shows moving in the other direction.
Other exercises in this chapter
Problem 21
Find the linearization of \(f(x, y)\) at the indicated point \(\left(x_{0}, y_{0}\right) .\) \(f(x, y)=\tan (x+y) ;(0,0)\)
View solution Problem 21
Find the global maxima and minima of $$f(x, y)=x^{2}+y^{2}+4 x-1$$ on the disk $$ D=\left\\{(x, y): x^{2}+y^{2} \leq 9\right\\} $$
View solution Problem 22
Find the indicated partial derivatives. \(g(v, w)=\frac{w^{2}}{v+w} ; g_{v}(1,1)\)
View solution Problem 22
Use the definition of continuity to show that $$f(x, y)=\sqrt{9+x^{2}+y^{2}}$$ is continuous at \((0,0)\).
View solution