Problem 9

Question

Find a unit vector in the direction in which \(f\) increases most rapidly at \(\mathbf{p} .\) What is the rate of change in this direction? \(f(x, y)=x^{3}-y^{5} ; \mathbf{p}=(2,-1)\)

Step-by-Step Solution

Verified
Answer
The unit vector is \( \left( \frac{12}{13}, \frac{-5}{13} \right) \) and the rate of change is 13.
1Step 1: Calculate the Gradient of f
The gradient of a function \( f(x, y) \) is given by the vector of its partial derivatives. We calculate: \[ abla f(x, y) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \] For \( f(x, y) = x^3 - y^5 \), the partial derivatives are \( \frac{\partial f}{\partial x} = 3x^2 \) and \( \frac{\partial f}{\partial y} = -5y^4 \). Thus, \[ abla f(x, y) = (3x^2, -5y^4) \]
2Step 2: Evaluate the Gradient at Point p
Substitute the point \( \mathbf{p} = (2, -1) \) into the gradient: \[ abla f(2, -1) = (3(2)^2, -5(-1)^4) = (12, -5) \]
3Step 3: Determine the Unit Vector
A unit vector \( \mathbf{u} \) in the direction of \( abla f(2, -1) = (12, -5) \) is found by dividing by its magnitude. First, compute the magnitude: \[ |abla f(2, -1)| = \sqrt{12^2 + (-5)^2} = \sqrt{144 + 25} = \sqrt{169} = 13 \] The unit vector is: \[ \mathbf{u} = \left( \frac{12}{13}, \frac{-5}{13} \right) \]
4Step 4: Find the Rate of Change in This Direction
The rate of change of \( f \) in the direction of the gradient \( abla f \) is given by the magnitude of the gradient at \( \mathbf{p} \). From Step 3, we calculated: \[ |abla f(2, -1)| = 13 \] Thus, the rate of change is 13.

Key Concepts

Partial DerivativesRate of ChangeUnit Vector
Partial Derivatives
Partial derivatives are a way of taking a derivative for functions of multiple variables. In simple terms, they measure how a function changes as one variable changes, while all other variables are held constant. If you think of a function as a surface over a plane, partial derivatives describe the slope of the surface in various directions.
  • The partial derivative with respect to a variable measures the rate of change of the function as that particular variable is slightly changed.
  • For instance, for a function of two variables, such as our example function \( f(x, y) = x^3 - y^5 \), the partial derivative with respect to \( x \) is \( \frac{\partial f}{\partial x} = 3x^2 \).
  • Similarly, the partial derivative with respect to \( y \) is \( \frac{\partial f}{\partial y} = -5y^4 \).
Understanding partial derivatives is key to using gradients, which tell us the direction of steepest ascent on a surface.
Rate of Change
The rate of change tells us how fast something is changing. In mathematics, this is often in relation to some function. For functions of several variables, this rate is directional, meaning that it depends on the path you take on the graph of the function.
  • In our exercise, the gradient vector \( abla f(x, y) \) is crucial as it points in the direction of the fastest increase of the function.
  • The magnitude of this gradient vector gives us the rate of change of the function at a specific point in that direction.
  • At point \( \mathbf{p} = (2, -1) \), the rate of change in the direction of the gradient was found to be 13, indicating how steeply the function rises there in its fastest direction.
This concept is widely applied in optimization problems where increasing or decreasing functions most efficiently is desired.
Unit Vector
A unit vector is a vector that has a magnitude (or length) of exactly one. Unit vectors are useful for representing directions without scaling anything in particular. They effectively indicate direction while keeping the vector's size standard.
  • In our problem, the unit vector in the direction of the gradient is vital to identify without altering the rate or direction.
  • This is done by dividing each component of the vector by the vector's magnitude.
  • For the gradient \( abla f(2, -1) = (12, -5) \), the magnitude was calculated as 13. Hence, the unit vector becomes \( \left( \frac{12}{13}, \frac{-5}{13} \right) \).
Unit vectors are used in physics and engineering to simplify complex vectors and make computations more standard and manageable.