Chapter 9: Problem 10
(a) Find the gradient of \(f\) . (b) Evaluate the gradient at the point \(P\) . (c) Find the rate of change of \(f\) at \(P\) in the direction of the vector \mathbf{u} . \(f(x, y)=y^{2} / x, \quad P(1,2), \quad \mathbf{u}=\left[ \begin{array}{ll}{\frac{2}{3}, \frac{1}{3} \sqrt{5}}\end{array}\right]\)
Short Answer
Step by step solution
Find the partial derivatives
Form the gradient vector
Evaluate the gradient at point P
Find the rate of change in the direction of \( \mathbf{u} \)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Multivariable Calculus
To analyze these functions, we consider how changes in input variables affect the output of the function. This is where concepts like gradients, partial derivatives, and directional derivatives become crucial tools.
Applications of multivariable calculus are vast, ranging from physics and engineering to economics and biological sciences. It enables us to model systems with many interacting variables, providing deeper insights into intricate real-world phenomena.
Partial Derivatives
Calculating partial derivatives involves treating all but one variable as constants, similar to calculating derivatives in single-variable calculus. This technique helps us understand how each variable individually influences the overall function.
For example, if \( f(x, y) = \frac{y^2}{x} \), the partial derivative with respect to \( x \) is \( f_x = -\frac{y^2}{x^2} \) and with respect to \( y \) is \( f_y = \frac{2y}{x} \). These results give us the rate at which \( f \) changes concerning each variable independently.
Rate of Change
The gradient, a vector, provides us with the direction and rate of the steepest ascent of the function. In simpler terms, it tells us which direction to move in, at a particular point, to increase the function's value most rapidly.
By evaluating the gradient at a specific point, as seen in the solution where the gradient at \( P(1, 2) \) is \( (-4, 4) \), we obtain the rate of change in both the \( x \) and \( y \) directions. This vector gives a convenient snapshot of the function's behavior around the point.
Directional Derivative
The formula for the directional derivative of a function \( f \) at a point \( P \) in direction \( \mathbf{u} \) is the dot product \( abla f(P) \cdot \mathbf{u} \). This computation gives the rate of change along the line in the preferred direction.
In the given problem, finding the directional derivative of \( f \) at point \( P \) in the direction of vector \( \mathbf{u} = \left( \frac{2}{3}, \frac{1}{3}\sqrt{5} \right) \) involved computing \( (-4, 4) \cdot \mathbf{u} \). The result, \( \frac{4\sqrt{5} - 8}{3} \), signifies how quickly the function's value changes in that specific direction.