Problem 9
Question
In Problems 9-12, find a unit vector in the direction in which \(f\) increases most rapidly at 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
Unit vector: \( \left(\frac{12}{13}, \frac{-5}{13}\right) \), Rate of change: 13.
1Step 1: Find the Gradient of the Function
The gradient of the function \( f(x, y) = x^3 - y^5 \) is a vector consisting of the first partial derivatives of \( f \). Calculate the partial derivative with respect to \( x \), which is \( \frac{\partial f}{\partial x} = 3x^2 \), and the partial derivative with respect to \( y \), which is \( \frac{\partial f}{\partial y} = -5y^4 \). Thus, the gradient is \( abla f(x, y) = (3x^2, -5y^4) \).
2Step 2: Evaluate the Gradient at Point \(\mathbf{p}\)
Substitute the coordinates of point \(\mathbf{p} = (2, -1)\) into the gradient to find the specific direction at that point. Calculate \( abla f(2, -1) = (3\cdot2^2, -5\cdot(-1)^4) = (12, -5) \).
3Step 3: Find the Unit Vector in Gradient Direction
The direction of maximum increase is given by the gradient vector \( (12, -5) \). To convert this to a unit vector, divide by its magnitude. The magnitude is \( \sqrt{12^2 + (-5)^2} = \sqrt{144 + 25} = \sqrt{169} = 13 \). Thus, the unit vector is \( \left(\frac{12}{13}, \frac{-5}{13}\right) \).
4Step 4: Determine the Rate of Change in this Direction
The rate of change in the direction of maximum increase is given by the magnitude of the gradient vector at \(\mathbf{p}\). As calculated, this magnitude is 13. Therefore, the rate of change is 13.
Key Concepts
Partial DerivativesUnit VectorRate of Change
Partial Derivatives
Partial derivatives are fundamental in understanding how functions change when their variables change. In a function of multiple variables, a partial derivative represents the rate of change of the function with respect to one of the variables while keeping the others constant.
Let's break it down with a friendlier analogy:
Suppose a hill represents a function where the height depends on both its position on the north-south and east-west directions. The partial derivative regarding the east-west direction would tell you how steep the hill is if you were to walk directly east or west, ignoring any change due to moving north or south.
The calculation of a partial derivative, such as \( \frac{\partial f}{\partial x} \), involves differentiating the function with respect to one variable while treating other variables as constants. In our example from the exercise, keeping \( y \) constant, \( \frac{\partial f}{\partial x} = 3x^2 \), reveals how the function changes along the \( x \) direction.
Let's break it down with a friendlier analogy:
Suppose a hill represents a function where the height depends on both its position on the north-south and east-west directions. The partial derivative regarding the east-west direction would tell you how steep the hill is if you were to walk directly east or west, ignoring any change due to moving north or south.
The calculation of a partial derivative, such as \( \frac{\partial f}{\partial x} \), involves differentiating the function with respect to one variable while treating other variables as constants. In our example from the exercise, keeping \( y \) constant, \( \frac{\partial f}{\partial x} = 3x^2 \), reveals how the function changes along the \( x \) direction.
Unit Vector
A unit vector is essentially a direction vector standardized to have a length of one. This means that while the direction of the vector remains unchanged, its magnitude does not affect calculations such as the rate of change.
Think of it as shrinking or growing the vector to fit perfectly inside a circle of radius one. This is useful because it helps us compare directions without worrying about how 'far' we go in those directions.
To compute a unit vector for a given vector, divide each component of the vector by its magnitude. In the exercise, the gradient vector \( (12, -5) \) is transformed into the unit vector \( \left(\frac{12}{13}, \frac{-5}{13}\right) \).
This keeps the direction but fits it to a uniform scale, making it easy to interpret and apply.
Think of it as shrinking or growing the vector to fit perfectly inside a circle of radius one. This is useful because it helps us compare directions without worrying about how 'far' we go in those directions.
To compute a unit vector for a given vector, divide each component of the vector by its magnitude. In the exercise, the gradient vector \( (12, -5) \) is transformed into the unit vector \( \left(\frac{12}{13}, \frac{-5}{13}\right) \).
This keeps the direction but fits it to a uniform scale, making it easy to interpret and apply.
Rate of Change
The rate of change in a given direction can provide much insight. It tells us how quickly the function increases or decreases as we move in that specific direction.
The steeper the incline in the direction of interest, the greater the rate of change. In mathematical terms, in context with the gradient vector, the rate of change at a given point is represented by the magnitude of the gradient.
This is because the gradient points in the direction of greatest increase. In our exercise, the direction of steepest ascent is precisely where the rate of change is maximum, calculated to be 13.
So, when we want to know 'how fast' things change in that optimal direction, the rate of change value gives us a clear, quantitative answer.
The steeper the incline in the direction of interest, the greater the rate of change. In mathematical terms, in context with the gradient vector, the rate of change at a given point is represented by the magnitude of the gradient.
This is because the gradient points in the direction of greatest increase. In our exercise, the direction of steepest ascent is precisely where the rate of change is maximum, calculated to be 13.
So, when we want to know 'how fast' things change in that optimal direction, the rate of change value gives us a clear, quantitative answer.
Other exercises in this chapter
Problem 8
Find the minimum distance between the origin and the plane \(x+3 y-2 z=4\).
View solution Problem 9
Use the total differential dz to approximate the change in \(z\) as \((x, y)\) moves from \(P\) to \(Q\). Then use a calculator to find the corresponding exact
View solution Problem 9
In Problems 7-12, find \(\partial w / \partial t\) by using the Chain Rule. Express your final answer in terms of \(s\) and \(t\). $$ w=e^{x^{2}+y^{2}} ; x=s \s
View solution Problem 9
\(f(x, y)=\cos x+\cos y+\cos (x+y)\); $$ 0
View solution