Problem 4
Question
Find the directional derivative of \(f\) at the point \(P\) in the direction of a. $$ f(x, y)=x-y^{2} ; P=(2,-3) ; \mathbf{a}=\mathbf{i}+2 \mathbf{j} $$
Step-by-Step Solution
Verified Answer
The directional derivative of \( f \) at point \( P \) in the direction of \( \mathbf{a} \) is \( \frac{13}{\sqrt{5}} \).
1Step 1: Calculate the Gradient of f
First, find the gradient of the function \( f \), which is a vector of its first partial derivatives. The partial derivative with respect to \( x \) is \( \frac{\partial f}{\partial x} = 1 \), and the partial derivative with respect to \( y \) is \( \frac{\partial f}{\partial y} = -2y \). Thus, the gradient is \( abla f = \langle 1, -2y \rangle \).
2Step 2: Evaluate the Gradient at Point P
Substitute the coordinates of point \( P=(2, -3) \) into the gradient. This gives \( abla f(2, -3) = \langle 1, -2(-3) \rangle = \langle 1, 6 \rangle \).
3Step 3: Normalize the Direction Vector a
The direction vector \( \mathbf{a} \) is given as \( \mathbf{i} + 2\mathbf{j} \), or \( \langle 1, 2 \rangle \). To normalize, find its magnitude: \( \| \mathbf{a} \| = \sqrt{1^2 + 2^2} = \sqrt{5} \). The normalized direction vector is \( \langle \frac{1}{\sqrt{5}}, \frac{2}{\sqrt{5}} \rangle \).
4Step 4: Calculate the Dot Product
The directional derivative is given by the dot product of the gradient at \( P \) and the normalized direction vector: \( abla f(P) \cdot \mathbf{u} \), where \( \mathbf{u} \) is the normalized vector. Therefore, compute \( \langle 1, 6 \rangle \cdot \langle \frac{1}{\sqrt{5}}, \frac{2}{\sqrt{5}} \rangle = \frac{1}{\sqrt{5}} (1) + \frac{12}{\sqrt{5}} = \frac{13}{\sqrt{5}} \).
Key Concepts
GradientPartial DerivativesNormalizationDot Product
Gradient
To understand the gradient, think of it as a tool that tells us how a function changes in different directions. Imagine you have a landscape with hills and valleys. The gradient points in the direction of the steepest ascent or descent. For a multivariable function like \( f(x, y) = x - y^2 \), the gradient is a vector combining the rates of change of \( f \) in the \( x \) and \( y \) directions. These rates of change are called partial derivatives.
To compute the gradient for our example function, we find the partial derivatives with respect to each variable: the partial derivative with respect to \( x \) is \( 1 \), and with respect to \( y \) is \(-2y\). This results in the gradient vector \( abla f = \langle 1, -2y \rangle \).
Understanding the gradient is crucial because it shows us the direction and rate of the fastest increase or decrease of the function at any given point.
To compute the gradient for our example function, we find the partial derivatives with respect to each variable: the partial derivative with respect to \( x \) is \( 1 \), and with respect to \( y \) is \(-2y\). This results in the gradient vector \( abla f = \langle 1, -2y \rangle \).
Understanding the gradient is crucial because it shows us the direction and rate of the fastest increase or decrease of the function at any given point.
Partial Derivatives
Partial derivatives measure how a function changes as we alter one variable while keeping others constant. Imagine examining a function's slope only in one direction at a time, like in the \( x \) or \( y \) directions. For the function \( f(x, y) = x - y^2 \), the partial derivative with respect to \( x \) is found by differentiating \( f \) while treating \( y \) as a constant: \( \frac{\partial f}{\partial x} = 1 \).
Similarly, the partial derivative with respect to \( y \) treats \( x \) as a constant, giving us \( \frac{\partial f}{\partial y} = -2y \).
These partial derivatives are essential for constructing the gradient, which combines them into a vector indicating how the function increases or decreases in both directions.
Similarly, the partial derivative with respect to \( y \) treats \( x \) as a constant, giving us \( \frac{\partial f}{\partial y} = -2y \).
These partial derivatives are essential for constructing the gradient, which combines them into a vector indicating how the function increases or decreases in both directions.
Normalization
Normalization involves scaling a vector so that its magnitude becomes 1. This process keeps the vector's direction unchanged. The main reason to normalize a vector is when you want only the direction, but not the length, to influence calculations. For the direction vector \( \mathbf{a} = \langle 1, 2 \rangle \), its magnitude is \( \sqrt{1^2 + 2^2} = \sqrt{5} \).
To normalize it, we divide each component of the vector by its magnitude, resulting in the normalized vector \( \langle \frac{1}{\sqrt{5}}, \frac{2}{\sqrt{5}} \rangle \).
This normalized vector is crucial for calculating the directional derivative, as it ensures that the movement in a particular direction is uniform.
To normalize it, we divide each component of the vector by its magnitude, resulting in the normalized vector \( \langle \frac{1}{\sqrt{5}}, \frac{2}{\sqrt{5}} \rangle \).
This normalized vector is crucial for calculating the directional derivative, as it ensures that the movement in a particular direction is uniform.
Dot Product
The dot product, also known as the scalar product, involves two vectors and provides a scalar output. It's calculated by multiplying corresponding components and summing the results. If \( \mathbf{v} = \langle v_1, v_2 \rangle \) and \( \mathbf{w} = \langle w_1, w_2 \rangle \), the dot product is \( v_1w_1 + v_2w_2 \). In our example, we find the directional derivative by computing the dot product of the gradient at point \( P \) and the normalized direction vector.
The gradient vector at \( P = (2, -3) \) is \( \langle 1, 6 \rangle \), and the normalized direction vector is \( \langle \frac{1}{\sqrt{5}}, \frac{2}{\sqrt{5}} \rangle \). The dot product is thus \( 1 \cdot \frac{1}{\sqrt{5}} + 6 \cdot \frac{2}{\sqrt{5}} \), calculated as \( \frac{13}{\sqrt{5}} \).
The dot product is a useful tool for determining how much one vector extends in the direction of another.
The gradient vector at \( P = (2, -3) \) is \( \langle 1, 6 \rangle \), and the normalized direction vector is \( \langle \frac{1}{\sqrt{5}}, \frac{2}{\sqrt{5}} \rangle \). The dot product is thus \( 1 \cdot \frac{1}{\sqrt{5}} + 6 \cdot \frac{2}{\sqrt{5}} \), calculated as \( \frac{13}{\sqrt{5}} \).
The dot product is a useful tool for determining how much one vector extends in the direction of another.
Other exercises in this chapter
Problem 4
Find the gradient of the function. $$ f(x, y)=\frac{x y-1}{x^{2}+y^{2}} $$
View solution Problem 4
Compute \(d z / d t\). $$ z=\tan ^{-1}\left(y^{2}-x^{2}\right) ; x=\sin t, y=\cos t $$
View solution Problem 4
Find the first partial derivatives of the function. $$ g(x, y)=x^{3} e^{2 y} $$
View solution Problem 4
Evaluate the limit. $$ \lim _{(x, y, z) \rightarrow(-1,2,0)}\left(x^{2}+3 y-4 z^{2}+2\right) $$
View solution