Problem 2 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the directional derivative of \(f\) at the point \(\mathbf{p}\) in the direction of \(\mathbf{a}\). \(f(x, y)=y^{2} \ln x ; \mathbf{p}=(1,4) ; \mathbf{a}=\mathbf{i}-\mathbf{j}\)

Step-by-Step Solution

Answer

The directional derivative is \( 8\sqrt{2} \).

Step 1: Find the gradient of f

The gradient of a function, \( f(x, y) \), is \( abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \). For \( f(x, y) = y^2 \ln x \), calculate the partial derivatives:- \( \frac{\partial f}{\partial x} = \frac{y^2}{x} \) (using the chain rule for \( \ln x \))- \( \frac{\partial f}{\partial y} = 2y \ln x \) (treat \( x \) as a constant and differentiate \( y^2 \))Thus, \[ abla f = \left( \frac{y^2}{x}, 2y \ln x \right) \].

Step 2: Evaluate the gradient at point \( \mathbf{p} = (1, 4) \)

Substitute \( x = 1 \) and \( y = 4 \) into the gradient:- \( \frac{y^2}{x} = \frac{4^2}{1} = 16 \)- \( 2y \ln x = 2(4) \ln(1) = 0 \) (since \( \ln(1) = 0 \))Therefore, \( abla f(1, 4) = (16, 0) \).

Step 3: Normalize the direction vector \( \mathbf{a} = \mathbf{i} - \mathbf{j} \)

The directional derivative requires a unit vector. Calculate the magnitude of \( \mathbf{a} = (1, -1) \):\[ \| \mathbf{a} \| = \sqrt{1^2 + (-1)^2} = \sqrt{2} \].Thus, the unit vector \( \mathbf{u} \) in the direction of \( \mathbf{a} \) is:\[ \mathbf{u} = \left( \frac{1}{\sqrt{2}}, \frac{-1}{\sqrt{2}} \right) \].

Step 4: Compute the directional derivative

The directional derivative \( D_{\mathbf{u}}f(\mathbf{p}) \) in the direction of a unit vector \( \mathbf{u} \) is given by:\[ D_{\mathbf{u}}f(\mathbf{p}) = abla f(\mathbf{p}) \cdot \mathbf{u} \].Substitute \( abla f(1, 4) = (16, 0) \) and \( \mathbf{u} = \left( \frac{1}{\sqrt{2}}, \frac{-1}{\sqrt{2}} \right) \) into the dot product:\[ (16, 0) \cdot \left( \frac{1}{\sqrt{2}}, \frac{-1}{\sqrt{2}} \right) = 16 \cdot \frac{1}{\sqrt{2}} + 0 \cdot \left( \frac{-1}{\sqrt{2}} \right) = 16 \times \frac{1}{\sqrt{2}} = \frac{16}{\sqrt{2}} \].Simplify using \( \sqrt{2} \approx 1.414 \):\[ \frac{16}{\sqrt{2}} = 8\sqrt{2} \].

Key Concepts

Gradient

The gradient of a function is a crucial concept when dealing with multivariable calculus. It tells us the direction of the steepest ascent of a function and consists of the partial derivatives of the function with respect to each of the variables. For any function like \( f(x, y) = y^2 \ln x \), the gradient \( abla f \) is expressed as:

• \( \frac{\partial f}{\partial x} \) - this represents the rate of change of \( f \) with respect to \( x \),
• \( \frac{\partial f}{\partial y} \) - this signifies the rate of change of \( f \) with respect to \( y \).

The gradient \( abla f \) is a vector pointing in a direction of fastest increase, combining the partial derivatives into a coherent description of change. In practical terms, it's like a compass in the terrain of functions, pointing uphill.

Partial Derivatives

Partial derivatives break down how a multivariable function changes as one individual variable changes, keeping others constant. If you look at each variable separately, it's like examining one dimension at a time rather than all at once. For the function \( f(x, y) = y^2 \ln x \), when taking the partial derivative:

• \( \frac{\partial f}{\partial x} \) shows how \( f \) changes as \( x \) changes, holding \( y \) constant.
• \( \frac{\partial f}{\partial y} \) reveals how \( f \) shifts with changes in \( y \), holding \( x \) constant.

Partial derivatives are fundamental building blocks for constructing the gradient, providing insight into how a surface stretches or contracts in different directions. They are used to form judgments about the behavior of functions near specific points.

Unit Vector

A unit vector is essentially a directional pointer, describing a direction in a space without any consideration of magnitude or length. To find the unit vector in the direction of another vector, divide each component of the original vector by its magnitude. For example, with vector \( \mathbf{a} = \mathbf{i} - \mathbf{j} = (1, -1) \), its magnitude is:\[\| \mathbf{a} \| = \sqrt{1^2 + (-1)^2} = \sqrt{2} \]and thus, the unit vector \( \mathbf{u} \) becomes:\[\mathbf{u} = \left( \frac{1}{\sqrt{2}}, \frac{-1}{\sqrt{2}} \right)\]In calculations such as finding the directional derivative, using the unit vector ensures you're moving in the correct direction but without altering the magnitude of change implied by the gradient.

Dot Product

The dot product is a way of multiplying two vectors to return a scalar, which is a measure of how much one vector goes in the direction of another. It's mathematically defined as the sum of the products of their corresponding components. If \( \mathbf{v} = (a, b) \) and \( \mathbf{w} = (c, d) \), the dot product \( \mathbf{v} \cdot \mathbf{w} \) is:\[\mathbf{v} \cdot \mathbf{w} = ac + bd\]In context, to find the directional derivative, perform the dot product between the function's gradient at a point and the unit vector of the desired direction. The result represents how much the function changes in that specific direction. This concept effectively combines directions from both the gradient and the unit vector to yield an actionable scalar value.