Problem 24 - Calculus Early Transcendentals

Question

Compute the directional derivative of the following functions at the given point P in the direction of the given vector. Be sure to use a unit vector for the direction vector. $$h(x, y)=e^{-x-y} ; P(\ln 2, \ln 3) ;\langle 1,1\rangle$$

Step-by-Step Solution

Answer

Answer: The directional derivative is $$-\frac{1}{3\sqrt{2}}$$.

Step 1: 1. Calculate the gradient of the function

To calculate the gradient of the function $$h(x, y) = e^{-x-y}$$, we need to find its partial derivatives with respect to x and y. $$ \frac{\partial h}{\partial x} = - e^{-x-y} \\ \frac{\partial h}{\partial y} = - e^{-x-y} $$ So the gradient of the function is $$\nabla h(x,y) = \langle - e^{-x-y}, - e^{-x-y}\rangle$$.

Step 2: 2. Normalize the direction vector

We have the direction vector $$\langle 1, 1 \rangle$$. To find its unit vector, we need to divide the vector by its magnitude. The magnitude of the vector is: $$\sqrt{1^2 + 1^2} = \sqrt{2}$$ Thus, the unit vector for the direction is: $$\frac{1}{\sqrt{2}}\langle 1, 1 \rangle = \langle \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \rangle$$.

Step 3: 3. Compute the directional derivative

Let's compute the directional derivative of the function at point P(\(\ln2, \ln3\)) in the direction of the unit vector, using the dot product: $$ D_{\vec{u}}h(x,y) = \nabla h(x,y) \cdot \vec{u} $$ $$ D_{\vec{u}}h(\ln 2, \ln 3) = \langle - e^{-(\ln 2 + \ln 3)}, - e^{-(\ln 2 + \ln 3)}\rangle \cdot \langle \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \rangle$$ $$ D_{\vec{u}}h(\ln 2, \ln 3) = - e^{-(\ln 6)}\left(\frac{1}{\sqrt{2}}\right)- e^{-(\ln 6)}\left(\frac{1}{\sqrt{2}}\right) $$ Now, we know that $$6=2*3$$, so we can simplify using some logarithmic properties $$D_{\vec{u}}h(\ln 2, \ln 3) = -\frac{1}{6}\left(\frac{1}{\sqrt{2}}\right)- \frac{1}{6}\left(\frac{1}{\sqrt{2}}\right)$$ $$D_{\vec{u}}h(\ln 2, \ln 3) = -\frac{1}{3\sqrt{2}}$$ So, the directional derivative of the function $$h(x, y) = e^{-x-y}$$ at point P(\(\ln 2, \ln 3\)) in the direction of the vector $$\langle 1, 1 \rangle$$ is $$-\frac{1}{3\sqrt{2}}$$.

Key Concepts

Gradient

The gradient is a vector that contains all the partial derivatives of a given function. It points in the direction of the greatest rate of increase of the function. For a function of two variables, such as \( h(x, y) = e^{-x-y} \), the gradient \( abla h(x,y) \) is computed by finding the partial derivatives with respect to each variable, \( x \) and \( y \).

In this example, both partial derivatives are the same:

• \( \frac{\partial h}{\partial x} = - e^{-x-y} \)
• \( \frac{\partial h}{\partial y} = - e^{-x-y} \)

Hence, the gradient is \( abla h(x,y) = \langle - e^{-x-y}, - e^{-x-y}\rangle \). This vector tells you how quickly and in which direction \( h \) increases the most.

Partial Derivative

Partial derivatives represent the rate of change of a multivariable function with respect to one of those variables while holding the others constant. For instance, the partial derivative of \( h(x, y) = e^{-x-y} \) with respect to \( x \) is \( \frac{\partial h}{\partial x} = - e^{-x-y} \).

This measures how \( h \) changes as \( x \) changes, keeping \( y \) the same. Likewise, \( \frac{\partial h}{\partial y} = - e^{-x-y} \) tells you how \( h \) changes as \( y \) is altered, while \( x \) remains fixed. Partial derivatives are fundamental in finding the gradient, which is crucial for calculating the directional derivative.

Unit Vector

A unit vector has a magnitude of 1 and indicates direction. To compute a directional derivative, the direction vector must be converted into a unit vector. For example, consider the direction vector \( \langle 1, 1 \rangle \).

The magnitude of this vector is \( \sqrt{1^2 + 1^2} = \sqrt{2} \). To make it a unit vector, divide the original vector by its magnitude:

• \( \frac{1}{\sqrt{2}}\langle 1, 1 \rangle = \langle \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \rangle \)

The resulting unit vector keeps the same direction as \( \langle 1, 1 \rangle \) but ensures that its length is 1. This is essential for accurately evaluating directional derivatives.

Dot Product

The dot product (also known as the scalar product) is an algebraic operation that takes two equal-length sequences of numbers (usually coordinate vectors), and returns a single number. For two vectors \( \mathbf{a} = \langle a_1, a_2 \rangle \) and \( \mathbf{b} = \langle b_1, b_2 \rangle \), the dot product is calculated as:

• \( \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 \)

In the context of directional derivatives, we use the dot product to find the derivative of a function in the direction of a unit vector \( \mathbf{u} \). For \( h(x, y) \) at the specific point \( P(\ln 2, \ln 3) \) with the unit vector \( \langle \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \rangle \), the directional derivative is:

• \( D_{\mathbf{u}}h = abla h \cdot \mathbf{u} \)

This effectively measures how much the function changes when moving in the direction of \( \mathbf{u} \). In our solution, this results in \( -\frac{1}{3\sqrt{2}} \), indicating how steeply the function \( h \) decreases in that specific direction.