Problem 11 - Calculus Early Transcendentals

Question

\(11-17\) Find the directional derivative of the function at the given point in the direction of the vector \(\mathbf{v}\) . $$f(x, y)=1+2 x \sqrt{y}, \quad(3,4), \quad \mathbf{v}=\langle 4,-3\rangle$$

Step-by-Step Solution

Answer

The directional derivative is 2.3.

Step 1: Find the Gradient

First, we need to find the gradient of the function \( f(x, y) = 1 + 2x\sqrt{y} \). The gradient is a vector of partial derivatives: \( abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \). This gives us:\( \frac{\partial f}{\partial x} = 2\sqrt{y} \) and \( \frac{\partial f}{\partial y} = \frac{x}{\sqrt{y}} \). So, \( abla f = \langle 2\sqrt{y}, \frac{x}{\sqrt{y}} \rangle \).

Step 2: Evaluate the Gradient at the Point

Now we evaluate the gradient at the given point \((3, 4)\). Plug \(x = 3\) and \(y = 4\) into \(abla f = \langle 2\sqrt{y}, \frac{x}{\sqrt{y}} \rangle \). Thus, \(2\sqrt{4} = 4\) and \(\frac{3}{\sqrt{4}} = \frac{3}{2}\). The gradient at \((3,4)\) is \(\langle 4, \frac{3}{2} \rangle \).

Step 3: Normalize the Direction Vector

The directional derivative requires a unit direction vector, so we normalize \(\mathbf{v} = \langle 4, -3 \rangle \). The magnitude of \(\mathbf{v}\) is \(\sqrt{4^2 + (-3)^2} = 5\). The unit vector in the direction of \(\mathbf{v}\) is \(\langle \frac{4}{5}, -\frac{3}{5} \rangle \).

Step 4: Calculate the Directional Derivative

Finally, compute the directional derivative \(D_\mathbf{v}f\) by taking the dot product of the gradient at \((3,4)\) and the unit direction vector: \(D_\mathbf{v}f = \langle 4, \frac{3}{2} \rangle \cdot \langle \frac{4}{5}, -\frac{3}{5} \rangle \). This is \(4\times\frac{4}{5} + \frac{3}{2}\times (-\frac{3}{5}) = \frac{16}{5} - \frac{9}{10} = \frac{32}{10} - \frac{9}{10} = \frac{23}{10} = 2.3\).

Key Concepts

Gradient

The gradient is a fundamental concept in multivariable calculus represented by the symbol \( abla f \). It is essentially a vector composed of the partial derivatives of a function. For a given function \( f(x, y) \), the gradient \( abla f \) provides the direction and rate of the steepest ascent from any given point.

• To compute the gradient, calculate the partial derivative of the function with respect to each independent variable. This yields the components of the gradient vector.
• In the context of the problem, for the function \( f(x, y) = 1 + 2x\sqrt{y} \), the gradient is \( abla f = \langle 2\sqrt{y}, \frac{x}{\sqrt{y}} \rangle \).
• Each component represents the rate of change of the function with respect to the corresponding variable, holding the other variable constant.

Once you have the gradient, evaluate it at the specific point of interest to understand how the function behaves locally at that point.

Partial Derivatives

Partial derivatives are the derivatives of multivariable functions with respect to one variable at a time, treating all other variables as constant.

• For a function \( f(x, y) \), the partial derivative with respect to \( x \) is denoted as \( \frac{\partial f}{\partial x} \), and similarly, with respect to \( y \), it's \( \frac{\partial f}{\partial y} \).
• In essence, partial derivatives tell us how the function changes if we only vary one of the variables slightly.
• In our exercise, \( \frac{\partial f}{\partial x} = 2\sqrt{y} \) shows the sensitivity of the function to changes in \( x \) and \( \frac{\partial f}{\partial y} = \frac{x}{\sqrt{y}} \) shows it with respect to \( y \).

These computations are crucial as they form the components of the gradient and pave the way to finding directional derivatives.

Dot Product

The dot product, sometimes called the scalar product, is a way to multiply two vectors that results in a scalar, or real 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 computed as \( a_1b_1 + a_2b_2 \).
• The dot product is particularly useful when determining the component of one vector in the direction of another.
• In the context of the directional derivative, the dot product of the gradient vector and a unit vector gives you the rate of change of the function in the specified direction.

For this particular exercise, the dot product helps us evaluate how the function \( f \) changes in the direction of the vector \( \langle 4, -3 \rangle \), which has been normalized into a unit vector.

Unit Vector

A unit vector is a vector with a magnitude of 1, used to indicate direction. In mathematics, unit vectors are essential for computing directional derivatives.

• To convert any vector \( \mathbf{v} = \langle a, b \rangle \) into a unit vector, divide each component by the vector's magnitude \( ||\mathbf{v}|| \).
• The magnitude of a vector \( \mathbf{v} \) is found with the formula \( ||\mathbf{v}|| = \sqrt{a^2 + b^2} \).
• In our solution, for the vector \( \langle 4, -3 \rangle \), the magnitude is 5, giving the unit vector \( \langle \frac{4}{5}, -\frac{3}{5} \rangle \).

Using a unit vector is critical for finding directional derivatives because it ensures the direction of the change is consistent without altering the magnitude of what is being computed.