Problem 19 - 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. $$f(x, y)=10-3 x^{2}+\frac{y^{4}}{4} ; P(2,-3) ;\left\langle\frac{\sqrt{3}}{2},-\frac{1}{2}\right\rangle$$

Step-by-Step Solution

Answer

The directional derivative of the function at the given point and direction is $$-6\sqrt{3} + \frac{27}{2}$$.

Step 1: Compute the gradient of the function

The gradient of a function is a vector composed of the partial derivatives of the function with respect to each variable. For the given function $$f(x, y) = 10 - 3x^2 + \frac{y^4}{4}$$, we calculate the partial derivatives with respect to x and y respectively: Partial derivative with respect to x: $$\frac{\partial}{\partial x}f(x, y) = -6x$$ Partial derivative with respect to y: $$\frac{\partial}{\partial y}f(x, y) = y^3$$ So, the gradient of the function is given by: $$\nabla f(x, y) = \left\langle -6x, y^3 \right\rangle$$

Step 2: Ensure the direction vector is a unit vector

We are given the direction vector $$\left\langle \frac{\sqrt{3}}{2}, -\frac{1}{2} \right\rangle$$. We need to ensure that this is a unit vector. A vector is a unit vector if its magnitude equals 1. The magnitude of this vector is: $$\lVert \left\langle \frac{\sqrt{3}}{2}, -\frac{1}{2} \right\rangle \rVert = \sqrt{\left(\frac{\sqrt{3}}{2}\right)^2 + \left(-\frac{1}{2}\right)^2} = \sqrt{\frac{3}{4} + \frac{1}{4}} = \sqrt{1} = 1$$ Since the magnitude of the direction vector is 1, it is already a unit vector.

Step 3: Compute the directional derivative

To compute the directional derivative of the function at the given point $$P(2, -3)$$ in the direction of the unit vector, we need to find the dot product of the gradient at P with the unit direction vector. The gradient at P is: $$\nabla f(2, -3) = \left\langle -6(2), (-3)^3 \right\rangle = \left\langle -12, -27 \right\rangle$$ Now, we find the dot product of the gradient at P with the unit direction vector: $$D_{\text{unit vector}}f(2, -3) = \left\langle -12, -27 \right\rangle \cdot \left\langle \frac{\sqrt{3}}{2}, -\frac{1}{2} \right\rangle = -12 \cdot \frac{\sqrt{3}}{2} - 27 \cdot \left(-\frac{1}{2}\right)$$ Simplifying the expression: $$D_{\text{unit vector}}f(2, -3) = -6\sqrt{3} + \frac{27}{2}$$ Therefore, the directional derivative of the given function at the point $$P(2, -3)$$ in the direction of the given unit vector is $$-6\sqrt{3} + \frac{27}{2}$$.

Key Concepts

Gradient Vector

The gradient vector is a central concept when dealing with directional derivatives. It gives us the best direction in which to move a function to see the greatest rate of increase. In our function, the gradient is composed of the partial derivatives with respect to each variable:

• The partial derivative with respect to \( x \) is \( -6x \)
• The partial derivative with respect to \( y \) is \( y^3 \)

This means the gradient vector for the function \( f(x, y) = 10 - 3x^2 + \frac{y^4}{4} \) is \( abla f(x, y) = \langle -6x, y^3 \rangle \).
At point \( P(2, -3) \), substituting the values of \( x \) and \( y \), we have the gradient \( abla f(2, -3) = \langle -12, -27 \rangle \).
The gradient tells us how the function is changing at any point in space, and is fundamental in calculating the directional derivative.

Partial Derivatives

Partial derivatives are the building blocks for the gradient vector. They measure how a function changes as you vary one variable while keeping others constant. For the function \( f(x, y) \), we compute two partial derivatives:

• For \( x \), the derivative is \( \frac{\partial}{\partial x}f(x, y) = -6x \).
• For \( y \), the derivative is \( \frac{\partial}{\partial y}f(x, y) = y^3 \).

These derivatives provide critical information. The derivative with respect to \( x \) tells us how the function changes as \( x \) changes, while keeping \( y \) fixed. Similarly, the derivative with respect to \( y \) tells how \( y \) impacts the function keeping \( x \) constant. Combined, these help form the gradient which is used to find the directional derivative.

Unit Vector

A unit vector defines the direction we are interested in. It is crucial because we need to ensure our vector has a magnitude of 1 to correctly measure rates of change in given directions.
The exercise provides us the direction vector \( \langle \frac{\sqrt{3}}{2}, -\frac{1}{2} \rangle \). To check if it's a unit vector, calculate its magnitude:

• Magnitude = \( \sqrt{\left(\frac{\sqrt{3}}{2}\right)^2 + \left(-\frac{1}{2}\right)^2} = \sqrt{\frac{3}{4} + \frac{1}{4}} = \sqrt{1} = 1 \)

Because its magnitude equals 1, it is indeed a unit vector. This confirmation is essential before using it in directional derivative calculations, ensuring an accurate measure of change in the function's value in the given direction.

Dot Product

The dot product operation is essential when finding directional derivatives. It combines two vectors to yield a scalar, which represents how much one vector "projects" onto another.
To find the directional derivative, we calculate the dot product of the gradient and the unit vector. Here, the calculation for the dot product is:

• Gradient at point \( P \): \( abla f(2, -3) = \langle -12, -27 \rangle \)
• Unit direction vector: \( \langle \frac{\sqrt{3}}{2}, -\frac{1}{2} \rangle \)

Dot product: \[ D_{\text{unit vector}}f(2, -3) = \left\langle -12, -27 \right\rangle \cdot \left\langle \frac{\sqrt{3}}{2}, -\frac{1}{2} \right\rangle = -12 \cdot \frac{\sqrt{3}}{2} - 27 \cdot \left(-\frac{1}{2}\right) \]Simplifying: \[ -6\sqrt{3} + \frac{27}{2} \]
This value is the directional derivative of the function at \( P \) in the specified direction, indicating how the function behaves when moving precisely in that direction.