Problem 23 - 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. $$f(x, y)=3 x^{2}+2 y+5 ; P(1,2) ;\langle-3,4\rangle$$

Step-by-Step Solution

Answer

Answer: The directional derivative is -2.

Step 1: Find the gradient of the function

To find the gradient (or sometimes called the "nabla" represented by the symbol "∇") of the function $$f(x, y) = 3x^2 + 2y + 5$$, we need to compute the partial derivatives with respect to both x and y. The partial derivative of f with respect to x is: $$\frac{\partial f}{\partial x} = 6x$$ The partial derivative of f with respect to y is: $$\frac{\partial f}{\partial y} = 2$$ Therefore, the gradient of f is given by: $$\nabla f = \langle \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \rangle = \langle 6x, 2 \rangle$$

Step 2: Normalize the direction vector

We are given the direction vector $$\langle -3, 4 \rangle$$. To find the unit vector (a vector with magnitude 1) in the same direction, we first find the magnitude of the given vector. The magnitude of a vector $$\langle a, b \rangle$$ is given by: Magnitude = $$\sqrt{a^2 + b^2}$$ For our vector, the magnitude is: Magnitude = $$\sqrt{(-3)^2 + (4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5$$ Now, we can find the unit vector by dividing each component of the given vector by the magnitude: Unit vector = $$\frac{\langle -3, 4 \rangle}{5} = \langle -\frac{3}{5}, \frac{4}{5} \rangle$$

Step 3: Compute the dot product of the gradient and the normalized direction vector

The directional derivative is the dot product of the gradient and the unit vector in the given direction. First, let's find the gradient at point P(1, 2): $$\nabla f(1, 2) = \langle 6(1), 2 \rangle = \langle 6, 2 \rangle$$ Now, we can compute the dot product of the gradient and the unit vector: Directional Derivative = $$\langle 6, 2 \rangle \cdot \langle -\frac{3}{5}, \frac{4}{5} \rangle = 6 \cdot -\frac{3}{5} + 2 \cdot \frac{4}{5} = -\frac{18}{5} + \frac{8}{5} = -\frac{10}{5} = -2$$ Therefore, the directional derivative of the function $$f(x, y) = 3x^2 + 2y + 5$$ at point P(1, 2) in the direction of the vector $$\langle -3, 4 \rangle$$ is -2.

Key Concepts

Gradient Vector

The gradient vector provides a multi-dimensional rate of change for a function. It is essentially a vector comprised of all the partial derivatives of a function. For a two-variable function like the example here, \(f(x, y) = 3x^2 + 2y + 5\), the gradient vector \(abla f\) calculates the steepest ascent direction in relation to both \(x\) and \(y\).
For any function \(f(x, y)\), the gradient is computed by:

• Finding the partial derivative with respect to \(x\), in this case, \(6x\).
• Finding the partial derivative with respect to \(y\), resulting in \(2\).

The complete gradient vector is then \(abla f = \langle 6x, 2 \rangle\). At any specific point \((x, y)\), you evaluate the gradient by substituting \(x\) and \(y\) into these partial derivatives.
This concept is essential because the gradient points in the direction of the greatest rate of increase of the function.

Partial Derivatives

Partial derivatives refer to how a function changes as you vary one of its input variables, while keeping the others constant. This concept lives at the heart of multivariable calculus. For the given function \(f(x, y) = 3x^2 + 2y + 5\), the partial derivative with respect to \(x\) and \(y\) informs us how the function changes in those directions.
For the partial derivative of \(f\) with respect to \(x\):
\[ \frac{\partial f}{\partial x} = 6x \]
This tells us that the rate of change of \(f\) with respect to changes in the \(x\)-axis is proportional to \(x\).
For the partial derivative of \(f\) with respect to \(y\):
\[ \frac{\partial f}{\partial y} = 2 \]
This indicates that the function changes at a constant rate of 2 as you move along the \(y\)-axis direction.
Understanding partial derivatives helps you analyze how multidimensional changes affect a function's outcome.

Unit Vector

A unit vector indicates direction and has a magnitude of one. It is often used to specify directions without implying changes in scale. When dealing with directional derivatives, it's important to convert any given direction vector into a unit vector, ensuring that only the direction affects the derivative, not the length.
For the vector \(\langle -3, 4 \rangle\), we calculate its magnitude to turn it into a unit vector. The magnitude is calculated as:
\[ \sqrt{(-3)^2 + (4)^2} = \sqrt{25} = 5 \]
Dividing each component of the vector by this magnitude gives the unit vector:

• \( -3/5 \) for the x-component
• \( 4/5 \) for the y-component

So, the unit vector becomes \(\langle -\frac{3}{5}, \frac{4}{5} \rangle\).
Using a unit vector ensures that the calculated directional derivative solely reflects the direction's effect.

Dot Product

The dot product is a method of multiplying two vectors which results in a scalar. In the context of directional derivatives, it's used to combine the gradient vector with the unit direction vector to quantify the function's directional rate of change.
The dot product of two vectors \(\langle a, b \rangle\) and \(\langle c, d \rangle\) is computed as:
\[ a \cdot c + b \cdot d \]
In our solution, the gradient at point \(P(1, 2)\), \(\langle 6, 2 \rangle\), and the unit vector \(\langle -\frac{3}{5}, \frac{4}{5} \rangle\) yield a dot product of:
\[ 6 \cdot -\frac{3}{5} + 2 \cdot \frac{4}{5} = -2 \]
The resulting value \(-2\) represents the rate of change of the function in the given direction.
The dot product concept ties together the gradient and direction seamlessly, producing a coherent measure of directional change.