Problem 20 - 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. $$g(x, y)=\sin \pi(2 x-y) ; P(-1,-1) ;\left\langle\frac{5}{13},-\frac{12}{13}\right\rangle$$

Step-by-Step Solution

Answer

In summary, we calculated the directional derivative of the function \(g(x, y) = \sin \left(\pi(2x - y)\right)\) at point \(P(-1, -1)\) in the direction of the vector \(\left\langle\frac{5}{13},-\frac{12}{13}\right\rangle\). The directional derivative is \(-\frac{7\pi}{13}\).

Step 1: Calculate the partial derivatives

Let's find the partial derivatives of the function \(g(x, y) = sin(\pi(2x - y))\). Partial derivative with respect to x: $$\frac{\partial g}{\partial x} = \frac{\partial}{\partial x} (\sin (\pi(2x - y))) = 2\pi \cos (\pi(2x - y))$$ Partial derivative with respect to y: $$\frac{\partial g}{\partial y} = \frac{\partial}{\partial y} (\sin (\pi(2x - y))) = -\pi \cos (\pi(2x - y))$$

Step 2: Compute the Gradient

Now, we will find the gradient of the function, which is the vector consisting of these partial derivatives: $$\nabla g = \left\langle \frac{\partial g}{\partial x}, \frac{\partial g}{\partial y} \right\rangle = \left\langle 2\pi \cos (\pi(2x - y)), -\pi \cos (\pi(2x - y)) \right\rangle$$

Step 3: Evaluate Gradient at Point P

We are given the point \(P(-1, -1)\). Let's evaluate the gradient at this point: $$\nabla g(-1, -1) = \left\langle 2\pi \cos (\pi(2(-1) - (-1))), -\pi \cos (\pi(2(-1) - (-1))) \right\rangle = \left\langle 2\pi, \pi \right\rangle$$

Step 4: Find the Unit Direction Vector

We are given the direction vector \(\left\langle\frac{5}{13},-\frac{12}{13}\right\rangle\). We note that it is already a unit vector since its magnitude is 1 (\(\frac{1}{13^2}(5^2+(-12)^2)=1\)), so we don't need to normalize it.

Step 5: Compute the Dot Product of Gradient and Unit Direction Vector

Now, we will find the directional derivative by taking the dot product of the gradient with the given unit direction vector: $$D_{\vec{u}}g(-1, -1) = \left\langle 2\pi, \pi \right\rangle \cdot \left\langle\frac{5}{13},-\frac{12}{13}\right\rangle = 2\pi \left(\frac{5}{13}\right) + \pi \left(-\frac{12}{13}\right) = -\frac{7\pi}{13}$$ Thus, the directional derivative of the given function \(g(x, y) = \sin \left(\pi(2x - y)\right)\) at point \(P(-1, -1)\) in the direction of the vector \(\left\langle\frac{5}{13},-\frac{12}{13}\right\rangle\) is \(-\frac{7\pi}{13}\).

Key Concepts

Partial Derivatives

Partial derivatives address how functions change as we tweak one of their variables while keeping others constant. If you think of a function as a landscape of peaks and valleys, then partial derivatives show the slope or grade of the landscape when moving along parallel lines to each axis.

• With Respect to x: For any function, like our example with \(g(x, y) = \sin(\pi(2x - y))\), the partial derivative with respect to \(x\) is calculated by treating \(y\) as a constant and differentiating the function like it's a single-variable function.
• With Respect to y: Similarly, the partial derivative with respect to \(y\) is found by treating \(x\) as a constant and differentiating accordingly.

In our exercise, these were calculated as:

• \( \frac{\partial g}{\partial x} = 2\pi \cos(\pi(2x - y)) \)
• \( \frac{\partial g}{\partial y} = -\pi \cos(\pi(2x - y)) \)

Understanding partial derivatives helps us comprehend the behavior of multivariable functions as each variable changes independently.

Gradient Vector

The gradient vector is an intelligent tool that combines all partial derivatives into one vector. It provides critical information about how a function behaves in multiple dimensions.

The gradient is noted as \(abla g\) and consists of each partial derivative. For our example function, the gradient was:\[ abla g = \left\langle 2\pi \cos(\pi(2x - y)), -\pi \cos(\pi(2x - y)) \right\rangle \]

• Useful Insights: The gradient points in the direction of the steepest increase of the function.
• At a Specific Point: By evaluating the gradient at a specific point, such as point \(P(-1, -1)\), we discern the local behavior of the function there. The calculated gradient at that point was \(\langle 2\pi, \pi \rangle\).

The gradient is immensely handy for optimizing functions and comprehending their directional changes.

Unit Vector

A unit vector is a vector with a magnitude of one. It’s used to specify a direction without altering the size of the function response.

In the context of directional derivatives, the unit vector tells us exactly which direction we are analyzing the slope without enhancing or minimizing it.

• Calculating Magnitude: The magnitude of a vector \(\langle a, b \rangle\) is found using \(\sqrt{a^2 + b^2}\).
• Normalization: If a given vector is not a unit vector, we can divide each component of the vector by its magnitude to convert it into a unit vector. However, in our exercise, \(\langle\frac{5}{13}, -\frac{12}{13}\rangle\) was already a unit vector. Its magnitude is 1, confirming it as a unit vector, simplifying our computations.

Utilizing unit vectors ensures we are not swaying our results by the vector's size, focusing purely on direction.

Dot Product

The dot product is an algebraic operation that combines two vectors to produce a scalar (single number). It is vital in numerous fields, including physics and engineering, for establishing angles and projections.

It can be visualized as measuring how much one vector points in the direction of another vector. In our exercise, it combines the gradient and the unit vector to enable us to calculate the directional derivative.

• Calculation: To calculate the dot product of two vectors \(\langle a, b \rangle\) and \(\langle c, d \rangle\), use \(a \cdot c + b \cdot d\).
• Directional Derivative: In our context, the dot product between the gradient \(\langle 2\pi, \pi \rangle\) and the unit vector \(\langle \frac{5}{13}, -\frac{12}{13}\rangle\) was \(D_{\vec{u}}g(-1, -1) = -\frac{7\pi}{13}\). This result is the directional derivative, a real number, representing the rate of change of our function \(g\) at \(P\) in the direction of the given unit vector.

The dot product is an essential step in finding out how a function adjusts in any given direction.