Problem 17 - 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. $$f(x, y)=x^{2}-y^{2} ; P(-1,-3) ;\left\langle\frac{3}{5},-\frac{4}{5}\right\rangle$$

Step-by-Step Solution

Answer

Answer: The directional derivative is -6.

Step 1: Calculate the gradient of the function

First, we need to find the gradient of the function \(f(x, y) = x^2 - y^2\). The gradient is given by the partial derivatives of the function, that is, \(\nabla f(x, y) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right)\). So, let's compute the partial derivatives: $$\frac{\partial f}{\partial x} = 2x$$ $$\frac{\partial f}{\partial y} = -2y$$ Now we can form the gradient vector as follows: $$\nabla f(x, y) = \left\langle 2x, -2y \right\rangle$$

Step 2: Verify the direction vector is a unit vector

We are given the direction vector as \(\left\langle\frac{3}{5},-\frac{4}{5}\right\rangle\). It's easy to check that this is a unit vector: $$\left\|\left\langle\frac{3}{5},-\frac{4}{5}\right\rangle\right\| = \sqrt{\left(\frac{3}{5}\right)^2 + \left(-\frac{4}{5}\right)^2} = \sqrt{\frac{9}{25} + \frac{16}{25}} = \sqrt{1} = 1$$ Since its magnitude is 1, it is indeed a unit vector.

Step 3: Compute the directional derivative

Our goal is to compute the directional derivative, which is obtained by taking the dot product of the gradient of the function at point P and the unit direction vector: $$D_u f(P) = \nabla f(P) \cdot u$$ First, we'll find the gradient of the function at point P(-1, -3): $$\nabla f(-1, -3) = \left\langle 2(-1), -2(-3) \right\rangle = \left<-2, 6\right>$$ Now, let's compute the dot product: $$D_u f(P) = \left<-2, 6\right> \cdot \left\langle\frac{3}{5},-\frac{4}{5}\right\rangle = -2\left(\frac{3}{5}\right) + 6\left(-\frac{4}{5}\right) = -\frac{6}{5} - \frac{24}{5} = -\frac{30}{5} = -6$$ Therefore, the directional derivative of the function \(f(x, y) = x^2 - y^2\) at point P(-1, -3) in the direction of the vector \(\left\langle \frac{3}{5}, -\frac{4}{5} \right\rangle\) is -6.

Key Concepts

Gradient

The concept of a gradient is crucial in multivariable calculus. It allows us to determine the direction and rate of change of a function. For the function \(f(x, y) = x^2 - y^2\), the gradient, denoted as \(abla f(x, y)\), is a vector composed of the function's partial derivatives. Here, it is \(\left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right)\). This means that to form the gradient, we calculate

• \(\frac{\partial f}{\partial x} = 2x\) – the rate of change of \(f\) with respect to \(x\), holding \(y\) constant.
• \(\frac{\partial f}{\partial y} = -2y\) – the rate of change of \(f\) with respect to \(y\), holding \(x\) constant.

So, the gradient vector of the function is \(abla f(x, y) = \langle 2x, -2y \rangle\). The gradient points in the direction of the greatest rate of increase of the function and its magnitude signifies the rate of the steepest ascent.

Partial Derivatives

Partial derivatives are an extension of the concept of derivatives to functions of several variables. They provide insights into how the function changes as one particular variable changes, while keeping the others constant. In the function \(f(x, y) = x^2 - y^2\), partial derivatives are used to determine how changes in \(x\) and \(y\) affect \(f\):

• \(\frac{\partial f}{\partial x} = 2x\): This derivative gives the rate of change of \(f\) as only \(x\) changes, which shows how sensitive the function is to variations in \(x\).
• \(\frac{\partial f}{\partial y} = -2y\): This derivative shows the effect on \(f\) when only \(y\) changes. It offers a view into the impact of variations in \(y\) on the function.

By computing these derivatives, we can construct the gradient and thus evaluate the directional derivative, leading us to understand how a function behaves in multiple directions.

Unit Vector

A unit vector is a vector with a magnitude of 1, used primarily to indicate direction without influencing magnitude. In computing directional derivatives, it's essential because it ensures that only the way of direction is considered. For example, given the vector \(\left\langle \frac{3}{5}, -\frac{4}{5} \right\rangle\), to confirm it's a unit vector, compute its magnitude:

• \(\sqrt{\left(\frac{3}{5}\right)^2 + \left(-\frac{4}{5}\right)^2} = \sqrt{\frac{9}{25} + \frac{16}{25}} = \sqrt{1} = 1\).

This ensures that when using this vector in calculations such as the directional derivative, it purely represents the direction, not altering the computation scale by size. Using a unit vector is crucial for keeping calculations standardized and meaningful.

Dot Product

The dot product is a scalar result of two vectors that indicates how much one vector extends in the direction of another. In directional derivative computations, the dot product of the gradient and a unit vector gives the rate of change of the function in a specific direction. To calculate, take the example of:

• Gradient at point \(P(-1, -3): abla f(-1, -3) = \langle -2, 6 \rangle\).
• Unit direction vector: \(\left\langle \frac{3}{5}, -\frac{4}{5} \right\rangle\).

The dot product, \(abla f(P) \cdot u\), is:

• \(-2\left(\frac{3}{5}\right) + 6\left(-\frac{4}{5}\right) = -\frac{6}{5} - \frac{24}{5} = -\frac{30}{5} = -6\)

This results in the directional derivative, telling us specifically how the function \(f(x, y)\) changes at point \(P\) in the direction given by the unit vector. The dot product's importance lies in simplifying the directional variation of multivariable functions.