Problem 24 - Calculus Early Transcendentals: Pearson New International Edition

Question

Given that \(f_{x}(2,4)=-3\) and \(f_{y}(2,4)=8\), find the directional derivative of \(f\) at \((2,4)\) in the direction toward \((5,0)\).

Step-by-Step Solution

Answer

The directional derivative is \(-\frac{41}{5}\).

Step 1: Understand the Direction Vector

To find the directional derivative, we need the direction vector. The point given for direction is \((5,0)\) and we want the direction from the point \((2,4)\) to \((5,0)\). The vector in the direction is \((5-2, 0-4) = (3, -4)\).

Step 2: Normalize the Direction Vector

The directional derivative requires a unit vector in the given direction. To normalize the vector \((3, -4)\), we calculate its magnitude: \( \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = 5 \). The unit vector is \( \left( \frac{3}{5}, \frac{-4}{5} \right) \).

Step 3: Use the Definition of the Directional Derivative

To find the directional derivative \( D_\mathbf{u} f \) in the direction of the unit vector \( \mathbf{u} = \left( \frac{3}{5}, \frac{-4}{5} \right) \), we apply the formula:\[D_\mathbf{u} f = f_x(2,4) \cdot \frac{3}{5} + f_y(2,4) \cdot \frac{-4}{5}\]

Step 4: Plug in Known Values

Substitute the given partial derivatives into the formula:\[D_\mathbf{u} f = (-3) \cdot \frac{3}{5} + 8 \cdot \frac{-4}{5}\]

Step 5: Perform Calculations

Carry out the multiplication and addition:\[D_\mathbf{u} f = \left(-\frac{9}{5}\right) + \left(-\frac{32}{5}\right) = \frac{-41}{5}\]

Step 6: Conclusion

The directional derivative of \(f\) at \((2,4)\) in the direction toward \((5,0)\) is \(-\frac{41}{5}\).

Key Concepts

Partial Derivatives

Partial derivatives are crucial when dealing with functions of multiple variables, like in the exercise above. Essentially, a partial derivative represents the rate at which a function changes as one of the variables is altered, while the others are kept constant. For a function \( f(x, y) \), the partial derivative \( f_x \) would measure how \( f \) changes with respect to \( x \), keeping \( y \) constant. Similarly, \( f_y \) gauges the change concerning \( y \), keeping \( x \) constant.
In our exercise, we are given the values of partial derivatives at a specific point, \( f_x(2,4) = -3 \) and \( f_y(2,4) = 8 \). These values tell us how the function \( f \) is changing horizontally (along the \( x \)-axis) and vertically (along the \( y \)-axis) at the point \( (2, 4) \). Understanding these changes helps us determine the overall directional derivative in any specified direction.

Gradient Vector

The gradient vector is a very powerful tool in multivariable calculus, particularly when working with directional derivatives. This vector combines all the partial derivatives of a function into a single vector, which essentially points in the direction of the steepest ascent of the function. For a function \( f(x, y) \), the gradient vector \( abla f \) is defined as:

• \( abla f = (f_x, f_y) \)

At the point \( (2, 4) \) in our example, the gradient vector will be \((f_x(2,4), f_y(2,4)) = (-3, 8)\). This vector embodies the direction in which the function \( f \) increases most rapidly from that point. However, when we want to find the rate of change in a different direction, as in this exercise, we must adjust this information using a unit vector that represents the desired direction.

Unit Vector

A unit vector is essentially a vector that has a magnitude of 1, providing a standardized way to indicate direction. When we use vectors to show direction, it often does not matter how long the vector is. Thus, normalizing the vector to become a unit vector is essential, especially for calculating directional derivatives.
In the problem, the direction vector from point \((2,4)\) to \((5,0)\) is found as \((3, -4)\). To convert this vector into a unit vector, we need to divide each component by its magnitude. The magnitude of the vector \((3, -4)\) is \(5\), calculated using the formula \( \sqrt{3^2 + (-4)^2} = 5 \). Thus, the unit vector becomes \( \left( \frac{3}{5}, \frac{-4}{5} \right) \).
This unit vector plays a crucial role when calculating the directional derivative, as it precisely aligns the function's change rate with the desired orientation, ensuring correct application of the formula derived from the dot product of the gradient and this unit vector.