Problem 31 - Calculus for Biology and Medicine

Question

Compute the directional derivative of \(f(x, y)\) at the point \(P\) in the direction of the point \(Q\). $$ f(x, y)=2 x^{2} y-3 x, P=(2,1), Q=(3,2) $$

Step-by-Step Solution

Answer

The directional derivative at point \( P \) in the direction of \( Q \) is \( \frac{13}{\sqrt{2}} \).

Step 1: Compute the Gradient of f

The gradient of a function \( f(x, y) \) is a vector of its partial derivatives. First, find \( \frac{\partial f}{\partial x} \) and \( \frac{\partial f}{\partial y} \). For \( f(x, y) = 2x^2y - 3x \), the partial derivative with respect to \( x \) is \( \frac{\partial f}{\partial x} = 4xy - 3 \), and with respect to \( y \) is \( \frac{\partial f}{\partial y} = 2x^2 \). Thus, the gradient is \( abla f = \langle 4xy - 3, 2x^2 \rangle \).

Step 2: Evaluate the Gradient at Point P

Substitute the point \( P = (2, 1) \) into the gradient \( abla f \). Thus, \( abla f(P) = \langle 4(2)(1) - 3, 2(2)^2 \rangle = \langle 8 - 3, 8 \rangle = \langle 5, 8 \rangle \).

Step 3: Determine the Unit Direction Vector

Find the direction vector from \( P \) to \( Q \) by subtracting the coordinates of \( P \) from \( Q \). This gives \( \mathbf{d} = Q - P = (3 - 2, 2 - 1) = \langle 1, 1 \rangle \). To convert this to a unit vector, find its magnitude: \( \| \mathbf{d} \| = \sqrt{1^2 + 1^2} = \sqrt{2} \). The unit vector is \( \mathbf{u} = \frac{1}{\sqrt{2}} \langle 1, 1 \rangle = \langle \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \rangle \).

Step 4: Compute the Directional Derivative

The directional derivative of \( f \) at point \( P \) in the direction of vector \( \mathbf{u} \) is given by \( D_{\mathbf{u}} f = abla f(P) \cdot \mathbf{u} \). Calculate the dot product: \( abla f(P) \cdot \mathbf{u} = \langle 5, 8 \rangle \cdot \langle \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \rangle = 5\frac{1}{\sqrt{2}} + 8\frac{1}{\sqrt{2}} \). Simplifying, we get \( \frac{5+8}{\sqrt{2}} = \frac{13}{\sqrt{2}} \).

Key Concepts

Gradient

When dealing with functions involving several variables, like our example function \( f(x, y) = 2x^2y - 3x \), it's useful to understand the concept of a gradient. The gradient is essentially a vector that points in the direction of the greatest rate of increase of the function. It consists of the partial derivatives of the function.

This is because each component of the gradient represents the rate at which the function changes as we vary one of its variables, while keeping the others constant. For a function \( f(x, y) \), the gradient is expressed as \( abla f = \langle \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \rangle \). In the solution above, the computed gradient is \( \langle 4xy - 3, 2x^2 \rangle \).

• The first component, \( 4xy - 3 \), represents how the function changes with respect to \( x \).
• The second component, \( 2x^2 \), represents how the function changes with respect to \( y \).

After calculating these components, substitute the specific point \( P = (2, 1) \) to find the gradient at this location, yielding \( \langle 5, 8 \rangle \). When working through problems involving directional derivatives, **compute the gradient** first to gain insights into the direction that maximizes the function's rate of change.

Partial derivatives

Partial derivatives are fundamental in understanding how a multivariable function changes with respect to its input variables. They represent the rate of change of the function along the axis of just one variable, while all other variables are held constant.

For example, if we revisit our function \( f(x, y) = 2x^2y - 3x \):

• \( \frac{\partial f}{\partial x} = 4xy - 3 \) describes how \( f \) changes if we modify \( x \) but keep \( y \) constant.
• \( \frac{\partial f}{\partial y} = 2x^2 \) explains how \( f \) varies when \( y \) changes, maintaining \( x \) at its current value.

By learning these derivatives, you can discern which variable's changes affect the function's value the most. In real-world applications, this helps to optimize or minimize functions by understanding variable sensitivity. Remember, the partial derivatives are the building blocks of the gradient, making it critical for calculating it in multivariable scenarios.

Unit vector

Unit vectors are vectors with a magnitude of 1. They are essential in defining directions in space since they maintain only the direction characteristics, without affecting magnitude. In the context of directional derivatives, a unit vector helps in measuring the function's rate of change in a specified direction.

To find a unit vector, consider our direction vector \( \mathbf{d} = \langle 1, 1 \rangle \) from point \( P \) to \( Q \). First, calculate the magnitude of \( \mathbf{d} \), which is \( \sqrt{2} \). Then, the unit vector is \( \mathbf{u} = \langle \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \rangle \).

• This normalized vector keeps the direction intact but scales the vector to a length of 1, allowing us to focus solely on direction.
• It becomes extremely useful when computing directional derivatives since it isolates the direction of interest.

In summary, it's crucial to transform any direction vector into a unit vector when finding directional derivatives, ensuring that the measurement of change isn't inflated by vector length.