Problem 15 - Calculus for Biology and Medicine

Question

In Problems 15-18, 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 is \( \frac{13\sqrt{2}}{2} \).

Step 1: Find the Gradient

To find the directional derivative, we first calculate the gradient of the function \( f(x, y) = 2x^2y - 3x \). The gradient \( abla f(x, y) \) is found by taking the partial derivatives with respect to \( x \) and \( y \). 1. Partial derivative with respect to \( x \):\[ \frac{\partial f}{\partial x} = \frac{\partial}{\partial x}(2x^2y - 3x) = 4xy - 3 \]2. Partial derivative with respect to \( y \):\[ \frac{\partial f}{\partial y} = \frac{\partial}{\partial y}(2x^2y - 3x) = 2x^2 \]So, the gradient is:\[ abla f(x, y) = \langle 4xy - 3, 2x^2 \rangle \]

Step 2: Evaluate the Gradient at Point P

Now, substitute the coordinates of point \( P = (2, 1) \) into the gradient to find \( abla f \) at \( P \).\[ abla f(2, 1) = \langle 4 \times 2 \times 1 - 3, 2 \times 2^2 \rangle = \langle 8 - 3, 8 \rangle = \langle 5, 8 \rangle \]

Step 3: Determine the Direction Vector

The direction vector is determined by subtracting point \( P \) from point \( Q \). If \( Q = (3, 2) \), then the direction vector \( \mathbf{v} \) is:\[ \mathbf{v} = Q - P = \langle 3 - 2, 2 - 1 \rangle = \langle 1, 1 \rangle \]

Step 4: Normalize the Direction Vector

Normalize the direction vector to calculate the unit vector in the direction of \( Q \).The magnitude of \( \mathbf{v} \) is:\[ \| \mathbf{v} \| = \sqrt{1^2 + 1^2} = \sqrt{2} \]Thus, the unit vector \( \mathbf{u} \) in the direction of \( Q \) is:\[ \mathbf{u} = \frac{1}{\sqrt{2}} \langle 1, 1 \rangle = \langle \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \rangle \]

Step 5: Compute the Directional Derivative

The directional derivative \( D_{\mathbf{u}}f \) at point \( P \) in the direction of \( \mathbf{u} \) is given by the dot product:\[ D_{\mathbf{u}}f = abla f(2, 1) \cdot \mathbf{u} = \langle 5, 8 \rangle \cdot \langle \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \rangle \]Calculating this gives:\[ D_{\mathbf{u}}f = 5 \times \frac{1}{\sqrt{2}} + 8 \times \frac{1}{\sqrt{2}} = \frac{5}{\sqrt{2}} + \frac{8}{\sqrt{2}} = \frac{13}{\sqrt{2}} \]Rationalize the denominator, if required:\[ D_{\mathbf{u}}f = \frac{13\sqrt{2}}{2} \]

Key Concepts

Gradient

The gradient of a function is a vector that points in the direction of the greatest rate of increase of the function. When we have a function like \( f(x, y) = 2x^2y - 3x \), the gradient is a way to combine the effects of changes in both the \( x \) and \( y \) directions.
For this function, the gradient \( abla f(x, y) \) is found by computing the partial derivatives with respect to each variable. These are:

• \( \frac{\partial f}{\partial x} = 4xy - 3 \)
• \( \frac{\partial f}{\partial y} = 2x^2 \)

Therefore, the gradient is \( abla f(x, y) = \langle 4xy - 3, 2x^2 \rangle \).
The gradient can be visualized as an arrow that points in the direction of increasing function values, and its magnitude signifies the rate of increase.

Partial Derivatives

Partial derivatives are a fundamental concept used to understand multivariable functions. They describe how a function changes as one of the input variables is varied while all other variables are kept constant.
For example, for the function \( f(x, y) = 2x^2y - 3x \), the partial derivative with respect to \( x \) is: \[\frac{\partial f}{\partial x} = 4xy - 3\]
Here, \( y \) is treated as a constant, and only \( x \) is varied.
Similarly, for \( y \), the partial derivative is:\[\frac{\partial f}{\partial y} = 2x^2\]
In this case, \( x \) is treated as a constant.
Partial derivatives help in constructing the gradient vector, which is crucial for determining the Directional Derivative.

Unit Vector

A unit vector is a vector with a magnitude of 1. It essentially serves to indicate a direction without changing the magnitude of a vector.
In the context of directional derivatives, we often convert a given direction vector into a unit vector. This makes calculating the directional derivative straightforward.
For example, starting with a direction vector \( \mathbf{v} = \langle 1, 1 \rangle \) from point \( P \) to \( Q \), the magnitude of \( \mathbf{v} \) is:\[\| \mathbf{v} \| = \sqrt{1^2 + 1^2} = \sqrt{2}\]
To convert this into a unit vector, divide each component by the magnitude:

• \( \mathbf{u} = \frac{1}{\sqrt{2}} \langle 1, 1 \rangle = \langle \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \rangle \)

This helps in the calculation of the dot product for the directional derivative, maintaining direction while ensuring the vector length is 1.

Dot Product

The dot product, also known as the scalar product, is a way of multiplying two vectors to produce a scalar, or single number. It measures the extent to which two vectors are pointing in the same direction.
To compute the dot product of two vectors \( \mathbf{a} = \langle a_1, a_2 \rangle \) and \( \mathbf{b} = \langle b_1, b_2 \rangle \), you calculate:\[\mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 \]
In finding a directional derivative, the gradient vector \( abla f \) is dotted with a unit vector \( \mathbf{u} \) in the direction of choice. For example, with \( abla f(2, 1) = \langle 5, 8 \rangle \) and \( \mathbf{u} = \langle \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \rangle \) from our problem:

• \( D_{\mathbf{u}}f = 5 \times \frac{1}{\sqrt{2}} + 8 \times \frac{1}{\sqrt{2}} = \frac{13}{\sqrt{2}} \)

The final directional derivative can be expressed as a cleaner fraction by rationalizing, giving \( \frac{13\sqrt{2}}{2} \). This result shows how the rate of change of the function aligns with the chosen direction.