Problem 32 - Calculus for Biology and Medicine

Question

Compute the directional derivative of \(f(x, y)\) at the point \(P\) in the direction of the point \(Q\). $$ f(x, y)=4 x y+y^{2}, P=(-1,1), Q=(3,2) $$

Step-by-Step Solution

Answer

The directional derivative is \(\frac{14}{\sqrt{17}}\).

Step 1: Find the Gradient of the Function

The first step is to find the gradient of the function \(f(x, y)\), which is the vector of its first partial derivatives. The gradient \(abla f\) is given by:\[ abla f(x, y) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \].First, compute \(\frac{\partial f}{\partial x} = 4y\).Next, compute \(\frac{\partial f}{\partial y} = 4x + 2y\).Thus, the gradient is \(abla f(x, y) = (4y, 4x + 2y)\).

Step 2: Evaluate the Gradient at Point P

Now, substitute the coordinates of point \(P\) into the gradient. Given \(P = (-1, 1)\), evaluate \(abla f(-1, 1)\):\[ abla f(-1, 1) = (4 \cdot 1, 4 \cdot (-1) + 2 \cdot 1) = (4, -4 + 2) = (4, -2) \].Thus, the gradient at point \(P\) is \((4, -2)\).

Step 3: Find the Direction Vector

The direction vector \(\mathbf{d}\) is determined by the difference between the coordinates of point \(Q\) and point \(P\). Since \(Q = (3, 2)\) and \(P = (-1, 1)\), compute the direction vector:\[ \mathbf{d} = (3 - (-1), 2 - 1) = (4, 1) \].

Step 4: Normalize the Direction Vector

To find the unit vector in the direction of \(\mathbf{d}\), normalize the vector. The magnitude \(\|\mathbf{d}\|\) is:\[ \|\mathbf{d}\| = \sqrt{4^2 + 1^2} = \sqrt{16 + 1} = \sqrt{17} \].The unit vector \(\mathbf{u}\) is:\[ \mathbf{u} = \left( \frac{4}{\sqrt{17}}, \frac{1}{\sqrt{17}} \right) \].

Step 5: Compute the Directional Derivative

The directional derivative of \(f\) at \(P\) in direction of \(\mathbf{u}\) is given by the dot product of \(abla f(P)\) and \(\mathbf{u}\):\[ D_{\mathbf{u}}f = abla f(P) \cdot \mathbf{u} = (4, -2) \cdot \left( \frac{4}{\sqrt{17}}, \frac{1}{\sqrt{17}} \right) \].Calculate the dot product:\[ D_{\mathbf{u}}f = 4 \cdot \frac{4}{\sqrt{17}} + (-2) \cdot \frac{1}{\sqrt{17}} = \frac{16}{\sqrt{17}} - \frac{2}{\sqrt{17}} = \frac{14}{\sqrt{17}} \].

Key Concepts

Gradient

In mathematics, the gradient vector is a crucial concept when dealing with functions of multiple variables. Imagine the gradient as a kind of compass showing the direction of the steepest increase of the function. For a function like \(f(x, y)\), the gradient \(abla f\) is found by computing the partial derivatives for each variable involved.

The partial derivative with respect to \(x\) measures how the function changes as \(x\) changes while keeping \(y\) constant. Similarly, the partial derivative with respect to \(y\) measures the change of the function with changes in \(y\), holding \(x\) fixed.

Thus, the gradient is expressed as:

• \( abla f(x, y) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \)

For \(f(x, y)=4xy+y^2\), we compute the partial derivatives:

• \(\frac{\partial f}{\partial x} = 4y\)
• \(\frac{\partial f}{\partial y} = 4x + 2y\)

This results in a gradient vector of \(abla f(x, y) = (4y, 4x + 2y)\), telling us how the function changes with small movements in \(x\) and \(y\).

Partial Derivatives

Partial derivatives are essential in finding the gradient of a function. They represent the rate of change of the function as one variable changes, with all other variables held constant.

Consider the function \(f(x, y) = 4xy + y^2\). To compute its partial derivatives, you treat each variable in turn as a single variable function.

• For \(x\): \(\frac{\partial f}{\partial x} = 4y\) – here, \(y\) is treated like a constant.
• For \(y\): \(\frac{\partial f}{\partial y} = 4x + 2y\) – here, \(x\) is treated like a constant.

These derivatives are steps toward understanding how a multi-variable function behaves as each variable is tweaked slightly. They come together to form the gradient, which is particularly useful for assessing how small changes impact the function.

Unit Vector

A unit vector is crucial in defining directions in mathematical problems, like finding the directional derivative. It is a vector with a magnitude of 1, indicating a specific direction without scaling it.

To convert any vector into a unit vector, you divide it by its magnitude. For instance, if you have a vector \(\mathbf{d} = (4, 1)\):

• First, calculate the magnitude: \(\| \mathbf{d} \| = \sqrt{4^2 + 1^2} = \sqrt{17}\).
• Then, divide each component by this magnitude to get the unit vector: \(\mathbf{u} = \left( \frac{4}{\sqrt{17}}, \frac{1}{\sqrt{17}} \right)\).

This transformation gives a vector measuring exactly 1 unit long, ensuring we only influence direction - not distance. In context, a unit vector helps in calculating directional derivatives, indicating how much the function changes in a specified direction.

Dot Product

The dot product is a central operation in vector algebra, especially when finding directional derivatives. It connects vectors by projecting one onto another, allowing us to measure both the magnitude and direction of influence.

A simple formula guides its calculation for two vectors \(\mathbf{a} = (a_1, a_2)\) and \(\mathbf{b} = (b_1, b_2)\):

• \( \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 \)

This operation results in a single number, indicating the extent of overlap between the vectors in terms of direction.

In our example, calculating the directional derivative of the function entails finding the dot product between the gradient at point \(P\), \((4, -2)\), and the unit vector \(\left( \frac{4}{\sqrt{17}}, \frac{1}{\sqrt{17}} \right)\).

• This results in: \( D_{\mathbf{u}}f = 4 \cdot \frac{4}{\sqrt{17}} + (-2) \cdot \frac{1}{\sqrt{17}} = \frac{16}{\sqrt{17}} - \frac{2}{\sqrt{17}} = \frac{14}{\sqrt{17}} \).

This expression tells us exactly how the function value changes as we move in the specified direction from point \(P\).