Problem 18 - 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 \(Q .\) $$ f(x, y)=e^{x-y}, P=(2,2), Q=(1,-1) $$

Step-by-Step Solution

Answer

The directional derivative at point \( P = (2, 2) \) in the direction of point \( Q = (1, -1) \) is \( \frac{\sqrt{10}}{5} \).

Step 1: Find the Gradient of f

The gradient of a function \( f(x, y) \) is given by \( abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \). For \( f(x, y) = e^{x-y} \), compute the partial derivatives: \( \frac{\partial f}{\partial x} = e^{x-y} \) and \( \frac{\partial f}{\partial y} = -e^{x-y} \). Therefore, \( abla f = (e^{x-y}, -e^{x-y}) \).

Step 2: Evaluate the Gradient at Point P

Substitute \( x = 2 \) and \( y = 2 \) into the gradient \( abla f \). This gives \( abla f(2,2) = (e^{2-2}, -e^{2-2}) = (1, -1) \).

Step 3: Determine the Direction Vector

The direction vector from point \( P \) to point \( Q \) is calculated as \( \overrightarrow{PQ} = (1 - 2, -1 - 2) = (-1, -3) \).

Step 4: Normalize the Direction Vector

The direction vector must be a unit vector for calculating the directional derivative. The magnitude of \( \overrightarrow{PQ} = (-1, -3) \) is \( \sqrt{(-1)^2 + (-3)^2} = \sqrt{10} \). The unit vector is \( \frac{1}{\sqrt{10}}(-1, -3) = \left(-\frac{1}{\sqrt{10}}, -\frac{3}{\sqrt{10}}\right) \).

Step 5: Compute the Directional Derivative

The directional derivative \( D_u f(P) \) is found by taking the dot product of \( abla f(2,2) \) and the unit vector \( \left(-\frac{1}{\sqrt{10}}, -\frac{3}{\sqrt{10}}\right) \). Therefore, \( D_u f(2,2) = (1, -1) \cdot \left(-\frac{1}{\sqrt{10}}, -\frac{3}{\sqrt{10}}\right) = -\frac{1}{\sqrt{10}} + \frac{3}{\sqrt{10}} = \frac{2}{\sqrt{10}} \).

Step 6: Simplify the Result

The expression \( \frac{2}{\sqrt{10}} \) can be simplified by rationalizing the denominator: \( \frac{2}{\sqrt{10}} \times \frac{\sqrt{10}}{\sqrt{10}} = \frac{2\sqrt{10}}{10} = \frac{\sqrt{10}}{5} \).

Key Concepts

Gradient

When dealing with multivariable functions, the concept of the gradient becomes crucial. The gradient is a vector that points in the direction of the greatest increase of a function. It is denoted by \( abla f \) and is comprised of the partial derivatives of the function with respect to each variable. The gradient of a function \( f(x, y) \) is expressed as:\[abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right)\]If you're considering a function like \( f(x, y) = e^{x-y} \), the gradient would be:

• \( \frac{\partial f}{\partial x} = e^{x-y} \)
• \( \frac{\partial f}{\partial y} = -e^{x-y} \)

Thus, the gradient is \( abla f = (e^{x-y}, -e^{x-y}) \). This gradient vector not only tells you the direction of the greatest ascent but its magnitude gives you the rate of change in that direction.

Partial Derivatives

Partial derivatives are a core component of multivariable calculus. They offer insights into how functions change as one specific variable changes, while keeping the other variables constant.When you have a function like \( f(x, y) \), the partial derivative with respect to \( x \) means examining how \( f \) changes as \( x \) changes, with \( y \) held constant. Similarly, finding the partial derivative with respect to \( y \) looks at changes in \( f \) as \( y \) varies, holding \( x \) still.For example:- Given \( f(x, y) = e^{x-y} \): - The partial derivative with respect to \( x \) is \( \frac{\partial f}{\partial x} = e^{x-y} \). - The partial derivative with respect to \( y \) is \( \frac{\partial f}{\partial y} = -e^{x-y} \).By analyzing these derivatives, you understand the rate at which each variable influences the function when other variables don't change. This concept is crucial in fields like economics and physics, where understanding variable interactions is key.

Unit Vector

A unit vector is a vector with a magnitude of one. In the context of directional derivatives, the unit vector is crucial because it ensures that the derivative is measured consistently, without any scaling by the vector's length. Let's say you have a direction vector \( \overrightarrow{PQ} \). To convert it into a unit vector, you find its magnitude and then divide each component of the vector by this magnitude.For example, if your direction vector is \( (-1, -3) \):

• Calculate its magnitude: \( \sqrt{(-1)^2 + (-3)^2} = \sqrt{10} \)
• Create the unit vector: \( \left(-\frac{1}{\sqrt{10}}, -\frac{3}{\sqrt{10}}\right) \)

With a unit vector, you ensure that any directional calculations, such as directional derivatives, are normalized and not influenced by the arbitrary length of the direction vector.

Direction Vector

The concept of a direction vector is fundamental when you want to determine how a function changes not just randomly, but in a specific direction. It essentially points from one point to another in space, indicating the path along which you wish to examine a function's change.For instance, if you start at point \( P \) and want to go toward a point \( Q \), the direction vector \( \overrightarrow{PQ} \) is calculated by subtracting the coordinates of \( P \) from \( Q \). Using points \( P(2, 2) \) and \( Q(1, -1) \):

• Calculate \( \overrightarrow{PQ} = (1 - 2, -1 - 2) = (-1, -3) \)

This direction vector doesn't necessarily have a length of one, which is why it's often normalized into a unit vector for calculations such as directional derivatives. This ensures a consistent measure of how much a function is changing in a particular direction.