Problem 26 - Calculus for Biology and Medicine

Question

Find a unit vector that is normal to the level curve of the function $$ f(x, y)=x y $$ at the point \((2,3)\).

Step-by-Step Solution

Answer

The unit vector normal to the level curve at (2, 3) is \( \left( \frac{3}{\sqrt{13}}, \frac{2}{\sqrt{13}} \right) \).

Step 1: Find the Gradient

First, we need to find the gradient, \( abla f \), of the function \( f(x, y) = xy \). The gradient gives us the direction in which the function increases most rapidly, and it is also normal to the level curve at any given point. The gradient is calculated as the vector of partial derivatives with respect to \( x \) and \( y \).\[ abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) = (y, x) \]So the gradient of \( f \) at any point \( (x, y) \) is \( (y, x) \).

Step 2: Evaluate the Gradient at the Point

Next, substitute the point \( (2, 3) \) into the gradient. This will give us the actual vector that is normal to the level curve at \( (2,3) \).\[ abla f(2,3) = (3, 2) \]

Step 3: Calculate the Magnitude of the Gradient

Now, calculate the magnitude of the gradient vector \((3, 2)\). This is necessary to convert the gradient into a unit vector.\[ \| abla f(2,3) \| = \sqrt{3^2 + 2^2} = \sqrt{9 + 4} = \sqrt{13} \]

Step 4: Normalize the Gradient to Find the Unit Vector

To find the unit vector, divide each component of the gradient by its magnitude. This gives the unit vector normal to the level curve at the point \( (2, 3) \).\[ \text{Unit vector} = \left( \frac{3}{\sqrt{13}}, \frac{2}{\sqrt{13}} \right) \]

Key Concepts

Partial Derivatives

In calculus, when dealing with functions of multiple variables, partial derivatives are used to understand the rate of change of the function with respect to one variable while keeping others constant. Suppose you have a function like \( f(x, y) = xy \). To find the partial derivative with respect to \( x \), you treat \( y \) as a constant. Consequently, the partial derivative of \( f(x, y) \) with respect to \( x \) is \( \frac{\partial f}{\partial x} = y \). Similarly, treating \( x \) as a constant, the partial derivative of \( f(x, y) \) with respect to \( y \) becomes \( \frac{\partial f}{\partial y} = x \).
These derivatives provide the basis for the gradient vector in multivariable calculus, which is the theme of the solution to this problem.

• The function's behavior changes in different directions, indicated by these derivatives.
• Partial derivatives give local approximation and are crucial for optimizing the function.

Understanding partial derivatives is the building block for compound concepts like gradients and level curves.

Level Curve

A level curve is a curve that represents all points where a function of two variables, like \( f(x, y) \), has the same constant value. In simpler terms, it’s like drawing a contour line on a topographic map showing areas of equal elevation.
For example, consider the function \( f(x, y) = xy \). A level curve is given by an equation of the form \( xy = c \), where \( c \) is a constant. Each different \( c \) results in a different level curve.
In the context of the exercise, for the point \( (2, 3) \), the level curve would be \( 2 \times 3 = 6 \), which is a curve consisting of all points \( (x, y) \) such that \( xy = 6 \).

• Level curves provide a visual and analytical way of understanding the behavior of a function in two dimensions.
• They help identify where the function has the same value across different input values.

Recognizing level curves is foundational in visualizing multivariable functions and in applications such as physics and engineering.

Unit Vector

A unit vector is a vector that has a magnitude of 1, used primarily to indicate direction rather than magnitude.
It's important in the context of gradients and level curves because the gradient vector points in the direction of the steepest ascent of a function. To focus on direction, we convert the gradient to a unit vector by dividing the gradient by its magnitude.
For instance, in the given problem, we first derived the gradient as \( (3, 2) \) at the point \( (2, 3) \). To find the unit vector, we computed the magnitude of this gradient: \[ \| abla f(2,3) \| = \sqrt{13} \] Following this, the unit vector is: \[ \left( \frac{3}{\sqrt{13}}, \frac{2}{\sqrt{13}} \right) \]

• The unit vector maintains the direction of the original vector.
• It simplifies problems involving directions into a standardized scale.

Grasping the concept of a unit vector is essential for navigation and direction-focused applications in physics and engineering.