Problem 40 - Calculus for Biology and Medicine

Question

Find a unit vector that is normal to the level curve of the function $$f(x, y)=x^{2}+\frac{y^{2}}{9}$$ at the point \((1,3)\).

Step-by-Step Solution

Answer

The unit vector normal to the level curve at \((1,3)\) is \(\left( \frac{3\sqrt{10}}{10}, \frac{\sqrt{10}}{10} \right)\).

Step 1: Compute the Gradient Vector

The gradient vector \(abla f(x, y)\) for a function \(f(x, y) = x^2 + \frac{y^2}{9}\) is given by \(( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} )\). Compute the partial derivatives:\[ \frac{\partial f}{\partial x} = 2x, \quad \frac{\partial f}{\partial y} = \frac{2y}{9} \]Thus, the gradient vector is \(abla f(x, y) = (2x, \frac{2y}{9})\).

Step 2: Evaluate the Gradient at the Point

Substitute the point \((1, 3)\) into the gradient vector:\[ abla f(1, 3) = (2 \cdot 1, \frac{2 \cdot 3}{9}) = (2, \frac{6}{9}) = (2, \frac{2}{3}) \]

Step 3: Find the Magnitude of the Gradient Vector

To convert the gradient vector to a unit vector, first find its magnitude:\[ \|abla f(1, 3)\| = \sqrt{2^2 + \left(\frac{2}{3}\right)^2} = \sqrt{4 + \frac{4}{9}} = \sqrt{\frac{36}{9} + \frac{4}{9}} = \sqrt{\frac{40}{9}} = \frac{\sqrt{40}}{3} = \frac{2\sqrt{10}}{3} \]

Step 4: Normalize the Gradient Vector

Divide each component of the gradient vector by its magnitude to obtain the unit vector:\[ \mathbf{u} = \left( \frac{2}{\frac{2\sqrt{10}}{3}}, \frac{\frac{2}{3}}{\frac{2\sqrt{10}}{3}} \right) = \left( \frac{3}{\sqrt{10}}, \frac{1}{\sqrt{10}} \right) \]

Step 5: Simplify the Unit Vector

Rationalize the denominator to finalize the unit vector:\[ \mathbf{u} = \left( \frac{3\sqrt{10}}{10}, \frac{\sqrt{10}}{10} \right) \]

Step 6: Verify the Unit Vectors' Length

Verify that the obtained vector \(\mathbf{u}\) is indeed a unit vector by checking its length:\[ \|\mathbf{u}\| = \sqrt{\left(\frac{3\sqrt{10}}{10}\right)^2 + \left(\frac{\sqrt{10}}{10}\right)^2} = \sqrt{\frac{90}{100} + \frac{10}{100}} = \sqrt{1} = 1 \]The length is 1, confirming \(\mathbf{u} = \left( \frac{3\sqrt{10}}{10}, \frac{\sqrt{10}}{10} \right)\) is a unit vector.

Key Concepts

Partial Derivatives

Partial derivatives are fundamental in understanding how a function changes in multi-dimensional spaces. In calculus, when you have a function involving more than one variable, partial derivatives help us see how the function changes as one variable changes, while keeping the other variables constant.
In our example, the function is defined as \( f(x, y) = x^2 + \frac{y^2}{9} \). We need to find how the function changes with respect to \(x\) and \(y\):

• The partial derivative with respect to \(x\), denoted \( \frac{\partial f}{\partial x} \), considers \(y\) to be constant. The derivative results in \(2x\).
• Similarly, the partial derivative with respect to \(y\), \( \frac{\partial f}{\partial y} \), treats \(x\) as constant and evaluates to \( \frac{2y}{9} \).

Together, these partial derivatives form the gradient vector \((2x, \frac{2y}{9})\). This vector indicates the direction and rate of change of the function. Understanding and computing partial derivatives is essential for tasks like optimizing functions and analyzing changes across different dimensions.

Unit Vector

A unit vector is a vector that has a magnitude (length) of exactly 1. Unit vectors are crucial in various mathematical and physical applications because they provide a direction without affecting magnitude, acting merely as directional indicators.To find a unit vector, we often start with a given vector and modify its components by dividing each by the vector's magnitude.
In our solution, we computed the gradient vector \((2, \frac{2}{3})\) at point \((1, 3)\). The magnitude of this vector is found using the formula:\[\|abla f(1, 3)\| = \sqrt{2^2 + \left(\frac{2}{3}\right)^2} = \frac{2\sqrt{10}}{3}\]Then, each component of the vector is divided by this magnitude to achieve the unit vector:\[\mathbf{u} = \left( \frac{3}{\sqrt{10}}, \frac{1}{\sqrt{10}} \right)\]Rationalizing the components yields:\[\mathbf{u} = \left( \frac{3\sqrt{10}}{10}, \frac{\sqrt{10}}{10} \right)\]We also verify its magnitude: \(\|\mathbf{u}\| = 1\). This confirms our vector is truly a unit vector, illustrating how unit vectors denote direction effectively without changing a vector's inherent magnitude.

Level Curve

In mathematics, a level curve is a two-dimensional slice through a three-dimensional surface. It represents points on a plane where the function takes a constant value. Understanding level curves helps in visualizing how a function behaves across different regions.Given the function \( f(x, y) = x^2 + \frac{y^2}{9} \), a level curve is formed by setting \( f(x, y) = c \), where \(c\) is a constant.
For example, if \( f(x, y) = 1 \), the resulting equation describes a set of points that form an ellipse in the \(xy\)-plane.

• Level curves show the topology or shape of a surface, such as hills and valleys in physical terrain.
• They are particularly useful in optimization, as they allow for understanding where peaks, valleys, or saddle points occur.

In the context of this problem, our goal is to find a normal vector to the level curve at \((1, 3)\). This normal vector is orthogonal to the curve and is represented by the gradient vector, emphasizing the crucial role of the gradient in characterizing level curves.