Problem 25 - Calculus for Biology and Medicine

Question

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

Step-by-Step Solution

Answer

The unit vector normal to the curve at \((1, 3)\) is \( \left( \frac{2}{\sqrt{733}}, \frac{-27}{\sqrt{733}} \right) \).

Step 1: Find the Gradient Vector

The gradient vector, denoted as \( abla f \), provides a direction that is normal to the level curves of the function. To compute it, differentiate the function \( f(x, y) = x^2 - y^3 \) with respect to \( x \) and \( y \). The partial derivative with respect to \( x \) is \( \frac{\partial f}{\partial x} = 2x \), and with respect to \( y \) is \( \frac{\partial f}{\partial y} = -3y^2 \). Therefore, the gradient vector is \( abla f = (2x, -3y^2) \).

Step 2: Evaluate the Gradient Vector at the Point

We need to evaluate \( abla f \) at the given point \((1, 3)\). Substitute \( x = 1 \) and \( y = 3 \) into the gradient vector: \[ abla f(1, 3) = (2 \cdot 1, -3 \cdot 3^2) = (2, -27) \]. Thus, the gradient vector at the point \((1, 3)\) is \((2, -27)\).

Step 3: Compute the Magnitude of the Gradient Vector

The magnitude \( \| abla f \| \) of the gradient vector \((2, -27)\) is found using the formula: \[ \| abla f \| = \sqrt{2^2 + (-27)^2} \].Calculate it as \( \sqrt{4 + 729} = \sqrt{733} \).

Step 4: Obtain the Unit Normal Vector

To find the unit vector normal to the level curve at \((1, 3)\), divide the gradient vector by its magnitude. The unit vector \( \mathbf{u} \) is given by: \[ \mathbf{u} = \left( \frac{2}{\sqrt{733}}, \frac{-27}{\sqrt{733}} \right) \].This is a unit vector normal to the level curve at the point \((1, 3)\).

Key Concepts

Gradient Vector

The gradient vector is a fundamental concept in calculus that helps us understand how a function changes as we move through its input space.
In the context of a multivariable function like \( f(x, y) \), the gradient vector, denoted as \( abla f \), points in the direction of the greatest rate of increase of the function. This characteristic makes it perpendicular, or normal, to the level curves or surfaces of the function.
To find the gradient of a function \( f(x, y) = x^2 - y^3 \), we compute the partial derivatives with respect to \( x \) and \( y \).

• The partial derivative with respect to \( x \), denoted \( \frac{\partial f}{\partial x} \), is obtained by differentiating the function while treating \( y \) as a constant. For our function, \( \frac{\partial f}{\partial x} = 2x \).
• Similarly, the partial derivative with respect to \( y \), \( \frac{\partial f}{\partial y} \), is found by differentiating with respect to \( y \) while keeping \( x \) constant, resulting in \( \frac{\partial f}{\partial y} = -3y^2 \).

Thus, the gradient vector is expressed as \( abla f = (2x, -3y^2) \).
By evaluating it at a specific point, like \((1, 3)\), we find the vector \((2, -27)\), indicating the direction and steepness of the slope of \( f \) at that point.

Partial Derivatives

Partial derivatives are used to understand the behavior of functions with multiple variables by examining how a function changes as one variable changes while keeping others constant.
These derivatives are crucial when dealing with multivariable calculus and are especially helpful in determining the gradient vector.
In a function \( f(x, y) \), the partial derivative with respect to \( x \), \( \frac{\partial f}{\partial x} \), shows how \( f \) changes as \( x \) changes.

• For \( f(x, y) = x^2 - y^3 \), the partial derivative with respect to \( x \) is \( 2x \), indicating that the output of \( f \) changes proportionally to \( 2x \) when \( x \) changes by a small amount.
• Similarly, the partial derivative with respect to \( y \), \( \frac{\partial f}{\partial y} = -3y^2 \), indicates that changing \( y \) affects \( f \) proportionally to \(-3y^2\).

Understanding these derivatives helps in interpreting how different parts of a multivariable function contribute to its overall behavior. This process aids in visualizing surfaces and curves represented by these functions.

Unit Vector

A unit vector is a vector of length one, which is particularly useful in mathematics as it provides direction without concern for magnitude.
To transform any vector into a unit vector, we divide by its magnitude.
In the problem of finding a unit vector normal to a level curve, like the function \( f(x, y) = x^2 - y^3 \), we start with its gradient vector.
After determining the gradient vector \( (2, -27) \) at the point \( (1, 3) \), we recognize it as being normal to the level curve.

• The magnitude of the gradient vector is calculated as \( \| abla f \| = \sqrt{2^2 + (-27)^2} = \sqrt{733} \).
• Dividing the components of the gradient by this magnitude gives the unit normal vector \( \mathbf{u} = \left( \frac{2}{\sqrt{733}}, \frac{-27}{\sqrt{733}} \right) \).

This unit vector effectively represents the direction normal to the level curve at the specified point, aiding in geometrical interpretations and further applications in physics and engineering.