Problem 38 - Calculus for Biology and Medicine

Question

In what direction does \(f(x, y)=\ln \left(x^{2}+y^{2}\right)\) increase most rapidly at \((1,1) ?\)

Step-by-Step Solution

Answer

The function increases most rapidly in the direction \(\left( \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right)\) at \((1,1)\).

Step 1: Identify the Gradient

The direction in which a function increases most rapidly is given by its gradient. The gradient of a function \(f(x, y)\) is a vector consisting of its partial derivatives: \[abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right).\]

Step 2: Compute Partial Derivatives

For the given function \(f(x, y) = \ln(x^2 + y^2)\), compute the partial derivatives with respect to \(x\) and \(y\). - The partial derivative with respect to \(x\) is given by: \[\frac{\partial f}{\partial x} = \frac{2x}{x^2+y^2}.\]- The partial derivative with respect to \(y\) is given by: \[\frac{\partial f}{\partial y} = \frac{2y}{x^2+y^2}.\]

Step 3: Evaluate the Gradient at (1,1)

Substitute \((x, y) = (1, 1)\) into the partial derivatives to compute the gradient at this point:\[abla f(1, 1) = \left( \frac{2 \cdot 1}{1^2 + 1^2}, \frac{2 \cdot 1}{1^2 + 1^2} \right) = \left( \frac{2}{2}, \frac{2}{2} \right) = (1, 1).\]

Step 4: Determine the Direction of Most Rapid Increase

The gradient vector at \((1,1)\) is \((1, 1)\), which indicates that the function increases most rapidly in the direction of the vector \((1, 1)\) from the point \((1, 1)\). This direction should be expressed as a unit vector:\[\text{Unit vector} = \frac{1}{\sqrt{2}} (1, 1) = \left( \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right).\]

Key Concepts

Partial Derivatives

Partial derivatives are fundamental in understanding how a function changes with respect to one of its variables, while keeping the other variables constant. In the context of multivariable calculus, partial derivatives are crucial as they isolate the effect of one variable at a time.
\(\partial f/\partial x\) and \(\partial f/\partial y\) represent the rate of change of the function \(f(x, y)\) with respect to \(x\) and \(y\), respectively. These derivatives give us invaluable information about the slope or rate of change along each axis.

• The general formula to find a partial derivative with respect to \(x\) for a function \(f(x, y)\) is to differentiate \(f\) as if \(y\) is constant.
• Similarly, differentiate \(f\) with respect to \(y\) treating \(x\) as constant to get \(\partial f/\partial y\).

By calculating partial derivatives, you form the components of the gradient vector, which is essential for identifying the directions where the function changes most significantly.

Direction of Most Rapid Increase

The direction of most rapid increase in a function at a given point is described by its gradient vector at that point. The gradient vector is a vector field consisting of all the partial derivatives of a function.
For example, for the function \(f(x, y) = \ln(x^2 + y^2)\), the gradient at \((1, 1)\) is calculated as \((1, 1)\). This means the function will increase most steeply along the path in the same direction as \((1, 1)\).

• This direction provides insight into the steepest ascent, guiding us to the peak in a surface, much like climbing a hill.
• At any point on the graph of a function, following the gradient points you in the direction of that hill's steepest slope.

The calculation of the gradient involves taking the partial derivatives, which represent the rates of change along the axes, effectively summing them into a vector pointing to the steepest path.

Unit Vector

A unit vector is a vector with a magnitude of 1, used to indicate direction without considering magnitude. In many mathematical contexts, especially when dealing with directions derived from vectors, normalizing to a unit vector helps focus solely on the angle or orientation.
To convert any given vector into a unit vector, you divide each component of the vector by the vector’s magnitude (its length).
For a vector \((x, y)\), its magnitude is calculated using the formula \(\sqrt{x^2 + y^2}\).

• In our example, the gradient vector \((1, 1)\) is normalized to \(\left( \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right)\) because the magnitude of \((1, 1)\) is \(\sqrt{1^2 + 1^2} = \sqrt{2}\).

Normalizing to a unit vector ensures that we are left with a direction that one can easily apply to any calculation or representation that requires directional information without the influence of magnitude.