Problem 3 - Calculus for Biology and Medicine

Question

3\. Let \(f(x, y)=\sqrt{x^{2}+y^{2}}\) with \(x(t)=t\) and \(y(t)=\sin t\). Find the derivative of \(w=f(x, y)\) with respect to \(t\) when \(t=\pi / 3\).

Step-by-Step Solution

Answer

The derivative \( \frac{dw}{dt} \) at \( t = \pi/3 \) is approximately calculated by substituting the expressions into the formula given above.

Step 1: Understand the Problem

We need to find the derivative of the function \(w = f(x, y) = \sqrt{x^2 + y^2}\) with respect to \(t\). Here, \(x(t) = t\) and \(y(t) = \sin t\). The task is to calculate \(\frac{dw}{dt}\) at \(t = \frac{\pi}{3}\).

Step 2: Apply the Chain Rule

The chain rule for functions of several variables states that \(\frac{dw}{dt} = \frac{\partial f}{\partial x} \cdot \frac{dx}{dt} + \frac{\partial f}{\partial y} \cdot \frac{dy}{dt}\). We'll need to find \(\frac{\partial f}{\partial x}\), \(\frac{\partial f}{\partial y}\), \(\frac{dx}{dt}\), and \(\frac{dy}{dt}\).

Step 3: Compute Partial Derivatives and Time Derivatives

First, compute partial derivatives of \(f\):- \(\frac{\partial f}{\partial x} = \frac{x}{\sqrt{x^2 + y^2}}\),- \(\frac{\partial f}{\partial y} = \frac{y}{\sqrt{x^2 + y^2}}\).Second, compute the time derivatives:- \(\frac{dx}{dt} = 1\),- \(\frac{dy}{dt} = \cos t\).

Step 4: Substitute and Simplify

Substitute the derivatives into the chain rule formula:\[ \frac{dw}{dt} = \frac{x}{\sqrt{x^2 + y^2}} \cdot 1 + \frac{y}{\sqrt{x^2 + y^2}} \cdot \cos t \]This simplifies to:\[ \frac{dw}{dt} = \frac{x + y \cos t}{\sqrt{x^2 + y^2}} \]

Step 5: Evaluate at t = \(\frac{\pi}{3}\)

Substitute \(t = \frac{\pi}{3}\) into the expressions for \(x\) and \(y\):\- \(x = \frac{\pi}{3}\)- \(y = \sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}\)Substitute these into the derivative:\[ \frac{dw}{dt} = \frac{\frac{\pi}{3} + \frac{\sqrt{3}}{2} \cdot \frac{1}{2}}{\sqrt{\left(\frac{\pi}{3}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2}} \]

Step 6: Simplify the Expression Numerically

Numerically simplify the expression above:- \( \frac{dw}{dt} = \frac{\frac{\pi}{3} + \frac{\sqrt{3}}{4}}{\sqrt{\left(\frac{\pi}{3}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2}} \).Calculate under the square root:- \((\frac{\pi}{3})^2 = \frac{\pi^2}{9}\) and \((\frac{\sqrt{3}}{2})^2 = \frac{3}{4}\).Summing these gives a denominator of \(\sqrt{\frac{\pi^2}{9} + \frac{3}{4}}\).Finally, compute this to find \( \frac{dw}{dt} \) numerically.

Key Concepts

Partial Differentiation

Partial differentiation is a technique used when dealing with functions of multiple variables. This allows us to focus on how the function changes with respect to one variable, while keeping others constant. In the case of our exercise, the function is \(f(x, y) = \sqrt{x^2 + y^2}\). When we partially differentiate \(f\) with respect to \(x\), we obtain \(\frac{\partial f}{\partial x} = \frac{x}{\sqrt{x^2 + y^2}}\). Similarly, for \(y\), we have \(\frac{\partial f}{\partial y} = \frac{y}{\sqrt{x^2 + y^2}}\). These are essential steps in finding the derivative of \(w\) with respect to \(t\) using the chain rule.

• Key Point: Partial derivatives tell us the rate of change of the function as one variable changes while others remain constant.
• Application: This is useful in scenarios involving multiple factors, such as physics problems with several moving parts.

The ability to compute partial derivatives is a foundational skill in multivariable calculus, as it sets the stage for more complex operations like the chain rule used in this exercise.

Parametric Equations

Parametric equations allow us to express a set of related quantities as functions of an independent variable, usually \(t\). In this exercise, \(x(t) = t\) and \(y(t) = \sin t\). These equations provide a way to describe curves in a plane, or even more complex shapes in space.

• Utility: Parametric equations are valuable because they can model various phenomena, including time-varying physical systems.
• Example: Consider the motion of a particle: its path could be explained in terms of time, which impacts both its x- and y-coordinates.

With these equations, we can compute their time derivatives, i.e., \(\frac{dx}{dt} = 1\) and \(\frac{dy}{dt} = \cos t\), which are crucial for applying the chain rule effectively in determining \(\frac{dw}{dt}\). This transformational understanding offers flexibility in how functions and curves are analyzed, allowing for a richer exploration of calculus in practice.

Trigonometric Functions

Trigonometric functions are essential in calculus, especially when motions or oscillations are involved, such as in our problem with \(y(t) = \sin t\). These functions describe relationships between angles and sides in a right triangle, but they also model cycles and oscillations, like waves.

• Functionality: In this problem, \(\sin t\) describes the vertical position of a point as \(t\) changes, key for determining \(\frac{dy}{dt}\).
• Characteristics: With periodicity being a hallmark, \(\sin t\) performs a complete cycle every \(2\pi\) units. This characteristic makes it versatile in describing repetitive behaviors.

The cosine function, the derivative of sine, is equally crucial. Here, \(\frac{dy}{dt} = \cos t\), highlights how the slope of the \(\sin\) function evolves, guiding the rate of change for \(y(t)\). Understanding such trigonometric relationships enables us to navigate and solve more intricate calculus problems with confidence.