Problem 3 - Calculus for Biology and Medicine

Question

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 of \(w\) with respect to \(t\) at \(t=\pi/3\) is \(\frac{1}{\sqrt{\frac{\pi^2}{9} + \frac{3}{4}}} \cdot \left(\frac{\pi}{3} + \frac{\sqrt{3}}{4}\right)\).

Step 1: Express w as a Function of t

Given that \(x(t)=t\) and \(y(t)=\sin t\), we can write \(w = \sqrt{x(t)^2 + y(t)^2} = \sqrt{t^2 + \sin^2 t}\).

Step 2: Differentiate w with respect to t

To find \(\frac{dw}{dt}\), use the chain rule. The derivative of \(w = \sqrt{t^2 + \sin^2 t}\) with respect to \(t\) is:\[ \frac{dw}{dt} = \frac{1}{2\sqrt{t^2 + \sin^2 t}} \cdot \frac{d}{dt}(t^2 + \sin^2 t) \]

Step 3: Find the Derivative of the Inner Function

Compute \(\frac{d}{dt}(t^2 + \sin^2 t)\):- The derivative of \(t^2\) with respect to \(t\) is \(2t\).- The derivative of \(\sin^2 t\) with respect to \(t\) is \(2\sin t \cdot \cos t\) by using the chain rule.Thus, \(\frac{d}{dt}(t^2 + \sin^2 t) = 2t + 2\sin t \cdot \cos t\).

Step 4: Substitute the Derivative into the Chain Rule Expression

Substitute \(\frac{d}{dt}(t^2 + \sin^2 t) = 2t + 2\sin t \cdot \cos t\) into the expression obtained from the chain rule:\[ \frac{dw}{dt} = \frac{1}{2\sqrt{t^2 + \sin^2 t}} \cdot (2t + 2\sin t \cdot \cos t) \]

Step 5: Simplify the Expression

Factor out the common factor of 2:\[ \frac{dw}{dt} = \frac{1}{\sqrt{t^2 + \sin^2 t}} \cdot (t + \sin t \cdot \cos t) \]

Step 6: Evaluate at t = \(\pi/3\)

Substitute \(t = \pi/3\) into the expression:- \(\sin(\pi/3) = \frac{\sqrt{3}}{2}\) and \(\cos(\pi/3) = \frac{1}{2}\).- So, \(t + \sin t \cdot \cos t\) becomes \(\frac{\pi}{3} + \frac{\sqrt{3}}{2} \cdot \frac{1}{2}\).- Simplify this to \(\frac{\pi}{3} + \frac{\sqrt{3}}{4}\).Find \(\sqrt{(\frac{\pi}{3})^2 + (\frac{\sqrt{3}}{2})^2}\), which is the square root of \(\frac{\pi^2}{9} + \frac{3}{4}\).

Step 7: Final Calculation

Finally, we have:\[ \frac{dw}{dt} = \frac{1}{\sqrt{\frac{\pi^2}{9} + \frac{3}{4}}} \cdot \left(\frac{\pi}{3} + \frac{\sqrt{3}}{4}\right) \]Evaluate this expression to get the derivative value when \(t = \pi/3\).

Key Concepts

Derivative

In calculus, a derivative is a way to show rate of change. It tells us how a function changes as its input changes. Imagine driving a car; the speedometer shows how the speed changes with time—that's similar to a derivative. When we are dealing with functions that depend on more than one variable, derivatives can become more complex.

In the provided exercise, the derivative we are aiming to find is of the form \( \frac{dw}{dt} \), which calculates how swiftly the function \( w=f(x,y) \) changes as the parameter \( t \) changes. However, since \( w \) is expressed in terms of two variables, \( x \) and \( y \), which themselves are dependent on \( t \), we need advanced techniques such as the chain rule to find it.

Function of Multiple Variables

A function of multiple variables takes several inputs and produces an output. It's like a machine where you put in different ingredients, and get out a product. A common scenario for functions of multiple variables is in physics, where many quantities depend on more than one factor.

For our example, the function \( f(x, y) = \sqrt{x^2 + y^2} \) calculates the distance of point \( (x, y) \) from the origin \( (0, 0) \). Here, both \( x(t) \) and \( y(t) \) depend on the single variable \( t \). This is a perfect example of a function of multiple variables, but expressed in a way that shows how each variable influences the result.

Chain Rule

The chain rule is a crucial tool in calculus for dealing with composite functions. It helps us differentiate functions in the form of \( f(g(t)) \). In simple terms, it's like peeling an onion layer by layer, until you reach the core. Start with the innermost function and work outward to find the derivative.

In our problem, the chain rule is used because \( w \) is a composite function with \( t \) affecting both \( x \) and \( y \), which in turn affect \( w \). The rule states that to differentiate \( w = f(x,y) \) with respect to \( t \), you take the derivative of \( f \) with respect to its components \( x \) and \( y \), then multiply those derivatives by \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \) respectively. Finally, you add those results. This enables us to unravel the changes effectively.

Trigonometric Functions

Trigonometric functions like \( \sin \) and \( \cos \) are essential mathematical functions that relate the angles of a triangle to the lengths of its sides. They play a significant role in oscillating phenomena, like sound and light waves.

In this exercise, the function \( y(t) = \sin t \) adds a trigonometric component to the problem. When differentiating \( \sin t \) with respect to \( t \), its derivative is \( \cos t \). Moreover, since \( \sin^2 t \) is a part of the inner function in the square root, the chain rule is applied, requiring us to know derivatives like \( \frac{d}{dt} \sin^2 t = 2\sin t \cdot \cos t \). Understanding these derivatives is vital for solving complex problems involving oscillation or periodic changes.