Question

In Exercises \(21-24,\) find \(d w / d t\) using the appropriate Chain Rule. \(\begin{aligned} w &=x^{2}+y^{2} \\ x &=e^{t}, \quad y=e^{-t} \end{aligned}\)

Short answer

The derivative of \( w \) with respect to \( t \) is \( \frac{dw}{dt} = 2e^{2t} - 2e^{-2t}\).

Step-by-step solution

Find the Partial Derivatives of \( w \) with respect to \( x \) and \( y \)

The partial derivative of \( w \) with respect to \( x \) is given by \( \frac{\partial w}{\partial x} = 2x \). Similarly, the partial derivative of \( w \) with respect to \( y \) is given by \( \frac{\partial w}{\partial y} = 2y \).

Find the Derivatives of \( x \) and \( y \) with respect to \( t \)

The derivative of \( x \) with respect to \( t \) is given by \( \frac{dx}{dt} = e^t \). Similarly, the derivative of \( y \) with respect to \( t \) is given by \( \frac{dy}{dt} = -e^{-t} \).

Apply the Chain Rule

Now we can apply the chain rule to find the derivative of \( w \) with respect to \( t \). The chain rule states that \( \frac{dw}{dt} = \frac{\partial w}{\partial x}\frac{dx}{dt} + \frac{\partial w}{\partial y}\frac{dy}{dt} \). Substituting from Steps 1 and 2, we get \( \frac{dw}{dt} = 2xe^{t} + 2y(-e^{-t})\). Now, substituting \( x = e^t \) and \( y = e^{-t} \) into this equation, we get \( \frac{dw}{dt} = 2e^{2t} - 2e^{-2t}\).

Key concepts

Partial Derivatives

Understanding partial derivatives is crucial when dealing with functions of multiple variables. Imagine you're at a smoothie shop where you can add various ingredients to your smoothie. The change in the taste of your smoothie when adding a little more of one ingredient, keeping all others constant, is like taking a partial derivative. Mathematically, it measures the rate at which a function changes as one variable moves infinitesimally, while all other variables are held constant.

In our exercise, the function for w is like the smoothie, affected by both x and y which are the ingredients. By finding the partial derivatives, \( \frac{\partial w}{\partial x} = 2x \) and \( \frac{\partial w}{\partial y} = 2y \) as done in the first step, we determine how w changes as either x or y changes, while the other remains constant.

Exponential Functions

Let's dive into the realm of exponential functions, which are vital in understanding growth and decay processes; they're like the Interest Growth in a bank account over time. An exponential function has the general form \( f(x) = a^{x} \) where \( a \) is a constant and \( x \) is the exponent. These functions have the unique property where the rate of growth is directly proportional to the current value, generating a rapid increase or decrease.

In our context, the variables x and y from the exercise, defined as \( x = e^{t} \) and \( y = e^{-t} \) respectively, are exponential functions with base \( e \) (Euler's number). This underlines the significance of \( e \) in natural growth processes. These functions demonstrate how something increases (or decreases) at a rate that gets faster (or slower) over time.

Derivative with Respect to Time

The concept of taking a derivative with respect to time can be likened to tracking the speed of a runner throughout a race. It's not just about how fast the runner is going at a single moment; it's about how their velocity changes as the race progresses.

When a function depends on a variable that represents time (t), the derivative with respect to t reveals the instantaneous rate of change as time moves forward. For instance, in the exercise, derivatives \( \frac{dx}{dt} = e^{t} \) and \( \frac{dy}{dt} = -e^{-t} \) tell us how quickly x and y are changing at any given moment in time. The importance of this concept is profound, especially in physics and engineering where time-dependent changes are constantly analyzed.