Problem 55 - Calculus and its applications

Question

Find \(\frac{d y}{d t} \text { if } y=\frac{1}{u^{2}+u}\) and \(u=5+3 t\)

Step-by-Step Solution

Answer

The derivative \( \frac{dy}{dt} = -3 \frac{2u + 1}{(u^2 + u)^2} \) with \( u = 5 + 3t \).

Step 1: Differentiate y with respect to u

Given the function \( y = \frac{1}{u^2 + u} \), first find the derivative of \( y \) with respect to \( u \). We use the chain rule and the formula for differentiating a function of the form \( y = \frac{1}{g(u)} \), which is \( \frac{dy}{du} = -\frac{g'(u)}{(g(u))^2} \). Here, \( g(u) = u^2 + u \) so \( g'(u) = 2u + 1 \). Therefore, \( \frac{dy}{du} = -\frac{2u + 1}{(u^2 + u)^2} \).

Step 2: Differentiate u with respect to t

Given \( u = 5 + 3t \), differentiate \( u \) with respect to \( t \). The derivative is \( \frac{du}{dt} = 3 \).

Step 3: Apply the Chain Rule to find \( \frac{dy}{dt} \)

Using the chain rule, \( \frac{dy}{dt} = \frac{dy}{du} \cdot \frac{du}{dt} \). We previously found \( \frac{dy}{du} = -\frac{2u + 1}{(u^2 + u)^2} \) and \( \frac{du}{dt} = 3 \). Thus, substitute these values to get \( \frac{dy}{dt} = -\frac{2u + 1}{(u^2 + u)^2} \cdot 3 \).

Step 4: Substitute u back in terms of t

We know \( u = 5 + 3t \). Substitute this back into the expression for \( \frac{dy}{dt} \). Thus, \( \frac{dy}{dt} = -3 \frac{2(5 + 3t) + 1}{((5 + 3t)^2 + (5 + 3t))^2} \). Simplify to get the derivative in terms of \( t \).

Key Concepts

Chain Rule

The Chain Rule is a crucial tool in calculus for differentiating composite functions, which are functions formed by combining two or more functions.
Imagine you have a function within another, like nesting Russian dolls. The chain rule helps unravel these layers to find the derivative.

When you use the chain rule, you're actually combining different rates of change. Here's how it works:

• Identify the outer function and the inner function;
• Differentiation begins with the outer function, treating the inner function as a constant;
• Then, differentiate the inner function;
• The final derivative is obtained by multiplying these derivatives together.

In our exercise, we have a composition of functions where one function depends on another, such as the expression for y depending on u, which itself depends on t. The chain rule lets you figure out how y changes with t because it lets you traverse the path from y to u and then from u to t. This linked chain is crucial when dealing with real-world problems where variables depend on each other.

Function of a Function

A function of a function, also known as a composite function, represents situations where a function, let's call it f(g(x)), is created by applying one function to the results of another.
This type of function is handled effectively using the chain rule, as described.

In the problem, y is a function of u, which in turn is a function of t. These nested relationships mean that to find dy/dt, you must first compute dy/du and du/dt separately.

• First, treat g(u) = u^2 + u as a separate entity and differentiate it to get g'(u).
• This yields dy/du using the quotient rule to differentiate 1/(u^2 + u).
• Next, differentiate u = 5 + 3t directly to get du/dt.

By this method, you're differentiating at every stage of dependency among the functions.

Rates of Change

In mathematics, particularly in calculus, rates of change describe how one quantity changes in relation to another. This concept is central because it forms the basis of understanding the world in terms of change and motion.

In our exercise, we are interested in understanding how y changes as t changes. By conceptualizing each derivative we find, as a tiny step in a bigger journey, it becomes easier to see how connected everything is:

• dy/du tells us how y changes with u;
• du/dt informs us about how u changes with t;
• Combining these, dy/dt reveals how y depends on t through u.

The overall aim is to understand the direct link between the independent variable t and the dependent variable y. This kind of relationship is fundamental for applications in physics, economics, and beyond—anywhere change occurs.