Problem 16 - Calculus Early Transcendentals

Question

Use Theorem 7 to find the following derivatives. When feasible, express your answer in terms of the independent variable. $$d V / d t, \text { where } V=\frac{x-y}{y+z}, x=t, y=2 t, \text { and } z=3 t$$

Step-by-Step Solution

Answer

Answer: The derivative of the function $$V$$ with respect to time, $$t$$, is $$\frac{dV}{dt} = \frac{1}{5t}$$.

Step 1: Apply the Chain Rule to find the derivative of V

To find the derivative of $$V=\frac{x-y}{y+z}$$ with respect to \(t\), we first need to write the function as a composition of two functions, let's say $$u$$ and $$v$$, such that $$V=u(v)$$. Such functions are: $$u(v) = \frac{x - y}{v}$$ $$v(t) = y + z$$ Now, let's apply the Chain Rule: $$\frac{dV}{dt} = \frac{dV}{dv} \times \frac{dv}{dt}$$ First, compute the derivative of $$u(v)$$ with respect to $$v$$: $$\frac{d(u(v))}{d(v)} = -\frac{x - y}{v^2}$$ Second, compute the derivative of $$v(t)$$ with respect to $$t$$: $$\frac{d(v(t))}{dt} = \frac{d(y+z)}{dt} = \frac{dy}{dt} + \frac{dz}{dt} = 2 + 3 = 5$$ Now, substitute these values into the Chain Rule formula: $$\frac{dV}{dt} = -\frac{x - y}{v^2} \times 5$$

Step 2: Substitute the functions of t for x, y, and z

Now that we have the derivative of $$V$$ in terms of $$v$$, we need to substitute the functions of $$t$$ for $$x$$, $$y$$, and $$z$$ to express the derivative in terms of the independent variable: \(x = t\), \(y = 2t\), and \(z = 3t\); and \(v = y + z = 2t + 3t = 5t\). Plugging these values into the previous derivative formula: $$\frac{dV}{dt} = -\frac{t - 2t}{(5t)^2} \times 5$$

Step 3: Simplify the expression to get the final answer

Finally, we will simplify the expression for the derivative of $$V$$ with respect to $$t$$: $$\frac{dV}{dt} = -\frac{-t}{25t^2} \times 5$$ $$\frac{dV}{dt} = \frac{1}{5t}$$

Key Concepts

Derivative

A derivative is a fundamental concept in calculus that represents the rate at which a function changes. In simpler terms, it's how we understand the change of one quantity relative to another. For example, in this exercise, the derivative, \( \frac{dV}{dt} \), tells us how the variable \( V \) changes as \( t \) (our independent variable) changes. Calculating derivatives helps to comprehend the behaviors of functions and is widely used in fields like physics, economics, and biology.
- The derivative gives slope or rate of change at any point of the function.
- It provides a mathematical way to describe how quickly or slowly a function's output changes based on its inputs.
The process involves applying rules of differentiation, the most common being the Chain Rule, Product Rule, and Quotient Rule, depending on the function's composition.

Function Composition

Function composition is an essential concept when dealing with derivatives, especially when the function is more complex, like \( V = \frac{x-y}{y+z} \) in this exercise. Essentially, function composition involves two or more functions working together to produce a more complex function.
- To solve this, we decompose \( V \) into simpler functions, such as \( u(v) \) and \( v(t) \), and then apply calculus rules.

• \( u(v) = \frac{x - y}{v} \)
• \( v(t) = y + z \)

By splitting the function into parts, we make it possible to apply the Chain Rule more effectively. Recognizing when to use function composition can significantly simplify solving derivative problems and is crucial for efficiently finding derivatives of composed functions.

Independent Variable

The independent variable is the input or cause in a function, in our case, \( t \). It is significant because it defines the variable with respect to which we are differentiating. Understanding the role of independent variables is crucial because:

• They dictate how we apply rules of calculus like the Chain Rule.
• Determining the relationship between the independent and dependent variables helps in finding the derivative.
  In this exercise, we express everything in terms of \( t \) because it's our independent variable: \( x = t \), \( y = 2t \), and \( z = 3t \). This uniformity allows us to systematically substitute and solve the expression. Without clear identification of independent variables, the differentiation process can become confusing or lead to errors.

Simplification

Simplification is the final yet crucial step when finding derivatives. It's all about making the derivative expression as clean and straightforward as possible. In this exercise, after applying the Chain Rule and substituting all variables in terms of \( t \), we simplified:- Original derivative: \( \frac{dV}{dt} = -\frac{-t}{25t^2} \times 5 \)- Simplified final form: \( \frac{dV}{dt} = \frac{1}{5t} \).

Simplification ensures expressions are easy to interpret and use, whether in subsequent calculations or practical applications. It involves canceling common factors and rearranging terms for clarity. Simplification not only makes results more aesthetically pleasing but also increases their usefulness in real-world scenarios and further analysis.