Find The Derivative Ds/dt If S = T / (2t + 3).

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In the realm of calculus, derivatives play a pivotal role in understanding the rate at which a function's output changes with respect to its input. This concept finds widespread applications across various fields, from physics and engineering to economics and finance. In this comprehensive guide, we will embark on a step-by-step journey to determine the derivative of a specific function, s = t / (2t + 3), with respect to t. This process, often denoted as ds/dt, will reveal the instantaneous rate of change of s as t varies.

Unveiling the Quotient Rule: A Fundamental Tool

Before we delve into the specifics of our problem, it's crucial to introduce a fundamental tool in calculus known as the quotient rule. This rule provides a systematic approach to finding the derivative of a function that is expressed as the ratio of two other functions. In mathematical terms, if we have a function s(t) defined as:

s(t) = u(t) / v(t)

where u(t) and v(t) are differentiable functions, then the derivative of s(t) with respect to t, denoted as ds/dt, can be calculated using the following formula:

ds/dt = [v(t) * (du/dt) - u(t) * (dv/dt)] / [v(t)]^2

This formula might seem intimidating at first glance, but it's simply a structured way to handle the differentiation of quotients. Let's break it down into its key components:

  • v(t): The denominator of the original function.
  • (du/dt): The derivative of the numerator with respect to t.
  • u(t): The numerator of the original function.
  • (dv/dt): The derivative of the denominator with respect to t.
  • [v(t)]^2: The square of the denominator.

Now that we have the quotient rule in our arsenal, let's apply it to our specific problem.

Applying the Quotient Rule to s = t / (2t + 3)

In our case, we have the function s(t) defined as:

s(t) = t / (2t + 3)

To apply the quotient rule, we need to identify the numerator and denominator functions. Let's define:

  • u(t) = t (the numerator)
  • v(t) = 2t + 3 (the denominator)

Next, we need to find the derivatives of u(t) and v(t) with respect to t:

  • (du/dt) = d(t)/dt = 1 (the derivative of t with respect to t is 1)
  • (dv/dt) = d(2t + 3)/dt = 2 (the derivative of 2t + 3 with respect to t is 2)

Now that we have all the necessary components, we can plug them into the quotient rule formula:

ds/dt = [v(t) * (du/dt) - u(t) * (dv/dt)] / [v(t)]^2

ds/dt = [(2t + 3) * 1 - t * 2] / (2t + 3)^2

Simplifying the Expression: Unveiling the Derivative

Our next step is to simplify the expression we obtained in the previous section. This involves performing the arithmetic operations and algebraic manipulations to arrive at a more concise and interpretable form of the derivative.

Let's start by expanding the numerator:

(2t + 3) * 1 - t * 2 = 2t + 3 - 2t

Notice that the 2t terms cancel each other out, leaving us with:

3

Now, let's focus on the denominator. We have (2t + 3)^2. We can expand this expression using the formula (a + b)^2 = a^2 + 2ab + b^2:

(2t + 3)^2 = (2t)^2 + 2 * (2t) * 3 + 3^2

(2t + 3)^2 = 4t^2 + 12t + 9

Now, let's substitute the simplified numerator and denominator back into our expression for ds/dt:

ds/dt = 3 / (4t^2 + 12t + 9)

This is the derivative of s with respect to t. It tells us how the value of s changes as t changes. We can further simplify the denominator by recognizing it as a perfect square:

4t^2 + 12t + 9 = (2t + 3)^2

Therefore, we can write the final simplified form of the derivative as:

ds/dt = 3 / (2t + 3)^2

This expression represents the instantaneous rate of change of s with respect to t. It's a valuable tool for understanding the behavior of the function s(t) = t / (2t + 3).

Interpreting the Derivative: Understanding the Rate of Change

The derivative we found, ds/dt = 3 / (2t + 3)^2, provides us with crucial insights into how the function s(t) = t / (2t + 3) changes as t varies. Let's delve deeper into the interpretation of this derivative.

  • The Sign of the Derivative: The derivative is always positive because the numerator (3) is positive and the denominator (2t + 3)^2 is always non-negative (since it's a square). This tells us that s(t) is an increasing function. As t increases, s(t) also increases.

  • The Magnitude of the Derivative: The magnitude of the derivative indicates the steepness of the function's graph. A larger magnitude means a steeper slope, indicating a faster rate of change. Conversely, a smaller magnitude means a gentler slope, indicating a slower rate of change.

    • As t approaches infinity, the denominator (2t + 3)^2 becomes very large, causing the derivative to approach zero. This means that the rate of change of s(t) slows down as t becomes very large. The function approaches a horizontal asymptote.
    • As t approaches -3/2, the denominator (2t + 3)^2 approaches zero, causing the derivative to approach infinity. This means that the rate of change of s(t) becomes very large as t approaches -3/2. However, the function is not defined at t = -3/2, so this is a vertical asymptote.

  • Applications of the Derivative: The derivative has numerous applications in various fields:

    • Physics: It can represent velocity (the rate of change of position) or acceleration (the rate of change of velocity).
    • Engineering: It can be used to optimize designs, such as finding the maximum stress on a beam.
    • Economics: It can represent marginal cost or marginal revenue.
    • Finance: It can be used to calculate the rate of return on an investment.

Conclusion: Mastering the Art of Differentiation

In this comprehensive guide, we have successfully navigated the process of finding the derivative of the function s(t) = t / (2t + 3) with respect to t. We began by introducing the quotient rule, a fundamental tool for differentiating functions expressed as ratios. We then meticulously applied this rule to our specific problem, carefully identifying the numerator and denominator functions and their respective derivatives.

Through algebraic simplification, we arrived at the derivative: ds/dt = 3 / (2t + 3)^2. This expression provides us with valuable insights into the rate of change of s(t) as t varies. We interpreted the sign and magnitude of the derivative, and we explored its diverse applications in various fields.

By mastering the art of differentiation, we equip ourselves with a powerful tool for understanding and analyzing the dynamic behavior of functions. This skill is invaluable in various disciplines, empowering us to solve real-world problems and make informed decisions.

Remember, practice is key to solidifying your understanding of calculus concepts. Work through various examples and exercises to hone your skills and deepen your appreciation for the power of derivatives. Happy differentiating!