Differentiating G(t) = (8-5t)/(2t+3) Step-by-Step Solution

In the realm of calculus, differentiation stands as a cornerstone, enabling us to explore the rates at which functions change. This article delves into the process of differentiating a specific function, $g(t) = \frac{8-5t}{2t+3}$, providing a step-by-step guide suitable for students and enthusiasts alike. We will unravel the techniques required to find $g'(t)$, the derivative of $g(t)$, which represents the instantaneous rate of change of the function with respect to the variable t. Understanding differentiation is crucial not only in mathematics but also in various fields like physics, engineering, and economics, where it helps model and analyze dynamic systems. This exploration will not only enhance your understanding of calculus but also equip you with the skills to tackle similar problems with confidence. Let's embark on this mathematical journey, where we'll dissect the function $g(t)$ and reveal its derivative, $g'(t)$.

Understanding the Quotient Rule

Before we dive into the specifics of differentiating $g(t)$, it's essential to grasp the quotient rule, a fundamental concept in calculus. The quotient rule is specifically designed for finding the derivative of a function that is expressed as a ratio of two other functions. Mathematically, if we have a function $g(t)$ defined as $g(t) = \frac{u(t)}{v(t)}$, where both $u(t)$ and $v(t)$ are differentiable functions, then the derivative of $g(t)$, denoted as $g'(t)$, can be found using the following formula:

$g'(t) = \frac{u'(t)v(t) - u(t)v'(t)}{[v(t)]^2}$

In this formula:

• $u'(t)$ represents the derivative of the numerator function, $u(t)$.
• $v'(t)$ represents the derivative of the denominator function, $v(t)$.

The quotient rule essentially tells us that the derivative of a quotient is equal to the denominator times the derivative of the numerator, minus the numerator times the derivative of the denominator, all divided by the square of the denominator. It's crucial to remember the order of operations in the numerator to avoid errors. The quotient rule is a powerful tool in calculus, and mastering it is vital for differentiating a wide range of functions, especially those that appear in fractional form. In our specific case, $g(t) = \frac{8-5t}{2t+3}$, we will identify $u(t)$ as $8-5t$ and $v(t)$ as $2t+3$, and then apply the quotient rule to find $g'(t)$. This systematic approach ensures that we handle the differentiation process accurately and efficiently.

Applying the Quotient Rule to g(t) = (8-5t)/(2t+3)

Now, let's apply the quotient rule to our function, $g(t) = \frac{8-5t}{2t+3}$. As we established earlier, we can identify the numerator, $u(t)$, as $8-5t$ and the denominator, $v(t)$, as $2t+3$. To use the quotient rule, we first need to find the derivatives of both $u(t)$ and $v(t)$.

The derivative of $u(t) = 8 - 5t$, denoted as $u'(t)$, is found by applying the power rule and the constant multiple rule of differentiation. The derivative of a constant is zero, and the derivative of $-5t$ is $-5$. Therefore, $u'(t) = -5$.

Similarly, the derivative of $v(t) = 2t + 3$, denoted as $v'(t)$, is found using the same principles. The derivative of $2t$ is $2$, and the derivative of the constant $3$ is zero. Thus, $v'(t) = 2$.

Now that we have $u(t)$, $v(t)$, $u'(t)$, and $v'(t)$, we can plug these into the quotient rule formula:

$g'(t) = \frac{u'(t)v(t) - u(t)v'(t)}{[v(t)]^2}$

Substituting the values, we get:

$g'(t) = \frac{(-5)(2t+3) - (8-5t)(2)}{(2t+3)^2}$

This expression represents the derivative of $g(t)$. However, to obtain the simplest form, we need to simplify the numerator and potentially the denominator as well. In the next section, we will focus on simplifying this expression to arrive at the final, concise form of $g'(t)$. This process involves expanding the terms in the numerator, combining like terms, and potentially factoring to achieve the most simplified representation of the derivative.

Simplifying the Expression for g'(t)

Having applied the quotient rule, we arrived at the expression:

$g'(t) = \frac{(-5)(2t+3) - (8-5t)(2)}{(2t+3)^2}$

Now, our task is to simplify this expression to its most concise form. We begin by expanding the terms in the numerator. Distributing the $-5$ in the first term, we get $-10t - 15$. Distributing the $2$ in the second term, we get $16 - 10t$. So the numerator becomes:

$-10t - 15 - (16 - 10t)$

Next, we distribute the negative sign in front of the parentheses, which changes the signs of the terms inside:

$-10t - 15 - 16 + 10t$

Now, we combine like terms. We have $-10t$ and $+10t$, which cancel each other out. We also have $-15$ and $-16$, which combine to give $-31$. Therefore, the simplified numerator is $-31$.

The denominator is $(2t+3)^2$. We can leave it in this form or expand it. For the purpose of this simplification, leaving it in the squared form is often preferred, as it provides a clearer representation of the function's behavior, especially when analyzing critical points or asymptotes. However, for some applications, expanding the denominator might be necessary.

So, the simplified expression for $g'(t)$ is:

$g'(t) = \frac{-31}{(2t+3)^2}$

This is the final, simplified form of the derivative of $g(t)$. It tells us the instantaneous rate of change of the function $g(t)$ at any given point $t$. The negative sign indicates that the function is decreasing as t increases, and the denominator provides information about the function's behavior near the value where $2t+3 = 0$, which is $t = -\frac{3}{2}$.

The Final Result: g'(t) = -31/(2t+3)^2

After meticulously applying the quotient rule and simplifying the resulting expression, we have arrived at the derivative of the function $g(t) = \frac{8-5t}{2t+3}$. The final result is:

$g'(t) = \frac{-31}{(2t+3)^2}$

This concise expression encapsulates the rate of change of the function $g(t)$ with respect to the variable t. The negative sign in the numerator indicates that the function is decreasing as t increases. The denominator, $(2t+3)^2$, reveals important information about the function's behavior. Specifically, it tells us that the derivative is undefined when $2t+3 = 0$, which occurs at $t = -\frac{3}{2}$. This point is a critical point for the function, and it often indicates a vertical asymptote or a point of discontinuity in the original function $g(t)$.

Understanding the derivative is crucial for a variety of applications. For instance, we can use $g'(t)$ to find the slope of the tangent line to the graph of $g(t)$ at any point. We can also use it to determine the intervals where $g(t)$ is increasing or decreasing. Furthermore, the derivative is essential in optimization problems, where we seek to find the maximum or minimum values of a function.

In conclusion, the process of differentiating $g(t) = \frac{8-5t}{2t+3}$ not only provides us with its derivative, $g'(t) = \frac{-31}{(2t+3)^2}$, but also enhances our understanding of the behavior of the function and its rate of change. This exercise demonstrates the power of the quotient rule and the importance of simplification in calculus. Mastering these techniques is fundamental for tackling more complex problems in mathematics and related fields.

Applications of the Derivative g'(t)

Now that we have successfully found the derivative, $g'(t) = \frac{-31}{(2t+3)^2}$, it's important to understand its practical applications and how it can be used to gain further insights into the behavior of the original function, $g(t) = \frac{8-5t}{2t+3}$. The derivative, in essence, provides a powerful tool for analyzing the rate of change and the overall characteristics of a function.

1. Finding the Slope of the Tangent Line

One of the most fundamental applications of the derivative is in finding the slope of the tangent line to the graph of the function at a specific point. The derivative $g'(t)$ gives us the instantaneous rate of change of $g(t)$ at any value of t. Geometrically, this corresponds to the slope of the line that is tangent to the curve of $g(t)$ at the point (t, g(t)).

For example, if we want to find the slope of the tangent line at $t = 1$, we simply substitute $t = 1$ into the derivative:

$g'(1) = \frac{-31}{(2(1)+3)^2} = \frac{-31}{25}$

This means that at the point on the graph of $g(t)$ where $t = 1$, the slope of the tangent line is $- \frac{31}{25}$. This information can be used to write the equation of the tangent line, which is a linear approximation of the function near that point.

2. Determining Intervals of Increase and Decrease

The derivative also helps us determine the intervals over which the function is increasing or decreasing. A function is increasing where its derivative is positive and decreasing where its derivative is negative. In our case, $g'(t) = \frac{-31}{(2t+3)^2}$.

Notice that the numerator is always negative (-31), and the denominator, $(2t+3)^2$, is always positive (since it's a square) except when $2t+3 = 0$, which is at $t = -\frac{3}{2}$. Therefore, $g'(t)$ is always negative (or undefined at $t = -\frac{3}{2}$). This tells us that the function $g(t)$ is decreasing over its entire domain, except at the point $t = -\frac{3}{2}$, where it is undefined.

3. Identifying Critical Points and Local Extrema

Critical points are the points where the derivative is either zero or undefined. These points are crucial for finding local maxima and minima of a function. In our case, $g'(t)$ is undefined at $t = -\frac{3}{2}$, but it is never zero (since the numerator is a constant, -31). This means that $t = -\frac{3}{2}$ is a critical point.

To determine whether this critical point corresponds to a local maximum, a local minimum, or neither, we can use the first derivative test or the second derivative test. However, in this case, since the function is decreasing on both sides of $t = -\frac{3}{2}$ (as we determined in the previous section), this point is not a local extremum. Instead, it indicates a vertical asymptote in the original function $g(t)$.

4. Optimization Problems

The derivative is a key tool in optimization problems, where we want to find the maximum or minimum value of a function subject to certain constraints. While our specific function $g(t)$ doesn't have a maximum or minimum value in the traditional sense (it decreases continuously), the principles learned from finding its derivative are applicable to a wide range of optimization problems.

For example, if we had a related problem where we wanted to minimize the square of the derivative, we could use the techniques discussed here to find the value of t that achieves this minimum.

In summary, the derivative $g'(t) = \frac{-31}{(2t+3)^2}$ is not just an abstract mathematical expression; it is a powerful tool that provides valuable information about the function $g(t)$. It allows us to analyze the function's rate of change, identify intervals of increase and decrease, find critical points, and solve optimization problems. Mastering the concept of the derivative and its applications is essential for anyone studying calculus and its related fields.