Problem 28 - Single Variable Calculus

Question

Find the derivative of each function. \(g(t)=\frac{a t^{2}}{t^{2}+b}, \quad a, b\) constants

Step-by-Step Solution

Answer

The short answer to the question is: The derivative of the function \(g(t) = \frac{a t^2}{t^2 + b}\) is \(g'(t) = \frac{2abt}{(t^2+b)^2}\).

Step 1: Find the derivatives u'(t) and v'(t)

To find the derivatives of u(t) and v(t), we differentiate them with respect to t. For \(u(t) = at^2\), differentiating w.r.t t: \(u'(t) = 2at\) For \(v(t) = t^2 + b\), differentiating w.r.t t: \(v'(t) = 2t\)

Step 2: Apply the quotient rule

Now that we have all the values we need, we can apply the quotient rule formula. \(g'(t) = \frac{u'(t) v(t) - u(t) v'(t)}{[v(t)]^2}\) Plugging the values, we get: \(g'(t) = \frac{(2at)(t^2+b) - (at^2)(2t)}{(t^2+b)^2}\)

Step 3: Simplify the expression

Next, we will simplify the expression by expanding and simplifying the terms in the numerator. \(g'(t) = \frac{2at^3 + 2abt - 2at^3}{(t^2+b)^2}\) The \(2at^3\) terms cancel each other out: \(g'(t) = \frac{2abt}{(t^2+b)^2}\)

Step 4: Write the final answer

Now we have the derivative of the given function in its simplified form: \(g'(t) = \frac{2abt}{(t^2+b)^2}\)

Key Concepts

Quotient Rule

The quotient rule is a fundamental tool in calculus often employed when we have to find the derivative of a function that is the ratio of two differentiable functions. For example, in the given function
\( g(t)=\frac{a t^{2}}{t^{2}+b} \),
where \( a \) and \( b \) are constants, we have a numerator \( u(t) = at^2 \) and a denominator \( v(t) = t^2 + b \). To find the derivative of \( g(t) \), we apply the quotient rule, which is defined as:
\[ g'(t) = \frac{u'(t) v(t) - u(t) v'(t)}{[v(t)]^2} \].
This rule combines the derivatives of both the top and bottom parts of our fraction, denoted by \( u'(t) \) and \( v'(t) \) respectively. The application of the quotient rule simplifies the process by creating a formulaic approach to tackling complex ratios.

Differentiation

Differentiation is a core operation in calculus that measures how a function changes as its input changes. It's the process of finding a derivative, which gives the rate at which a function's value is changing at any given point. In this exercise, we are differentiating functions with respect to the variable \( t \). When we differentiate \( u(t) = at^2 \) and \( v(t) = t^2 + b \), we find that their derivatives are
\[ u'(t) = 2at \]
and
\[ v'(t) = 2t \]
respectively. These represent the slopes of the tangent lines to the curves of \( u \) and \( v \) at any point \( t \), which are crucial in the application of the quotient rule to find the derivative of the function \( g(t) \).

Calculus

Calculus is a vast field of mathematics that studies continuous change and encompasses a variety of operations and principles such as differentiation, integration, and limits. It's broken down into two main branches: differential calculus, which focuses on the concept of the derivative and deals with the rate of change; and integral calculus, which focuses on the concept of the integral and deals with accumulation. Scenarios that involve rate of change, like the one in our exercise, are where differential calculus and, by extension, the quotient rule come into play. Calculus has wide applications across science, engineering, economics, and many other fields.

Derivative Simplification

Simplifying the derivative is the final step in many differentiation problems. It ensures that the derivative is presented in the most reduced and understandable form. After applying the quotient rule, we often encounter a complex fraction that can typically be simplified by expanding and cancelling out similar terms. In our example, the initial application of the quotient rule yielded the expression
\[ g'(t) = \frac{(2at)(t^2+b) - (at^2)(2t)}{(t^2+b)^2} \].
Upon expansion and cancellation of like terms, we simplified the derivative to
\[ g'(t) = \frac{2abt}{(t^2+b)^2} \].
This is a much cleaner and more polished representation of the rate of change of \( g(t) \), which is easier to analyze and apply to further problems.