Problem 40 - Single Variable Calculus

Question

In Exercises \(25-40\), find the critical number \((s)\), if any, of the function. $$ g(t)=t^{2} \ln t $$

Step-by-Step Solution

Answer

The critical number of the function \(g(t) = t^2\ln t\) is \(t = e^{-\frac{1}{2}}\).

Step 1: Take the derivative using the product rule

First, we need to find the derivative of the function \(g(t)\). Since we have a product of two functions, we should use the product rule: \((fg)' = f'g + fg'\). In our case, we have \(f(t) = t^2\) and \(g(t) = \ln t\). Therefore, we need to find the derivatives of these two functions. For \(f(t)\), we have \(f'(t) = 2t\). For \(g(t)\), we have \(g'(t) = \frac{1}{t}\). Now we can apply the product rule to find the derivative of \(g(t)\): \(g'(t) = f'(t)g(t) + f(t)g'(t) = 2t(\ln t) + t^2\left(\frac{1}{t}\right)\).

Step 2: Simplify the derivative

Now, let's simplify the expression we found in step 1: \(g'(t) = 2t\ln t + t^2\left(\frac{1}{t}\right) = 2t\ln t + t\).

Step 3: Identify the critical numbers

Recall that critical numbers are the points where the derivative is either zero or does not exist. Let's analyze the derivative we found in step 2: \(g'(t) = 2t\ln t + t\) The natural logarithm function \(\ln t\) is not defined for \(t \leq 0\), so we only need to check for the zeros of the function, as the function is differentiable for all \(t > 0\). We want to solve the equation \(g'(t) = 0\) for \(t\): \(2t\ln t + t = 0\) To solve this equation, let's factor out a \(t\) from both terms: \(t(2\ln t + 1) = 0\) This equation has two solutions: \(t = 0\) and \(2\ln t + 1 = 0\). However, since the natural logarithm function is not defined for \(t \leq 0\), the first solution is not a critical number. Let's solve the second equation: \(2\ln t + 1 = 0 \implies \ln t = -\frac{1}{2}\) By the definition of the natural logarithm function, we have: \(t = e^{-\frac{1}{2}}\) So, we found one critical number: \(t = e^{-\frac{1}{2}}\). Thus, the critical number of the function \(g(t) = t^2\ln t\) is \(t = e^{-\frac{1}{2}}\).

Key Concepts

Product Rule

The product rule is a vital concept in calculus used when differentiating the product of two functions. It's particularly helpful in cases where a function is defined as the multiplication of two or more simpler functions. The rule states:

• If you have two functions, say, \( f(x) \) and \( g(x) \), the derivative of their product \(( fg(x) )'\) is given by: \( (fg)' = f'g + fg' \).

This formula tells us how to take the derivative of the product by finding:

• the derivative of the first function \( f' \) and multiplying it by the second function \( g \); and
• the first function \( f \) multiplied by the derivative of the second function \( g' \).

These two products are then added together. In our given problem, \( f(t) = t^2 \) and \( g(t) = \ln t \), allowing us to apply the product rule to find the derivative of the formation \( g(t) = t^2 \ln t \).

Derivative

In calculus, the derivative is a measure of how a function changes as its input changes. It represents the rate of change or the slope of the function at any point. For basic derivatives, there are common rules:

• The derivative of a constant is zero.
• The derivative of \( t^n \) is \( nt^{n-1} \).

For the problem \( g(t) = t^2 \ln t \), we find each part's derivative separately before combining them using the product rule:

• For \( t^2 \), the derivative is \( 2t \).
• For \( \ln t \), the derivative is \( \frac{1}{t} \).

These derivatives are then wisely composed using the product rule to obtain the overall derivative of the function \( g(t) \). Understanding derivatives is crucial, as they are foundational in calculus to solve many types of problems, including finding critical numbers.

Natural Logarithm

The natural logarithm, denoted as \( \ln t \), is a logarithm to the base \( e \), where \( e \) is an irrational number approximately equal to 2.71828. The natural logarithm is unique and important due to its properties and its role in the differentiation of exponential functions.

• One vital property is: \( \ln(e) = 1 \).
• The derivative of \( \ln t \) is \( \frac{1}{t} \).

An essential thing to note is that \( \ln t \) is only defined for \( t > 0 \). This is critical in the analysis of the function \( g(t) = t^2 \ln t \), as it impacts the domain where we search for critical numbers. Understanding \( \ln t \) equips us to tackle problems involving exponential growth and decay as well.

Differentiability

Differentiability refers to whether a function has a derivative at each point in its domain. A function is differentiable if it has a well-defined derivative for all values in its domain.

• A function is not differentiable at points where it creates a sharp corner or upright tangent.
• For \( g(t) = t^2 \ln t \), \( g(t) \) is differentiable in its domain, \( t > 0 \), since \( \ln t \) itself is only defined for positive \( t \).
• Within this context, we look for where the derivative equals zero or does not exist to find critical numbers.

It’s through these differentiability criteria that we can pinpoint potential critical numbers, helping us to determine where the function’s slope equals zero. Thus, differentiability serves as a cornerstone concept in analysis, especially when it comes to solving calculus problems.