Problem 46 - Applied Calculus

Question

If \(f(t)=t^{4}-3 t^{2}+5 t\), find \(f^{\prime}(t)\) and \(f^{\prime \prime}(t)\).

Step-by-Step Solution

Answer

The first derivative is \( f'(t) = 4t^3 - 6t + 5 \) and the second derivative is \( f''(t) = 12t^2 - 6 \).

Step 1: Apply Power Rule to Find the First Derivative

The function given is \( f(t) = t^4 - 3t^2 + 5t \). To find the first derivative \( f'(t) \), we'll use the power rule: if \( f(t) = at^n \), then \( f'(t) = nat^{n-1} \). 1. For the term \( t^4 \), the derivative is \( 4t^3 \).2. For \( -3t^2 \), the derivative is \( -6t \).3. For \( 5t \), the derivative is \( 5 \). Thus, \( f'(t) = 4t^3 - 6t + 5 \).

Step 2: Apply Power Rule to Find the Second Derivative

To find the second derivative \( f''(t) \), we differentiate the first derivative \( f'(t) = 4t^3 - 6t + 5 \).1. For the term \( 4t^3 \), apply the power rule to get \( 12t^2 \).2. For \( -6t \), the derivative is \( -6 \).3. The constant \( 5 \) has a derivative of \( 0 \).So, \( f''(t) = 12t^2 - 6 \).

Key Concepts

Power Rule

The power rule is a fundamental tool in calculus differentiation that simplifies finding the derivative of a power function. When you have a function in the form of \( f(t) = at^n \), the power rule allows you to find its derivative efficiently. The rule states:

• Multiply the exponent \( n \) by the coefficient \( a \).
• Subtract one from the exponent.

For instance, applying the power rule to \( t^4 \) involves multiplying \( 4 \) (the exponent) by \( 1 \) (the coefficient), resulting in \( 4t^3 \). This makes it easier to determine the rate at which functions change, which is essential for understanding behaviors like velocity and growth.

First Derivative

The first derivative of a function, often denoted as \( f'(t) \), provides crucial insight into the rate of change of the function. In simple terms, it tells us how fast or slow the function's value is changing at any given point. Using the power rule, let's break down the function \( f(t) = t^4 - 3t^2 + 5t \):

• The first term \( t^4 \) becomes \( 4t^3 \) after differentiation.
• \(-3t^2\) simplifies to \(-6t\).
• The linear term \(5t\) reduces to the constant \(5\).

Thus, the first derivative, \( f'(t) = 4t^3 - 6t + 5 \), reflects the changes in slopes or the steepness of the original function at different points.

Second Derivative

The second derivative, \( f''(t) \), is essentially the derivative of the first derivative. This means we apply the differentiation process to \( f'(t) \). The second derivative provides insights into the concavity of the function, helping us determine if the function is curving upwards or downwards. Let's look at the steps for our given function's first derivative \( f'(t) = 4t^3 - 6t + 5 \):

• The term \( 4t^3 \) becomes \( 12t^2 \) when differentiated again.
• The term \(-6t\) results in \(-6\).
• The constant \(5\) becomes \(0\), as constants do not change.

Therefore, the second derivative is \( f''(t) = 12t^2 - 6 \). This provides valuable information about the acceleration or deceleration of the rate of change itself.