Problem 47 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Find the first and second derivatives of the given function. \(h(t)=t^{4}-2 t^{3}+6 t^{2}-3 t+10\)

Step-by-Step Solution

Answer

The first derivative of the function \(h(t) = t^4 - 2t^3 + 6t^2 - 3t + 10\) is \(h'(t) = 4t^3 - 6t^2 + 12t - 3\), and the second derivative is \(h''(t) = 12t^2 - 12t + 12\).

Step 1: Use the Power Rule of Differentiation

The Power Rule of Differentiation states that if \(f(x) = x^n\), then \(f'(x) = nx^{n-1}\). So, to find the first derivative, we will differentiate the function component-wise. \(h'(t) = \frac{d}{dt}(t^4 - 2t^3 + 6t^2 - 3t + 10)\) Step 2: Differentiate each term of the function using the Power Rule

Step 2: Differentiation

Differentiate each term as follows: 1. \(\frac{d}{dt}(t^4) = 4t^3\) 2. \(\frac{d}{dt}(-2t^3) = -6t^2\) 3. \(\frac{d}{dt}(6t^2) = 12t\) 4. \(\frac{d}{dt}(-3t) = -3\) 5. \(\frac{d}{dt}(10) = 0\) Step 3: Combine the differentiated terms to obtain the first derivative

Step 3: First Derivative

The first derivative is the sum of the derivatives of the individual terms: \(h'(t) = 4t^3 - 6t^2 + 12t - 3\) Now, we will find the second derivative. Step 4: Find the second derivative of the function

Step 4: Use the Power Rule again

Once again, we use the Power Rule of Differentiation on the first derivative to obtain the second derivative: \(h''(t) = \frac{d^2}{dt^2}(4t^3 - 6t^2 + 12t - 3)\) Step 5: Differentiate each term of the first derivative using the Power Rule

Step 5: Differentiation

Differentiate each term as follows: 1. \(\frac{d^2}{dt^2}(4t^3) = 12t^2\) 2. \(\frac{d^2}{dt^2}(-6t^2) = -12t\) 3. \(\frac{d^2}{dt^2}(12t) = 12\) 4. \(\frac{d^2}{dt^2}(-3) = 0\) Step 6: Combine the differentiated terms to obtain the second derivative

Step 6: Second Derivative

The second derivative is the sum of the derivatives of the individual terms: \(h''(t) = 12t^2 - 12t + 12\) So, the first and second derivatives of the given function are: \(h'(t) = 4t^3 - 6t^2 + 12t - 3\) \(h''(t) = 12t^2 - 12t + 12\)

Key Concepts

Power Rule

The Power Rule is one of the most fundamental rules of differentiation in calculus. It allows us to easily find the derivative of polynomial expressions. According to the power rule, if you have a term in the form of \[f(x) = x^n,\] then the derivative is given by\[f'(x) = nx^{n-1}.\] This means you bring down the exponent as a coefficient, subtract one from the exponent, and write down the result as the new term.

For a function like \(h(t) = t^4 - 2t^3 + 6t^2 - 3t + 10,\) we apply the power rule to each term:

• The derivative of \(t^4\) is \(4t^3.\)
• For \(-2t^3,\) we have \(-6t^2.\)
• The term \(6t^2\) becomes \(12t.\)
• For the linear term \(-3t,\) it turns into \(-3.\)
• Finally, the constant \(10\) becomes \(0.\)

Understanding the power rule is key to easily tackling simple to complex polynomial functions.

First Derivative

Finding the first derivative of a function involves calculating how the function's output value changes as the input changes. It gives us the rate of change or the slope of the function at any point.

In the given function \(h(t) = t^4 - 2t^3 + 6t^2 - 3t + 10,\) to find the first derivative, we differentiate each term individually using the power rule (as discussed earlier). By differentiating:

• \(t^4\) becomes \(4t^3,\)
• \(-2t^3\) becomes \(-6t^2,\)
• \(6t^2\) becomes \(12t,\)
• \(-3t\) becomes \(-3,\)
• and \(10\) has no variable term, so its derivative is \(0.\)

Thus, the first derivative of the function is given by \(h'(t) = 4t^3 - 6t^2 + 12t - 3.\) This expression can be used to determine the slope of the original function at any point \(t.\)

Second Derivative

The second derivative goes further in understanding the behavior and characteristics of a function. While the first derivative gives the rate of change or slope, the second derivative provides information about the curvature or concavity of the function.

Finding the second derivative involves differentiating the first derivative. Meaning, we apply the power rule again on \(h'(t) = 4t^3 - 6t^2 + 12t - 3.\) Differentiating each term gives us:

• The derivative of \(4t^3\) is \(12t^2,\)
• \(-6t^2\) becomes \(-12t,\)
• the derivative of \(12t\) is \(12,\)
• and the constant \(-3\) differentiates to \(0.\)

Thus, the second derivative is \(h''(t) = 12t^2 - 12t + 12.\)

This tells us how the rate of change itself is changing. If you are analyzing the motion or path behavior, the second derivative can indicate acceleration or the nature of turning points, such as maxima or minima in a graph.