Problem 11 - Thomas Calculus

Question

Find the first and second derivatives. $$ r=\frac{1}{3 s^{2}}-\frac{5}{2 s} $$

Step-by-Step Solution

Answer

First derivative: \( \frac{-2}{3} s^{-3} - \frac{5}{2} s^{-2} \); Second derivative: \( 2 s^{-4} + 5 s^{-3} \).

Step 1: Apply the Power Rule for Each Term

First, let's rewrite the expression in a form that makes it easier to differentiate. We have \( r = \frac{1}{3} s^{-2} - \frac{5}{2} s^{-1} \). We will apply the power rule \( \frac{d}{ds} s^n = n s^{n-1} \) to each term.

Step 2: Differentiate the First Term

Using the power rule on the first term \( \frac{1}{3} s^{-2} \), we differentiate: \(-2 \cdot \frac{1}{3} s^{-3} = \frac{-2}{3} s^{-3} \).

Step 3: Differentiate the Second Term

Similarly, differentiate the second term \( \frac{5}{2} s^{-1} \) using the power rule: \(-1 \cdot \frac{5}{2} s^{-2} = \frac{-5}{2} s^{-2} \).

Step 4: Combine First Derivative Terms

Combine the differentiated terms to write the first derivative: \( \frac{dr}{ds} = \frac{-2}{3} s^{-3} - \frac{5}{2} s^{-2} \).

Step 5: Differentiate Again for the Second Derivative

Now, differentiate each term of the first derivative. Start with \( \frac{-2}{3} s^{-3} \), yielding \(-3 \cdot \frac{-2}{3} s^{-4} = 2 s^{-4} \).

Step 6: Differentiate the Remaining Term for Second Derivative

Differentiate the term \( \frac{-5}{2} s^{-2} \): \(-2 \cdot \frac{-5}{2} s^{-3} = 5 s^{-3} \).

Step 7: Combine Second Derivative Terms

Combine the terms to write the second derivative: \( \frac{d^2r}{ds^2} = 2 s^{-4} + 5 s^{-3} \).

Key Concepts

Power Rule

The power rule is a fundamental technique in calculus. It simplifies the process of differentiation, which is finding the derivative of a function. The rule states that if you have a term in the form of \(s^n\), its derivative is \(n \cdot s^{n-1}\).
This rule is incredibly useful when dealing with polynomials or any expression where terms have exponents.
For example:

• If you have \(s^3\), applying the power rule gives you \(3 \cdot s^{2}\).
• For \(s^{-2}\), it becomes \(-2 \cdot s^{-3}\).

By understanding the power rule, you can break down more complicated functions into simpler parts, making differentiation straightforward.

First Derivative

The first derivative of a function measures how the function's output changes as its input changes. In simpler terms, it's the function's rate of change.
For the expression given, \( r = \frac{1}{3} s^{-2} - \frac{5}{2} s^{-1} \), we need to find \( \frac{dr}{ds} \). Using the power rule, we calculate:

• The derivative of \(\frac{1}{3} s^{-2}\) is \(-\frac{2}{3} s^{-3}\).
• The derivative of \(\frac{5}{2} s^{-1}\) is \(-\frac{5}{2} s^{-2}\).

Combining these results, the first derivative is \( \frac{-2}{3} s^{-3} - \frac{5}{2} s^{-2} \). This expression gives the rate of change of \(r\) concerning \(s\).

Second Derivative

The second derivative provides information about the curvature of the function. It tells you how the rate of change itself is changing. This can indicate whether the function is accelerating or decelerating.
Starting from the first derivative \( \frac{-2}{3} s^{-3} - \frac{5}{2} s^{-2} \), we need to differentiate further:

• For \(\frac{-2}{3} s^{-3}\), the derivative is \(2 s^{-4}\).
• For \(\frac{-5}{2} s^{-2}\), the derivative is \(5 s^{-3}\).

Putting it all together, the second derivative is \( 2 s^{-4} + 5 s^{-3} \). This expression shows how quickly the slope of the original function changes.

Calculus

Calculus is the branch of mathematics that studies continuous change, much like geometry studies shape and algebra studies operations and their application.
It is divided into two main branches: differential calculus and integral calculus.

• Differential calculus focuses on the concept of a derivative, which represents the rate of change of a quantity.
• Integral calculus deals with the accumulation of quantities and the areas under and between curves.

The current problem is an example from differential calculus. We use differentiation to find the first and second derivatives of a function, giving us insight into the function's behavior and characteristics. Understanding the basics of calculus opens up a broader understanding of complex mathematical and real-world phenomena.