Problem 20 - Thomas Calculus

Question

Find the derivatives of the functions in Exercises \(19-38\) $$ q=\sqrt{2 r-r^{2}} $$

Step-by-Step Solution

Answer

The derivative is \( q' = \frac{1 - r}{\sqrt{2r - r^2}} \).

Step 1: Rewrite the Function

Before finding the derivative, rewrite the function to make it easier to differentiate. The function is \( q = \sqrt{2r - r^2} \). Rewrite this as \( q = (2r - r^2)^{1/2} \).

Step 2: Apply the Chain Rule

To find the derivative, apply the chain rule. The chain rule states that if \( y = (g(x))^n \), then \( \frac{dy}{dx} = n(g(x))^{n-1} \cdot g'(x) \). Here, \( g(x) = 2r - r^2 \) and \( n = 1/2 \).

Step 3: Differentiate the Outer Function

Apply the power rule to the outer function. According to the power rule, the derivative of \( (2r - r^2)^{1/2} \) is \( \frac{1}{2}(2r - r^2)^{-1/2} \).

Step 4: Differentiate the Inner Function

Find the derivative of the inner function, \( 2r - r^2 \). The derivative is \( 2 - 2r \).

Step 5: Combine Derivatives and Simplify

Combine the results from Steps 3 and 4 using the chain rule: \( q' = \frac{1}{2}(2r - r^2)^{-1/2} \cdot (2 - 2r) \). Simplify this expression to get: \( q' = \frac{2 - 2r}{2\sqrt{2r - r^2}} = \frac{1 - r}{\sqrt{2r - r^2}} \).

Key Concepts

Chain Rule

The chain rule is a fundamental technique used in calculus to find the derivative of a composition of functions. Consider it as a method to tackle more complex differentiation problems that involve nested functions. It essentially breaks down the differentiation process into manageable parts.

To apply the chain rule, we look at the outer and inner functions separately. For example, when differentiating something like \(y = (g(x))^n\), we treat \(g(x)\) as the inner function and \(n\) as how we handle the outer power. The formula given by the chain rule is:

• Outer Function Derivative: Differentiate the outer function but leave the inner function untouched.
• Inner Function Derivative: Differentiate the inner function.
• Combine: Multiply the results, following the chain rule formula \(\frac{dy}{dx} = n(g(x))^{n-1} \cdot g'(x)\).

In this exercise, our outer function was \( (2r - r^2)^{1/2}\), and after handling the outer component with the power rule, the chain rule allowed us to incorporate the inner function's derivative efficiently. This systematic approach ensures you handle even the most intricate functions smoothly.

Power Rule

The power rule is a straightforward and vital method used to find derivatives quickly and easily. It involves functions that are in the form \(x^n\), where \(n\) is any real number. The rule helps us take derivatives by reducing the power of the term by one and multiplying by the original power.

To apply the power rule, follow these simple steps:

• Identify the power of the expression you are differentiating.
• Bring down this power as a multiplier in front of the term.
• Decrease the original power by one to find the new exponent.

For example, in the given exercise, the outer function \( (2r - r^2)^{1/2}\) was differentiated using the power rule. The power, \(1/2\), came down in front, and the new exponent became \(-1/2\). This transformation underpins the concept because it simplifies the process of finding derivatives and is often paired with the chain rule.

Utilizing these steps makes differentiation a systematic and less daunting task, enhancing our calculus toolkit.

Calculus Exercises

Calculus exercises are indispensable for anyone looking to master the subject of calculus. They provide a hands-on approach to understanding the abstract concepts taught in the classroom or found in textbooks. Through regular practice, students deepen their comprehension of differentiation and integration and improve their problem-solving skills.

For the exercise discussed, rewriting the function as \(q = (2r - r^2)^{1/2}\) is a common first step. This step allows us to see where we can use calculus rules like the chain and power rules. Exercises like these train students to:

• Identify suitable techniques for various problems.
• Apply rules systematically rather than guesswork.
• Simplify expressions for clarity and accuracy.

Understanding how to transition from a complex function to its derivative equips students with the ability to tackle more challenging problems. Hence, regular practice with exercises is key to grasping calculus concepts effectively and preparing for exams or real-world applications.