Chapter 4: Problem 21

Find the hottest and coldest points on a bar of length 5 if \(T=4 x-x^{2}\), where \(x\) is the distance measured from the left end.

Short Answer

Expert verified

The hottest point is at \(x=2\) with \( T = 4 \), and the coldest point is at \(x=5\) with \( T = -5 \).

Step by step solution

Understand the Function

The function given is the temperature distribution along the bar: \[ T(x) = 4x - x^2 \]where \(x\) is the distance from the left end of the bar.

Find the Derivative

To find the critical points where the maximum and minimum temperatures occur, calculate the derivative of \(T(x)\): \[ T'(x) = \frac{d}{dx}(4x - x^2) = 4 - 2x \]

Set the Derivative to Zero

Find the critical points by setting the derivative equal to zero:\[4 - 2x = 0\]Solve for \(x\):\[ x = 2 \]

Evaluate the Temperature at Critical Points and End Points

Evaluate \( T(x) \) at the critical point \(x=2\) and at the endpoints \(x = 0\) and \(x = 5\):\[ T(0) = 4(0) - 0^2 = 0 \]\[ T(2) = 4(2) - 2^2 = 8 - 4 = 4 \]\[ T(5) = 4(5) - 5^2 = 20 - 25 = -5 \]

Determine the Hottest and Coldest Points

Compare the values: \[ T(0) = 0 \]\[ T(2) = 4 \]\[ T(5) = -5 \]The hottest point is at \(x=2\) with a temperature of 4. The coldest point is at \(x=5\) with a temperature of -5.

Unlock Step-by-Step Solutions & Ace Your Exams!

Full Textbook Solutions
Get detailed explanations and key concepts
Unlimited Al creation
Al flashcards, explanations, exams and more...
Ads-free access
To over 500 millions flashcards
Money-back guarantee
We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

derivative

In mathematics, the derivative of a function is a measure of how a function's output value changes as the input changes. It's a powerful concept in calculus used to find rates of change. For example, in our exercise, the temperature distribution function is given by RDF 4x - x^2. To find when this function reaches its hottest or coldest points, we need to compute its derivative. The derivative of the given function is found using the power rule for each term: T'(x) = 4 - 2x. This represents the rate of change of temperature concerning distance, 'x,' along the bar.

critical points

Critical points occur where the derivative of a function is zero or undefined. These points can indicate where a function changes direction, meaning transitions from increasing to decreasing, or vice versa. In our exercise, after finding the derivative T'(x) = 4 - 2x and setting it equal to zero, we get: RDF 4 - 2x = 0. Solving for 'x', we find: RDF x = 2. The value 'x = 2' is the critical point, where the function's slope shifts, potentially indicating a local maximum or minimum in the context of temperature along the bar.

endpoints

Endpoints are the boundary values of the function's domain. In the context of our exercise, the bar is of length 5, which means the endpoints are x = 0 and x = 5. To determine the overall hottest and coldest points, we must consider the function's values at these endpoints in addition to the critical points. For x = 0 and x = 5, we can evaluate the given temperature function as follows: RDF T(0) = 4(0) - 0^2 = 0. RDF T(5) = 4(5) - 5^2 = 20 - 25 = -5. Both of these values must be compared with the temperature at the critical point to identify the overall max and min temperatures.

maximum and minimum values

The maximum and minimum values of a function represent its highest and lowest points, respectively. To find them, we compare the function's value at critical points and endpoints. In our exercise: At x = 0, T(0) = 0, At x = 2, T(2) = 4, At x = 5, T(5) = -5. By comparing these values, we observe that the highest temperature (maximum value) is 4 at x = 2, and the lowest temperature (minimum value) is -5 at x = 5. Therefore, the hottest point on the bar is at x = 2, and the coldest point is at x = 5.