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Let \(f(t)\) be the temperature at time \(t\) where you live and sup- pose that at time \(t=3\) you feel uncomfortably hot. How do you feel about the given data in each case? \(\begin{array}{l}{\text { (a) } \mathrm{f}^{\prime}(3)=2, \quad \mathrm{f}^{\prime \prime}(3)=4 \quad \text { (b) } \mathrm{f}^{\prime}(3)=2, \quad \mathrm{f}^{\prime \prime}(3)=-4} \\ {\text { (c) } \mathrm{f}^{\prime}(3)=-2, \quad \mathrm{f}^{\prime \prime}(3)=4 \quad \text { (d) } \mathrm{f}^{\prime}(3)=-2, \quad \mathrm{f}^{\prime \prime}(3)=-4}\end{array}\)

Short Answer

Expert verified
(a) Getting hotter fast, (b) Hot but slowing, (c) Decreasing slowly, (d) Cooling quickly.

Step by step solution

01

Understanding Derivatives

First, let's identify what the first and second derivatives represent. The first derivative, \(f'(t)\), indicates the rate of change of temperature at time \(t\). A positive value means temperature is increasing, and a negative value means decreasing.The second derivative, \(f''(t)\), indicates the concavity of the temperature curve at time \(t\). A positive value means the temperature's rate of increase is accelerating, while a negative value means it is decelerating.
02

Analyze Case (a)

For case (a), \(f'(3) = 2\) and \(f''(3) = 4\):- \(f'(3) = 2\) means the temperature is rising at a rate of 2 units.- \(f''(3) = 4\) suggests that the rate of temperature increase is accelerating.The temperature is getting hotter, and it's accelerating quickly.
03

Analyze Case (b)

For case (b), \(f'(3) = 2\) and \(f''(3) = -4\):- \(f'(3) = 2\) indicates the temperature is currently rising, but \(f''(3) = -4\) tells that the rate of temperature increase is slowing down.It’s hot and getting hotter, but the increase is decelerating.
04

Analyze Case (c)

For case (c), \(f'(3) = -2\) and \(f''(3) = 4\):- \(f'(3) = -2\) means the temperature is decreasing.- \(f''(3) = 4\) shows that the rate of decrease is slowing down (i.e., it's getting less cold quickly).The temperature is decreasing, while the rate of decrease is slowing down, but it's still uncomfortably warm at the moment.
05

Analyze Case (d)

For case (d), \(f'(3) = -2\) and \(f''(3) = -4\):- \(f'(3) = -2\) indicates a decrease in temperature.- \(f''(3) = -4\) shows the rate of temperature decrease is increasing, further helping to cool down.The temperature is falling and picking up speed in cooling, moving towards a more comfortable zone quickly.

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Key Concepts

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

First Derivative
The first derivative, denoted as \( f'(t) \), plays a crucial role in calculus by indicating how a function is changing at any given point. In the context of temperature, it tells us the rate at which the temperature is changing over time at that specific moment.

When \( f'(t) \) is positive, it means the temperature is rising. For instance, a first derivative value of \( f'(3) = 2 \) suggests that the temperature increases by 2 units of temperature over the next unit of time.

Conversely, a negative \( f'(t) \) value indicates that the temperature is falling. Let's say \( f'(3) = -2 \), this would mean the temperature is decreasing by 2 units of temperature per unit of time.

Understanding the first derivative helps in predicting immediate changes in temperature and can provide insight into how the environment feels at that given moment.
Second Derivative
The second derivative, expressed as \( f''(t) \), provides insights into the behavior of the rate of change itself. It's like a detective looking deeper to tell more about the temperature curve's shape or its concavity.

If \( f''(t) \) is positive, it indicates that the rate of temperature change is accelerating. This could mean you not only feel the temperature rising but it's doing so faster with each passing moment.

For example, if \( f''(3) = 4 \), the rate of temperature increase is speeding up, making it hotter more rapidly.

On the other hand, a negative \( f''(t) \) shows that the rate of change is decelerating. So even if the temperature might still rise, it's happening more slowly. For instance, \( f''(3) = -4 \) would mean that while it's still getting hotter, the speed at which it becomes warmer is slowing down.
Rate of Change
Rate of change is a foundational concept in calculus and is depicted via the first derivative. In our scenario, it specifically concerns how the temperature level shifts over time.

Here are some essentials to consider about rate of change:
  • A positive rate of change (i.e., a positive first derivative) implies temperatures are climbing, so you might feel warmer with time.
  • A negative rate indicates temperatures are dropping, leading to cooler sensations.

This measure doesn’t only apply to temperature; it can be associated with any varying quantity such as speed or population growth. Knowing the rate of change lets you understand how swiftly or slowly changes are occurring, which is vital for modeling and anticipating future behaviors.
Temperature Modeling
Temperature modeling utilizes mathematical functions to predict temperature changes over time.

In any model, derivatives help us understand how temperature shifts:
  • The first derivative tells whether the temperature is increasing or decreasing at a certain point in time.
  • The second derivative sheds light on whether these changes are getting faster or slower.

These insights prove essential in scenarios like weather predictions, HVAC system controls, or even understanding climate change.

By analyzing cases like \( f'(3) = 2 \) and \( f''(3) = -4 \), we can predict not only what the temperature is doing at time \( t = 3 \) but also how the pattern might evolve shortly, providing a deeper understanding of thermal environments.

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Most popular questions from this chapter

A model rocket is fired vertically upward from rest. Its acceleration for the first three seconds is a(t) = \(60 t,\) at which time the fuel is exhausted and it becomes a freely "falling" body. Fourteen seconds later, the rocket's parachute opens, and the (downward) velocity slows linearly to \(-18 \mathrm{ft} / \mathrm{s}\) in 5 \(\mathrm{s}\) . The rocket then "floats" to the ground at that rate. (a) Determine the position function s and the velocity function \(v\) (for all times t). Sketch the graphs of s and v. (b) At what time does the rocket reach its maximum height, and what is that height? (c) At what time does the rocket land?

\(5-8\) Use Newton's method with the specified initial approximation \(x_{1}\) to find \(x_{3},\) the third approximation to the root of the given equation. (Give your answer to four decimal places.) \(x^{3}+2 x-4=0, \quad x_{1}=1\)

Show that for motion in a straight line with constant acceleration a, initial velocity \(v_{0},\) and initial displacement \(s_{0}\) , the dis- placement after time t is \(\mathrm{s}=\frac{1}{2} \mathrm{at}^{2}+v_{0} \mathrm{t}+\mathrm{s}_{0}\)

Use a computer algebra system to graph \(f\) and to find \(f^{\prime}\) and \(f^{\prime \prime} .\) Use graphs of these derivatives to estimate the intervals of increase and decrease, extreme values, intervals of concavity, and inflection points of \(f\). \(f(x)=\sqrt{x+5 \sin x}, \quad x \leqslant 20\)

(a) Apply Newton's method to the equation \(x^{2}-a=0\) to derive the following square-root algorithm (used by the ancient Babylonians to compute \(\sqrt{a} ) :\) \(\mathrm{x}_{\mathrm{n}+1}=\frac{1}{2}\left(\mathrm{x}_{\mathrm{n}}+\frac{\mathrm{a}}{\mathrm{x}_{\mathrm{n}}}\right)\) (b) Use part (a) to compute \(\sqrt{1000}\) correct to six decimal places.

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