Problem 52 - Brief Applied Calculus

Question

At time \(t=0\) a helicopter takes off gently and then 60 seconds later it lands gently. Let \(f(t)\) be its altitude above the ground at time \(t\) seconds. a. Will \(f^{\prime}(1)\) be positive or negative? Same question for \(f^{\prime \prime}(1)\) b. Will \(f^{\prime}(59)\) be positive or negative? Same question for \(f^{\prime \prime}(59)\).

Step-by-Step Solution

Answer

a. \(f'(1)\) is positive, \(f''(1)\) is positive; b. \(f'(59)\) is negative, \(f''(59)\) is positive.

Step 1: Understanding f(t)

To understand the altitude function \(f(t)\), we need to consider its behavior over time. The helicopter takes off gently from the ground at \(t=0\) and lands gently at \(t=60\). This suggests that \(f(t)\) starts at 0, increases to a maximum altitude, and then decreases back to 0.

Step 2: Determine f'(1)

The derivative \(f'(t)\) represents the rate of change of altitude with respect to time, or the helicopter's vertical speed. At \(t=1\), the helicopter is still in its initial ascent phase. Therefore, the rate of change \(f'(1)\) should be positive as the helicopter is climbing upwards.

Step 3: Determine f''(1)

The second derivative \(f''(t)\) provides information about the acceleration. At an early point like \(t=1\), since the helicopter is just starting to ascend and likely increasing its altitude more steeply, \(f''(1)\) would be positive, indicating an increasing rate of ascent.

Step 4: Determine f'(59)

At \(t=59\), the helicopter is nearing the end of its flight and is preparing to land. The altitude \(f(t)\) is decreasing, so the rate of change \(f'(59)\) is negative as the helicopter descends back toward the ground.

Step 5: Determine f''(59)

As the helicopter nears the ground, it should be decelerating its descent to land gently. This means the rate of descent is decreasing, so \(f''(59)\) will be positive, indicating a reduction in its negative speed.

Key Concepts

Derivative Analysis

When we analyze derivatives, we are essentially studying how a function changes over time. This involves examining both the first and second derivatives. The first derivative, \(f'(t)\), gives us the rate of change or the speed of the function at any given moment. Meanwhile, the second derivative, \(f''(t)\), tells us about the nature of this change—whether it's accelerating or decelerating.For a function like the altitude of a helicopter, \(f(t)\), this analysis provides insight into how quickly the helicopter ascends and descends:

• \(f'(t)\) shows whether the helicopter is climbing or descending by being positive or negative, respectively.
• If \(f''(t)\) is positive, it indicates the helicopter is increasing its speed while ascending or decreasing its speed while descending.

Understanding these derivatives helps predict and control the helicopter’s movement, ensuring it ascends and lands smoothly, without abrupt changes in speed.

First Derivative

The first derivative of a function, such as \(f'(t)\) for the helicopter problem, describes the rate at which the function's value is changing at any given instant. It's akin to the helicopter's instantaneous vertical speed. In the given scenario:

• At \(t=1\), the helicopter is early in its take-off, illustrating that \(f'(1)\) is positive. This positivity indicates an upward movement, as the helicopter is gaining altitude.
• Conversely, at \(t=59\), the helicopter approaches the ground, which makes \(f'(59)\) negative. This reflects the helicopter losing altitude as it prepares to land.

The first derivative is crucial for understanding how something is moving over time, whether it be an aircraft's ascent or descent, helping identify periods of climb and descent efficiently.

Second Derivative

The second derivative, \(f''(t)\), reveals how the rate of change itself is changing over time. It quantifies whether an object like a helicopter is accelerating or decelerating.For example, in our helicopter problem:

• At \(t=1\), \(f''(1)\) is positive, signifying increasing speed in gaining altitude. The helicopter is not just moving upwards, but it's climbing faster as it continues to ascend.
• As the helicopter approaches landing at \(t=59\), \(f''(59)\) remains positive. This positivity signals a decrease in descent speed, crucial for a gentle landing, indicating that the helicopter is decelerating as it approaches the ground.

The second derivative is an essential concept in calculus as it helps predict the future movement by considering the acceleration or deceleration, allowing for smoother transitions in motion for any moving object.