Problem 14 - Calculus Volume 1

Question

Sketch the situation if necessary and used related rates to solve for the quantities. A helicopter starting on the ground is rising directly into the air at a rate of 25 ft/sec. You are running on the ground starting directly under the helicopter at a rate of 10 ft/sec. Find the rate of change of the distance between the helicopter and yourself after 5 sec.

Step-by-Step Solution

Answer

The rate of change of distance is approximately 26.9 ft/s.

Step 1: Understanding the Problem

We have a helicopter rising vertically at 25 ft/s and a person on the ground moving horizontally at 10 ft/s. We need to find the rate of change of the distance between the two after 5 seconds.

Step 2: Setting Up the Coordinates

Consider the position of the helicopter at time \( t \) as \( (0, 25t) \) and the position of the person as \( (10t, 0) \). The person is moving horizontally, representing the x-axis, and the helicopter is moving vertically, representing the y-axis.

Step 3: Expressing the Distance Function

Let \( d \) be the distance between the person and the helicopter. This distance can be expressed using the Pythagorean theorem: \[ d = \sqrt{(10t)^2 + (25t)^2} = \sqrt{100t^2 + 625t^2} = \sqrt{725t^2} = \sqrt{725}t \].

Step 4: Finding the Rate of Change of Distance

To find the rate of change of distance, differentiate \( d \) with respect to \( t \):\[ \frac{dd}{dt} = \frac{d}{dt}(\sqrt{725}t) = \sqrt{725} \].

Step 5: Calculating the Rate at 5 Seconds

Since we want the rate at 5 seconds, substitute \( t = 5 \) into the derivative result:The rate of change after 5 seconds is still \( \sqrt{725} \).

Step 6: Final Calculation

Calculate \( \sqrt{725} \) to find the numerical value: \( \sqrt{725} \approx 26.9 \) ft/s.

Key Concepts

calculus problems

Calculus problems often involve rates of change and their real-world applications in multiple dimensions. In this exercise, we explored how calculus can be used to solve a problem involving moving objects, such as a helicopter and a runner. These problems typically require a solid understanding of how to set up mathematical models that represent real-world situations.

To set up these models, we often need to:

• Define the variables that represent the changing quantities, such as time, distance, and speed.
• Establish a coordinate system to help visualize the problem, making it easier to apply calculus tools.
• Apply relevant theorems, like the Pythagorean theorem in this exercise, to relate different variables in the system.

Once these models are in place, calculus techniques like differentiation are used to find rates of change, offering solutions that apply to a wide range of practical problems.

rate of change

The rate of change is a fundamental concept in calculus that describes how a quantity changes over time. In our problem, we are interested in the rate at which the distance between the helicopter and a person running on the ground changes with respect to time.

To solve this, we:

• First, establish a function that represents the distance between the two moving objects.
• Then, we differentiate this function with respect to time to find the rate of change.
• The derivative, in this case, tells us how fast the distance is changing at any point in time.

The concept of differentiation is key here, as it allows us to calculate instantaneous rates of change, giving insight into how quickly changes are occurring at any given moment. Understanding how to connect these changes back to real-world movements, like the rising helicopter, helps clarify the solution further.

distance between moving objects

When dealing with the distance between moving objects, like in this exercise, one must account for the directional movement in space. The distance between objects typically involves both horizontal and vertical components, especially when they are moving in perpendicular directions.

For the helicopter and person scenario, we:

• Use the coordinate positions to create a right triangle at any given time.
• Apply the Pythagorean theorem, which links these components through the equation \( d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2} \), where \( d \) is the distance.
• Understand this distance will change as each component changes over time, related back to their respective velocities.

This approach helps in visualizing how moving in different dimensions (one vertically, one horizontally) affects the overall distance between objects, a common task in problems involving dynamic changes.