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Magazine Chapter 3: Problem 43
A runner sprints around a circular track of radius 100 \(\mathrm{m}\) at a constant speed of 7 \(\mathrm{m} / \mathrm{s}\) . The runner's friend is standing at a distance 200 \(\mathrm{m}\) from the center of the track. How fast is the distance between the friends changing when the distance between them is 200 \(\mathrm{m} ?\)
Short Answer
Expert verified
The distance between the friends is changing at approximately 6.06 meters per second.
Step by step solution
01
Understand the Problem
You have a circular track with a radius of 100 meters. A runner is moving at 7 meters per second along this track. The runner's friend is standing 200 meters away from the center of the track. You need to find how fast the distance between the runner and the friend is changing when the distance between them is exactly 200 meters.
02
Draw a Diagram
Imagine a circle where the radius is 100 meters. Place the runner on the circumference of this circle. The friend's location forms a right triangle with the center of the circle and the point where the runner is located. Use this visualization to understand the relation between the runner, the friend, and the center of the circle.
03
Identify the Variables
Let \( R = 100 \, \mathrm{m} \) (radius of the circle), \( v = 7 \, \mathrm{m/s} \) (speed of the runner), and \( d(t) \) be the distance between the runner and their friend. Let \( \theta(t) \) be the angle, as measured from the center of the track between the friend's line and the runner's line of sight.
04
Use Related Rates
The distance between the runner and the friend forms the hypotenuse of a right triangle with the runner's path and the line connecting the center of the track to the friend. When \(d=200\), use the Pythagorean theorem to relate \(d\), \(R\), and the distance from the center to the friend: \[ d^2 = 100^2 + x^2 \] Differentiate with respect to time: \[ 2d \frac{dd}{dt} = 2x \frac{dx}{dt} \] Here, \(\frac{dx}{dt}\) is the velocity component in the direction of the friend.
05
Calculate \(x\) and Differentiate
For \(d=200\) meters, solve for \(x\): \[ 200^2 = 100^2 + x^2 \] Simplifying, \[ 40000 = 10000 + x^2 \] \[ x^2 = 30000 \] \[ x = \sqrt{30000} = 100\sqrt{3} \] Now substitute in the differentiation equation: \[ 400 \frac{dd}{dt} = 200\sqrt{3} \times 7 \] \[ \frac{dd}{dt} = \frac{200\sqrt{3} \times 7}{400} \] Finally, calculate \(\frac{dd}{dt}\): \[ \frac{dd}{dt} = \frac{7\sqrt{3}}{2} \approx 6.06 \] m/s.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Understanding Right Triangles
When we talk about right triangles, we refer to the geometric shape that consists of three sides, with one angle measuring exactly 90 degrees. In the context of this problem, the right triangle is part of the diagram that relates the runner, their friend, and the center of the circular track. This triangle helps visualize the spatial relationship between these points.
The specific right triangle here has:
The specific right triangle here has:
- One leg being the radius of the circular track, which is 100 meters.
- Another leg extending from the center of the circle to where the runner's friend stands, 200 meters away.
- The hypotenuse, which is the direct line connecting the runner and their friend, represents the distance we are concerned with.
The Pythagorean Theorem Applied
The Pythagorean theorem is a cornerstone of geometry. It states that in a right triangle, the square of the length of the hypotenuse (\(c\)) is equal to the sum of the squares of the other two sides (\(a\) and \(b\)): \[ c^2 = a^2 + b^2 \] . In this problem, the theorem helps us quantify relationships between the sides of our right triangle.
For our exercising runner:
For our exercising runner:
- \(a\) is the radius of the track, 100 meters.
- \(b\) is the remaining side, which needs to be solved.
- \(c\) is the hypotenuse, the direct distance between the runner and their friend.
Using Differentiation in Related Rates
Differentiation allows us to find how variables change in relation to each other. The method, called related rates, applies here to determine how quickly the distance between the friends changes. By setting our formula, derived from the Pythagorean theorem, for the runner's motion:\[ 2d \frac{dd}{dt} = 2x \frac{dx}{dt} \] , we see how differentiation plays a key role.
To tackle this:
To tackle this:
- \(\frac{dx}{dt}\) is the runner's speed contribution towards the change in distance.
- Plugging known quantities into the differentiated equation allows the calculation of \( \frac{dd}{dt} \), telling us how fast the distance between the runner and their friend is decreasing or increasing at the moment in question.
Circle Geometry Insights
Circle geometry helps unfold the relationship of a point's trajectory along a circle. With a fixed radius, like our circular track of 100 meters, and a running variable, like the runner's position, we can explore geometric principles that inform movement and distances.
Key points include:
Key points include:
- The circle's center acts as a fixed reference point from which all points on the circle's circumference maintain a constant distance, the radius (100 meters).
- By understanding a semi-circle, triangle, or segment of this circle, we can infer distances along these geometric moves, crucial for understanding time-bound changes in distances.
- The geometry of circles informs precise calculations around arcs, tangents, and segments, allowing for a better grasp of the runner and their friend鈥檚 spatial unfolding.
