Problem 71 - Calculus for Biology and Medicine

Question

Two people start biking from the same point. One bikes east at \(15 \mathrm{mph}\), the other south at \(18 \mathrm{mph}\). What is the rate at which the distance between the two people is changing after 20 minutes and after 40 minutes?

Step-by-Step Solution

Answer

The rate of change of distance at 20 minutes is 23.43 mph, and at 40 minutes, it is also 23.43 mph.

Step 1: Define the Scenario

Let's name the person riding east as Person A and the person riding south as Person B. We have the following information: Person A bikes at 15 mph, and Person B bikes at 18 mph. Both start from the same point.

Step 2: Calculate Distance Traveled

After time t (in hours), Person A's distance is \( 15t \) miles east, and Person B's distance is \( 18t \) miles south. For 20 minutes, \( t = \frac{20}{60} = \frac{1}{3} \) hours, and for 40 minutes, \( t = \frac{40}{60} = \frac{2}{3} \) hours.

Step 3: Express the Distance Formula

We need the rate of change of the distance between Person A and Person B. Apply the Pythagorean theorem: the distance d between them is \( d = \sqrt{(15t)^2 + (18t)^2} \).

Step 4: Differentiate with Respect to Time

Differentiate \( d = \sqrt{(15t)^2 + (18t)^2} \) with respect to time: \[ \frac{dd}{dt} = \frac{1}{2\sqrt{(15t)^2 + (18t)^2}} \cdot (2 \cdot 15t \cdot 15 + 2 \cdot 18t \cdot 18) \]Simplifying, we get: \[ \frac{dd}{dt} = \frac{15^2 t + 18^2 t}{\sqrt{(15t)^2 + (18t)^2}} \]

Key Concepts

Distance Formula

The distance formula is a key part of understanding how to determine the separating distance between two moving points. Imagine two bikers starting from the same location. As one moves east and the other south, they form a perpendicular angle with their paths. The distance between them is like the hypotenuse of a right triangle formed by their separate paths. This is where the Pythagorean theorem comes in handy. To find the distance between these two riders, you use the formula:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\]
For our bikers:

• The eastward distance: \(15t\)
• The southward distance: \(18t\)

So the distance formula becomes:
\[ d = \sqrt{(15t)^2 + (18t)^2}\]
This helps us calculate how far apart they are after a certain time.

Differentiation

Differentiation is the process of finding the rate at which something changes. In our biking problem, we want to find out how fast the distance between the two biking friends is increasing over time. Differentiation helps to determine this rate of change. It involves taking the derivative of a function with respect to a variable, which in this case is time.
We'll take the derivative of the distance function:
\[ \frac{dd}{dt} = \frac{1}{2\sqrt{(15t)^2 + (18t)^2}} \cdot (2 \cdot 15t \cdot 15 + 2 \cdot 18t \cdot 18)\]
Simplifying it gives:
\[ \frac{dd}{dt} = \frac{15^2 t + 18^2 t}{\sqrt{(15t)^2 + (18t)^2}}\]
This expression represents how quickly the distance between the bikers changes.

Biking Problem

Let's put our understanding of the distance formula and differentiation together in the context of this biking problem. When the bikers start from the same location, they form a right triangle as they move. The eastbound biker covers a distance of \(15t\) and the southbound biker \(18t\), after time \(t\) in hours. To solve this problem, we first calculate the distances they travel for the given times:

• At 20 minutes \(t = \frac{1}{3}\) hours
• At 40 minutes \(t = \frac{2}{3}\) hours

After calculating with the distance formula, we differentiate to find how fast the distance is changing, using the earlier derived differentiation formula. This provides valuable insight into such real-world scenarios where understanding rates of change is essential. This biking problem illustrates the practical application of calculus in daily life.