Math Problem Statement

You and a friend are riding your bikes to a restaurant that you think is east; your friend thinks the restaurant is north. You both leave from the same point, with you riding at 16 mph east and your friend riding 12 mph north. After you traveled 4 mi, at what rate is the distance between you changing?

Solution

This problem involves related rates, where we need to determine the rate at which the distance between two moving objects is changing over time. We can use the Pythagorean theorem and derivatives to solve this.

Step 1: Set up the coordinates

Let’s place your starting point at the origin (0,0)(0, 0) on a coordinate plane:

  • You are riding east at a speed of 16 mph.
  • Your friend is riding north at a speed of 12 mph.

After some time tt hours:

  • The distance you have traveled east is x(t)=16tx(t) = 16t miles.
  • The distance your friend has traveled north is y(t)=12ty(t) = 12t miles.

Step 2: Apply the Pythagorean theorem

The distance between you and your friend at time tt is the hypotenuse of a right triangle, where the legs are the distances you and your friend have traveled east and north, respectively. Let d(t)d(t) represent the distance between you and your friend:

d(t)=x(t)2+y(t)2d(t) = \sqrt{x(t)^2 + y(t)^2} d(t)=(16t)2+(12t)2=256t2+144t2=400t2=20td(t) = \sqrt{(16t)^2 + (12t)^2} = \sqrt{256t^2 + 144t^2} = \sqrt{400t^2} = 20t

Step 3: Differentiate with respect to time

Now, to find the rate at which the distance between you and your friend is changing, we need to differentiate d(t)d(t) with respect to time tt:

dd(t)dt=ddt(20t)=20\frac{dd(t)}{dt} = \frac{d}{dt}(20t) = 20

Thus, the distance between you and your friend is changing at a constant rate of 20 miles per hour.

Step 4: Find when t=416t = \frac{4}{16} (since you have traveled 4 miles)

Since you traveled 4 miles east, the time tt at that moment is:

t=416=0.25hours.t = \frac{4}{16} = 0.25 \, \text{hours}.

At this time, the rate at which the distance between you and your friend is changing remains 20 mph, as it is independent of tt.

Final Answer:

The distance between you and your friend is changing at a rate of 20 mph when you have traveled 4 miles.

Would you like more details or have any questions about the solution?

Here are 5 related questions to extend your understanding:

  1. How does the rate of change in distance vary if your speed changes?
  2. What would happen if both of you traveled in the same direction?
  3. Can you solve for the distance between you and your friend after 0.5 hours?
  4. How would the rate of change be affected if you traveled at 20 mph instead of 16 mph?
  5. What if your friend traveled in a direction 30° east of north?

Tip: Related rates problems often rely on the Pythagorean theorem when distances are involved, especially with right triangles.

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Math Problem Analysis

Mathematical Concepts

Related Rates
Differentiation
Pythagorean Theorem

Formulas

Pythagorean Theorem: d(t) = sqrt(x(t)^2 + y(t)^2)
Differentiation of distance: dd(t)/dt = 20

Theorems

Pythagorean Theorem

Suitable Grade Level

Grades 10-12