Two cars leave from the same point at the same time. One car an Learnin travels directly north and the other directly east. After 3 hours, the first car is 60 miles away from the Macm starting point and travels at 35 mph. At the same time, the second car is 80 miles to the east of the starting point and travels at 65 mph. At what rate is the distance between the two cars changing at this moment?

We are dealing with a related rates problem involving two cars moving away from a common starting point. One car travels north, and the other travels east. The goal is to determine how fast the distance between the two cars is changing at a specific moment.

### Step 1: Define Variables
- Let \( x \) represent the distance the eastbound car is from the starting point.
- Let \( y \) represent the distance the northbound car is from the starting point.
- Let \( z \) represent the distance between the two cars (the hypotenuse of the right triangle formed by their paths).
- We need to find \( \frac{dz}{dt} \), the rate at which the distance between the cars is changing.

We are given:
- After 3 hours, the first car (northbound) is 60 miles away, traveling at 35 mph. Thus, \( y(3) = 60 \) miles and \( \frac{dy}{dt} = 35 \) mph.
- After 3 hours, the second car (eastbound) is 80 miles away, traveling at 65 mph. Thus, \( x(3) = 80 \) miles and \( \frac{dx}{dt} = 65 \) mph.

### Step 2: Use the Pythagorean Theorem
The distance between the cars, \( z \), is given by:
\[
z^2 = x^2 + y^2
\]
Differentiating both sides of this equation with respect to time \( t \) using implicit differentiation:
\[
2z \frac{dz}{dt} = 2x \frac{dx}{dt} + 2y \frac{dy}{dt}
\]
Simplifying:
\[
z \frac{dz}{dt} = x \frac{dx}{dt} + y \frac{dy}{dt}
\]
We can solve for \( \frac{dz}{dt} \):
\[
\frac{dz}{dt} = \frac{x \frac{dx}{dt} + y \frac{dy}{dt}}{z}
\]

### Step 3: Calculate \( z \) at \( t = 3 \) hours
We know:
- \( x = 80 \) miles
- \( y = 60 \) miles

Using the Pythagorean theorem:
\[
z = \sqrt{x^2 + y^2} = \sqrt{80^2 + 60^2} = \sqrt{6400 + 3600} = \sqrt{10000} = 100 \text{ miles}
\]

### Step 4: Plug in Values
Now, substitute the known values into the equation for \( \frac{dz}{dt} \):
\[
\frac{dz}{dt} = \frac{80(65) + 60(35)}{100}
\]
\[
\frac{dz}{dt} = \frac{5200 + 2100}{100} = \frac{7300}{100} = 73 \text{ mph}
\]

### Final Answer:
The rate at which the distance between the two cars is changing after 3 hours is \( 73 \) mph.

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Would you like further details or have any questions about this solution?

Here are 5 related questions to expand your understanding:
1. What if the second car traveled northwest instead of east?
2. How would the result change if both cars traveled at constant speeds but along non-perpendicular paths?
3. How can you determine the exact position of each car after 4 hours?
4. What is the significance of the Pythagorean theorem in related rates problems?
5. What if the speeds of the cars increased after 3 hours? How would that affect the rate of change?

**Tip:** When dealing with related rates, always remember to carefully differentiate all variables with respect to time using implicit differentiation.

Two cars leave from the same point at the same time. One car travels directly north and the other directly east. After 3 hours, the first car is 60 miles away from the starting point and travels at 35 mph. At the same time, the second car is 80 miles to the east of the starting point and travels at 65 mph. At what rate is the distance between the two cars changing at this moment?

Learn how to solve a related rates problem involving two cars moving in perpendicular directions. After 3 hours, one car is 60 miles north traveling at 35 mph, and the other is 80 miles east traveling at 65 mph. Discover how fast the distance between the cars is changing using the Pythagorean theorem and implicit differentiation.

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