Related Rates Problem: Two cars start moving from the same point. One travels south at 40 mi/h and the other travels west at 30 mi/h. At what rate is the distance between the cars increasing three hours later?
To solve this related rates problem, we can use the Pythagorean theorem to determine the distance between the two cars.
Let's assume that the initial position of the cars is the origin (0,0) on a coordinate plane. After three hours, the car traveling south will be 40 * 3 = 120 miles south, and the car traveling west will be 30 * 3 = 90 miles west.
Let's represent the distance between the two cars as "d." According to the Pythagorean theorem, we have:
d^2 = (120)^2 + (90)^2
d^2 = 14400 + 8100
d^2 = 22500
d = sqrt(22500)
d = 150 miles
Now, let's differentiate both sides of the equation with respect to time to find the rate at which the distance between the cars is increasing:
2d * d/dt = 0
2 * 150 * d/dt = 0 (taking the square root of both sides)
300 * d/dt = 0
d/dt = 0/300
d/dt = 0 mi/h
Therefore, three hours later, the distance between the cars is not changing, so the rate at which the distance between the cars is increasing is 0 mi/h.
To solve this related rates problem, we need to find the rate at which the distance between the two cars is changing with respect to time.
Let's denote the position of the first car (traveling south) as x, and the position of the second car (traveling west) as y. Since they both start from the same point, we can consider this point as the origin.
Now, let's express the distances x and y as functions of time t. The first car is traveling south at a speed of 40 mi/h, so we can write x = 40t. Similarly, the second car is traveling west at a speed of 30 mi/h, so we can write y = 30t.
The distance between the two cars is given by the formula for the distance between two points in a Cartesian coordinate system, which is the square root of the sum of the squares of the differences in the x and y coordinates:
d = sqrt(x^2 + y^2)
Substituting the expressions for x and y, we get:
d = sqrt((40t)^2 + (30t)^2)
= sqrt(1600t^2 + 900t^2)
= sqrt(2500t^2)
= 50t
Now, we want to find the rate at which the distance d is changing with respect to time t. This is the derivative of d with respect to t:
dd/dt = d/dt (50t)
= 50
Therefore, the rate at which the distance between the two cars is increasing three hours later is 50 mi/h.
after t hours,
car One has traveled 40t miles
car Two has traveled 30t miles
Using the Pythagorean Theorem, the distance between the cars is 50t miles.
Looks like the distance is growing at a constant 50 mi/hr. No calculus even needed.