A car is traveling north towards an intersection at 60mph at the same time a truck is headed east toward the same intersection at 45mph. Find the rate of change of the distance between the car and truck when the car is 3 miles south of the intersection and the truck is 4 miles west of the intersection. Is this distance increasing or decreasing at that time?
Place the car on the negative y axis and label the distance to the origin y
Place the truck on the negative x axis and label the distance to the origin x
label the distance between them d
then
d^2 = x^2 + y^2
2d dd/dt = 2x dx/dt + 2y dy/dt
dd/dt =(x dx/dt + y dy/dt)/d
Given: when x = 4 , dx/dt = -45
when y = 3 , dy/dt = -60
d = 5 , (did you notice the 3-4-5 right-angled triangle ?)
dd/dt = (4(-45) + 3(-60)/5 = -72
At that moment the distance between them is decreasing at 72 mph
The distance between them is decreasing, since dd/dt is negative.
12
To find the rate of change of the distance between the car and the truck, we can use the concept of Pythagoras' theorem. Let's consider the car's position as the origin (0,0) and the intersection as the point (0,0). Using this, we can consider the position of the car as (0, -3) (3 miles south of the intersection), and the position of the truck as (-4, 0) (4 miles west of the intersection).
Now, we can calculate the distance between the car and the truck using the distance formula:
Distance = √((x2 - x1)^2 + (y2 - y1)^2),
where (x1, y1) represents the car's position and (x2, y2) represents the truck's position.
Distance = √((-4 - 0)^2 + (0 - (-3))^2)
= √(16 + 9)
= √25
= 5 miles.
Thus, when the car is 3 miles south of the intersection and the truck is 4 miles west of the intersection, the distance between them is 5 miles.
To find the rate of change of this distance, we can differentiate the distance equation with respect to time. Let's assume that the car and truck maintain their respective speeds of 60 mph and 45 mph.
Let t be the time elapsed since they passed their starting points.
x1 = 60t (equation for the car's horizontal position)
y1 = -3
x2 = -4
y2 = 45t (equation for the truck's vertical position)
Distance = √((60t - (-4))^2 + (-3 - 45t)^2)
Differentiating both sides of the equation with respect to t, we get:
d(Distance)/dt = (120t + 8)(60) / √((60t - (-4))^2 + (-3 - 45t)^2) + (-90t - 6)(45) / √((60t - (-4))^2 + (-3 - 45t)^2)
To determine if the distance is increasing or decreasing, we need to evaluate d(Distance)/dt at the given values of t.
When the car is 3 miles south of the intersection and the truck is 4 miles west of the intersection, we can calculate t using the car's position equation:
-3 = 60t
t = -3/60
t = -1/20 hour.
Plugging this value of t into d(Distance)/dt:
d(Distance)/dt = (120(-1/20) + 8)(60) / √((60(-1/20) - (-4))^2 + (-3 - 45(-1/20))^2) + (-90(-1/20) - 6)(45) / √((60(-1/20) - (-4))^2 + (-3 - 45(-1/20))^2)
Simplifying this expression will give us the rate of change of the distance between the car and truck at that specific time.