Car A is 30 mi apart from car b. Car A moves 30 mph due east at the same time car b moves 30 degrees south of east. how fast is the distance between them changing after an hour.
How are the two cars positioned relative to each other ?
e.g. is car A directly west of car B ?
is car A north of car B ?
I really don't know. We don't have any figure or anything. I just assumed there position in cartesian plane at (0,0) for Car B, and Car A at (30,0)
Ok, I will go with that assumption.
As I made my diagram, I also noticed that you did not state the speed of the car B.
I will also assume that the speeds are the same.
At a time of t hrs, let the distance between them be x miles
so by the cosine law:
x^2 = (30+30t)^2 + (30t)^2 - 2(30t)(30+30t)cos45°
then find dx/dt
Is there a typo in the last sentence?
"how fast is the distance between them changing after an hour." ?? , an hour ??? you mean 1 hour ?
if so, sub t = 1 to find x, then dx/dt
To find how fast the distance between the two cars is changing, we need to use the concept of related rates. Let's break down the problem and use trigonometry to solve it.
Let's assume that the position of Car A is the origin (0,0) and Car B is located at point B (30 miles apart from Car A). We need to find how fast the distance between Car A and Car B is changing after an hour.
The position of Car A after an hour can be calculated by using the formula: (x, y) = (initial position) + (velocity) * (time). In this case, Car A is moving 30 mph due east, so its position after an hour is (30, 0).
Car B is moving 30 degrees south of east. To find its position after an hour, we can use trigonometry. We can calculate its change in x-coordinate and y-coordinate.
Change in x-coordinate for Car B = distance * cos(angle)
= 30 miles * cos(30 degrees)
= 30 miles * √3 / 2
= 15√3 miles
Change in y-coordinate for Car B = distance * sin(angle)
= 30 miles * sin(30 degrees)
= 30 miles * 1/2
= 15 miles
So, the change in position of Car B after an hour is (∆x, ∆y) = (15√3, 15).
Now, let's find the distance between the two cars after 1 hour: distance = √((∆x)^2 + (∆y)^2)
distance = √((15√3)^2 + 15^2)
= √(675 + 225)
= √900
= 30 miles
Therefore, the distance between Car A and Car B after 1 hour is 30 miles.
To find how fast the distance between them is changing, we differentiate the distance function with respect to time.
d(distance) / dt = d/dt (√(900))
d(distance) / dt = (1/2) * (900)^(-1/2) * d(900) / dt
d(distance) / dt = (1/2) * (900)^(-1/2) * 0
Since the distance is not changing, the rate of change is zero. Therefore, after an hour, the distance between the two cars is not changing.