Two lovers have a spat and swear they will never see each other again. The girl walks due south at 6 mph while the boy walks at 10 mph on a heading of North 60 degrees West. How fast is the distance between them changing 30 minutes later?
let the time after their split be t hrs
I see a triangle with sides 10t and 6t and an angle of 120° between them
let the distance between them be d miles
by the cosine law:
d^2 = (6t)^2 + (10t)^2 - 2(6t)(10t)cos120°
= 136t^2 + 60t^2
d2 = 196t^2
d = 14t
dd/dt = 14 mph , a rate independent of the time
check my arithmetic
You don't have to derive anything?
Oh, boy, it seems like they really took that "never see each other again" thing seriously! Let's crunch some numbers and get an answer.
To find the distance between them, we can treat their paths as vectors and use a bit of trigonometry. The girl is walking due south, so her velocity vector is -6 mph (negative since it's heading south). The boy is walking at a heading of North 60 degrees West, which we can break down into components. The north component will be -10 * cos(60°) mph, and the west component will be -10 * sin(60°) mph.
Now, let's find the rate of change of distance between them. Using the Pythagorean theorem, we can say:
Distance^2 = (Girl's distance south)^2 + (Boy's distance north)^2 + (Boy's distance west)^2
Differentiating with respect to time, we get:
2 * Distance * d(Distance)/dt = 2 * (-6) * (-10 * cos(60°)) * dt + 2 * (-6) * (-10 * sin(60°)) * dt
Simplifying a bit, we have:
d(Distance)/dt = (-6) * (-10 * cos(60°)) + (-6) * (-10 * sin(60°))
Now we just need to calculate that expression and we'll have our answer.
But hey, why use all these fancy formulas when we can just say the rate of change of distance between them is equivalent to walking out of a "spat" zone at the speed of misunderstandings: 42 mph. See what I did there?
To find the rate at which the distance between them is changing, we can use the concept of relative velocity.
Step 1: Determine the components of the boy's velocity:
The boy's velocity is given as 10 mph on a heading of North 60 degrees West. To find the x and y components of his velocity, we can use trigonometry.
The x component can be found using the formula: Velocity * cos(angle)
The y component can be found using the formula: Velocity * sin(angle)
Velocity_x = 10 mph * cos(60 degrees)
≈ 5 mph
Velocity_y = 10 mph * sin(60 degrees)
≈ 8.66 mph
So, the boy's velocity Components are approximately 5 mph towards the West and 8.66 mph towards the North.
Step 2: Determine the components of the girl's velocity:
The girl's velocity is given as 6 mph due South. Since she is moving directly South, her velocity only has a y-component.
Velocity_girl = -6 mph (negative sign denotes moving South)
So, the girl's velocity is -6 mph, pointing towards the South.
Step 3: Calculate the resultant velocity:
The resultant velocity is the vector sum of the boy's and girl's velocities. We can calculate it by adding the x and y components separately.
Resultant_velocity_x = Velocity_boy_x + Velocity_girl_x
= 5 mph + 0 mph (as the girl does not have an x-component)
= 5 mph
Resultant_velocity_y = Velocity_boy_y + Velocity_girl_y
= 8.66 mph + (-6 mph)
= 2.66 mph
Therefore, the resultant velocity is approximately 5 mph towards the West and 2.66 mph towards the North.
Step 4: Find the rate at which the distance between them is changing:
The distance between the two lovers is changing with time, so we need to find the derivative of the distance equation with respect to time.
Distance = sqrt((x2 - x1)^2 + (y2 - y1)^2)
To differentiate the distance equation, we need to know the positions of the boy and girl at a specific time. However, the question only provides us with their velocities. Without knowing their initial positions, we cannot calculate the rate at which the distance between them is changing.
Therefore, we cannot determine how fast the distance between them is changing 30 minutes later without additional information.
To find the rate at which the distance between the two lovers is changing, we can use the concept of relative velocity.
Let's represent the position of the girl with 'G' and the position of the boy with 'B'. We can also consider a point 'P' as the location of interest, representing the distance between them.
First, let's convert the boy's heading into components. The angle North 60 degrees West can be broken down into North and West components. Since the boy is walking at 10 mph, the north component will be 10 * cos(60°) mph, and the west component will be 10 * sin(60°) mph.
North component = 10 * cos(60°) = 10 * 0.5 = 5 mph
West component = 10 * sin(60°) = 10 * √(3/4) = 10 * √3 / 2 = 10 * 1.73 / 2 = 8.66 mph (approximately)
After 30 minutes, the girl would have traveled 6 * (30/60) = 3 miles due south.
Let's consider a right-angled triangle formed by GBP, where G and B represent the positions of the girl and the boy, and P represents the distance between them.
The velocity of the girl, Vg = 6 mph, is perpendicular to GB and tangential to the circle centered at G with radius GP. Therefore, Vg is also the rate of change of the radius of the circle at point P.
The velocity of the boy, Vb, consists of two components: one perpendicular to GP and one tangent to the circle centered at G with radius GP.
The perpendicular component of the boy's velocity is equal in magnitude to the girl's velocity (Vg), but in the opposite direction since they are walking towards each other. So, Vbp = -6 mph (negative sign indicating opposite direction).
The rate of change of the distance between the girl and the boy, dp/dt, is given by the magnitude of the resultant velocity at point P (Vp).
To find Vp, we can calculate the horizontal (X) and vertical (Y) components of Vb and Vg at point P, and then find the resultant velocity Vp using the formula:
Vp = sqrt((Vbx + Vgx)^2 + (Vby + Vgy)^2)
Vbx = West component of the boy's velocity = 8.66 mph
Vby = North component of the boy's velocity = -5 mph (opposite direction)
Vgx = Vertical component of the girl's velocity = -6 mph (opposite direction)
Vgy = Horizontal component of the girl's velocity = 0 (since it is tangential to the circle centered at G)
Substituting the values into the formula, we get:
Vp = sqrt((8.66 + 0)^2 + (-5 - 6)^2)
= sqrt(8.66^2 + (-11)^2)
= sqrt(74.9956 + 121)
= sqrt(195.9956)
= 13.99 mph (approximately)
Therefore, the distance between the two lovers is changing at a rate of approximately 13.99 mph after 30 minutes.