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?
To find the rate at which the distance between the two lovers is changing, we need to use the concept of relative velocity.
Let's start by visualizing the situation. We can consider the starting point of the girl as the origin (0,0) on a coordinate plane. The girl walks due south at 6 mph, which means her velocity vector is (0, -6). The boy walks at 10 mph on a heading of North 60 degrees West, which can be represented by a vector of (-10 cos 60°, -10 sin 60°) using the angle and magnitude.
Now, let's find the position vectors of the girl and boy after 30 minutes (0.5 hours). Since distance traveled is given by speed multiplied by time, the girl's position vector is (0, -6 * 0.5) = (0, -3) and the boy's position vector is (-10 cos 60° * 0.5, -10 sin 60° * 0.5) = (-5, -5√3).
To find the distance between the two lovers, we can calculate the magnitude of the difference vector between their positions. The difference vector is ((0 - (-5))^2 + (-3 - (-5√3))^2) = (5^2 + (-3 + 5√3)^2) = (25 + (9 - 30√3 + 75)) = (109 - 30√3).
Now, let's find the derivative of this distance function with respect to time (t) to determine how fast the distance is changing. The derivative gives us the rate of change, which represents the speed at which the distance between them is changing.
By differentiating, we get d(distance)/dt = -30√3. Note that the negative sign indicates that the distance is decreasing as time passes.
Therefore, 30 minutes later, the rate at which the distance between the two lovers is changing is -30√3 mph.
the distance z is found using
z^2 = u^2 + v^2 - 2uv cos 2pi/3
what are u and v? The girl's and boy's distances, where
u = -6t
v = 10t
z^2 = 136t^2 - 60t^2
z^2 = 76t^2
so, now you know you have to find z(1/2), and then use
z dz/dt = 152 t