Cameron is located 2 miles due East of an intersection between two country roads. He decides to walk in a straight line (cross-country) toward a position 3 miles due North of the intersection; it takes Cameron one hour to reach that location.
(a) Find parametric equations for the location of Cameron at time t hours.
(b) Find the rate of change of the distance between Cameron and the intersection at time 15 minutes (t=1/4 hr).
is parametric equation same as the uniform linear motion? where x(t)=a+bt , y(t)c+dt would I use this to solve for a?
Yes.
X=2-2t
y=3t
To solve part (a) of the problem and find parametric equations for Cameron's location at time t hours, we can use two variables to represent his position: x for the distance traveled eastward from the intersection and y for the distance traveled northward.
Let's assume that at time t=0, Cameron is at the intersection between the two country roads. Since he walks in a straight line, his speed can be considered constant.
From the given information, we know that Cameron walks 2 miles due east and then 3 miles due north from the intersection. Since it takes him one hour to reach the final location, we can determine his speed as follows:
Speed = Distance/Time
Speed = (2 miles + 3 miles)/1 hour
Speed = 5 miles/hour
Now, let's find the parametric equations using the concepts of velocity and position. Velocity is the rate of change of position over time.
Given:
Initial position (x₀, y₀) = (0, 0) at t=0
Velocity (Vx, Vy) = (2 miles/hour, 3 miles/hour)
The parametric equations for Cameron's location at time t hours can be written as:
x = x₀ + Vx * t
y = y₀ + Vy * t
Substituting the given values, we have:
x = 0 + (2 miles/hour) * t
y = 0 + (3 miles/hour) * t
Therefore, the parametric equations for Cameron's location at time t hours are:
x = 2t
y = 3t
Now let's move on to part (b) and find the rate of change of the distance between Cameron and the intersection at time t=15 minutes (t=1/4 hr).
To find the rate of change, we need to calculate the time derivative of the distance function between Cameron and the intersection.
Distance = √(x² + y²)
Differentiating with respect to time, we get:
d(Distance)/dt = d(√(x² + y²))/dt
Using the chain rule, this simplifies to:
d(Distance)/dt = [(x * dx/dt) + (y * dy/dt)] / √(x² + y²)
Substituting the parametric equations we found earlier:
x = 2t
y = 3t
dx/dt = 2
dy/dt = 3
We can evaluate the rate of change at t=1/4 hr (15 minutes):
d(Distance)/dt = [(2t * 2) + (3t * 3)] / √((2t)² + (3t)²)
= (4t + 9t) / √(4t² + 9t²)
= (13t) / √(4t² + 9t²)
= 13t / √(13t²)
= 13t / (|t| √13)
Substituting t=1/4 hr:
d(Distance)/dt = 13 * (1/4) / (|1/4| √13)
= 13/4 / (√13/4)
= (13/4) * (4/√13)
= 13 / √13
Therefore, the rate of change of the distance between Cameron and the intersection at time t=15 minutes is 13 / √13 miles per hour.