At noon, Ship A is 100 km west of ship B. Ship A travels south at 35 km/h. Ship B travels North at 25 km/h. At 4 pm, how fast the distance between them change?
Make your diagram using an x-y grid
at noon, place ship B at the origin, and A on the left side of the x-axis
After some time of t hrs,
let B's position on the y-axis be Q, mark BQ=25t
draw a vertical line in the South direction and label A's position as P, marking AP=35t
Draw a horizontal line from P to hit the y-axis at R.
Now complete a right-angled triangle QPR, with PR = 100, QR = 25t+35t or 60t
We need d(PQ)/dt
PQ^2 = 100^2 + (60t)^2 = 10000+3600t^2
2 PQ d(PQ)/dt = 7200t
so at 4 pm, t = 4
PQ^@ = 100^2 +240^2
PQ = 260
2(260) d(PQ)/dt = 7200(4)
d(PQ)/dt = 55.38 km/h
check my arithmetic
To find how fast the distance between Ship A and Ship B is changing at 4 pm, you need to calculate the derivatives of their positions with respect to time, and then find the rate of change of the distance between them.
Let's start by calculating the positions of Ship A and Ship B at 4 pm.
At noon, Ship A is 100 km west of Ship B. So, at 4 pm (4 hours later), Ship A would have traveled south for 4 hours at a speed of 35 km/h, covering a distance of 4 * 35 = 140 km. Therefore, Ship A's position at 4 pm would be 100 km (west) - 140 km (south) = -40 km (west).
On the other hand, Ship B travels north at 25 km/h from its original position. So, at 4 pm (4 hours later), Ship B would have traveled north for 4 hours at a speed of 25 km/h, covering a distance of 4 * 25 = 100 km. Therefore, Ship B's position at 4 pm would be 100 km (north) from its original position.
Now, let's calculate the distance between Ship A and Ship B at 4 pm.
The distance between two objects in a coordinate plane is given by the Pythagorean theorem:
distance = √((x2 - x1)^2 + (y2 - y1)^2)
In this case, the distance between Ship A and Ship B at 4 pm is:
distance = √((-40 - 0)^2 + (0 - 100)^2)
distance = √(1600 + 10000)
distance = √11600
distance ≈ 107.68 km
To find how fast the distance between them is changing at 4 pm, we need to calculate the rate of change of the distance with respect to time.
Using the chain rule of calculus, we'll differentiate the above equation with respect to time.
Let's denote the distance between Ship A and Ship B at 4 pm as D(t), where t is the time in hours.
D(t) = √((-40 - 0)^2 + (0 - 100)^2)
D(t) = √1600 + 10000
D(t) = √11600
To find dD/dt (the rate of change of the distance), we differentiate D(t) with respect to t:
dD/dt = 1/(2√11600) * d(11600)/dt
dD/dt = 1/(2√11600) * 0
dD/dt = 0
Therefore, the rate at which the distance between Ship A and Ship B is changing at 4 pm is 0 km/h. This means that at 4 pm, the distance between them is not changing.