A port and a radar station are 2 mi apart on a straight shore running east and west. a ship leaves the port at noon traveling northeast at a rate of 15 mi/hr. if the ship maintains its speed and course, what is the rate of change of the tracking angle theta between the shore and the line between the radar station and the ship at 12:30 pm?

The rate of change of the tracking angle is 0.05599 rad/sec

Step-by-step explanation:

Here the ship is traveling at 15 mi/hr north east and

Port to Radar station = 2 miles

Distance traveled by the ship in 30 minutes = 0.5 * 15 = 7.5 miles

Therefore the ship, port and radar makes a triangle with sides

2, 7.5 and x

The value of x is gotten from cosine rule as follows

x² = 2² + 7.5² - 2*2*7.5*cos(45) = 39.04

x = 6.25 miles

By sine rule we have

Therefore,

α = Angle between radar and ship α

∴ α = 58.052

Where we put

to get

and differentiate to get

= 3.208 degrees/second = 0.05599 rad/sec.

x = central angle apc = measure of minor arc ac

the two angles x and 30 add to 180, so x+30 = 180 and x = 180-30 = 150. the measure of minor arc ac is 150 degrees.

what we need to do from here is find the circumference and then multiply it by x/360 since there are 360 degrees in a circle. in this case, x = 150, so x/360 = 150/360 = 15/36 = 5/12. the shortest distance along the curve from a to c is 5/12 of the distance around the entire circle (aka circumference)

c = circumference

c = pi*d

c = pi*24

c = 24pi

now multiply this by 5/12 to get

distance along minor arc ac = (5/12)*(circumference)

distance along minor arc ac = (5/12)*(24pi)

distance along minor arc ac = 10pi

the formula you can use is

l = (x/360)*pi*d

with

l = length of minor arc

x = central angle

d = diameter

which essentially is the steps we took though a bit differently.

the answer is:

mean: 6

mad: 2.14