Explanation:
It is given that,
A surface ship is moving in a straight line (horizontally) at 16 km/hr, [tex]\dfrac{dx}{dt}=16\ km/hr[/tex]
At the same time, an enemy submarine maintains a position directly below the ship while diving at an angle that is 35 degrees below the horizontal, [tex]\theta=35[/tex]
We need to find the rate of decrease of the submarine's altitude i.e. [tex]\dfrac{dy}{dt}[/tex]
Using trigonometry, [tex]cos(35)=\dfrac{x}{y}[/tex]
[tex]y=sec(35)\ x[/tex]
On differentiating the above equation we get :
[tex]\dfrac{dy}{dt}=sec(35).\dfrac{dx}{dt}[/tex]
[tex]\dfrac{dy}{dt}=sec(35).\times 16[/tex]
[tex]\dfrac{dy}{dt}=19.53\ km/hr[/tex]
So, the submarine's altitude is decreasing at the rate of 19.53 km/hr. Hence, this is the required solution.
