Rates of change question?

Assuming the cliff is vertical.

Let A be the angle of depression of the boat when it is at a horizontal distant, x from the cliff, then,

cotA=x/100, =>

x=100cotA

dx/dA= -100cosec^2 A, also

v=dx/dt= -50m/min.

By the chain rule formula for differentiation,

dx/dt=(dx/dA)*(dA/dt), thus,

dA/dt=(dx/dt)/(dx/dA) =>

dA/dt=(1/2)sin^2 A

=(1/2){(2-sqrt3)/4}

dA/dt=(2-sqrt3)/8 rad/min

tan x = height/distance from bottom of cliff

let distance from bottom of cliff=y height=h=100m

tan x=h/y

h is constant

differentiate with respect to time

sec^2(x)dx/dt=(-h/y^2) (dy/dt)

dy/dt= 50m per min=50/60 m per sec

x=15 degrees y can be found using tanx=h/y

substitute values. and get the answer

tan (theta) = 100/x where x is the distance of boat from bottom of the cliff

x = 100* cot (theta)

Differentiate with respect to time

dx/dt = 100(-cosec^2(theta) (d(theta)/dt)

50 = 100{cosec^2(15){d(theta)/dt)

1/2 =3.8637033051562845275 {d(theta)/dt}

d(theta)/dt = (1/2)(1/3.8637033051562845275)

= 0.12940952255126 angles/minute .....................Ans