Question

What is the rate of change of the angle of elevationin(rad/sec)when the balloon is 40m above the ground ?

Answer 1

We make a drawing to find the relationship between the angle and the sides

given
dh/dt=4m/sec

we get
tan(x)=h/40
where x is the angle and h is the height
take the derivitive of both sides
1/cos²(x)(x)dx/dt=$\frac{h'40-0}{40^2}$
1/cos²(x)dx/dt=$\frac{h(dh/dt)40}{1600}$
1/cos²(x)dx/dt=$\frac{h(dh/dt)}{40}$
solve for dx/dt

times both sides by cos²(x)
dx/dt=$\frac{h(dh/dt)cos²(x)}{40}$
remember that dh/dt=4
also, solve for the angle at that time

h=40 and base=40 so x=45 degrees (pi/4 radians)

input that
dx/dt=$\frac{4cos²(x)}{40}$
dx/dt=$\frac{cos²(x)}{10}$
dx/dt=0.05

A is the answer