Find the rate of change of α.?

Question

2. Anne and Bill are watching a hot air balloon as it rises straight up from its launching pad in a flat field. At 1:30 the bottom of the balloon gondola is 47 feet above the landing pad and rising at a rate of 54 feet per minute. Anne is standing still on top of a 12 foot high ridge and is 38 feet west of the path of ascent. Bill is in the field 29 feet north of the pad and walking toward it at a rate of 78 feet per minute.
Anne’s eye level is at 5 feet. Let α be the angle of elevation as
Anne looks at the bottom of the gondola. Bill’s eye level is at 6 feet. Let β be the angle of
elevation as Bill looks at the bottom of the gondola.

Answer 1

Looking at the geometry of the scenario (no pun intended).
using 1:30 as t=0

tan(α)=(30+54*t)/38

t is expressed in minutes

the height above Anne's eye level is 47-12-5+54*t feet
and the horizontal distance is 38 feet

the rate of change is the first derivative.
The function can be expressed as
α=atan((30+54*t)/38)
which has the form of
h(g(x))
where h=atan(g)
and g=(30+54*t)/38
the first derivative of atan(g) is
1/(1+g^2)
and the first derivative of g is
54/38
using the chain rule
dα/dt=(1/(1+(30+54*t)^2/38^2))*54/38
simplify
dα/dt=(38*54/(38^2+(30+54*t)^2))
dα/dt=(38*54/(38^2+(30+54*t)^2))

dα/dt=513/(586+810*t+729*t^2)

You didn't ask for the rate of change of β. It is similar, you have to add the quotient rule, though. Here's the set-up

tan(β)=(41+54*t)/(29-78*t)
where t is expressed in minutes

you still do the first derivative of atan as 1/(1+x^2)
and apply the chain rule. In this case, though
g(x)=f(x)/q(x), where f(x)=41+54*t, and q(x)=29-78*t
the quotient rule applied yields
g'(x)=(q(x)*f'(x)-q'(x)*f(x))/(q(x))^2

It's some messy algebra.

j

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