q-7 How would i make an equation?

Question

A balloon is rising at a rate of 4 meters per second from a point on the ground 53 meters from an observer. Find the rate of change of the angle of elevation from the observer to the balloon when the balloon is 27 meters above the ground.

Answer 1

construct a right triangle with y=height= 27, x = base = 53

y'=8 ft/s

dT/dt = T' wgeb t - 27

tanT = y/53

y = 53tanT

y'= *69tanT)'= 60sec^2(T)(T')

4=53sec^(T)(T')

T' = (4/53)cos^(T) = (4cos^(T)/53

base = 53, height = 27

hypotenuse = c = sqr(27^2+53^2) = about 59.48

when y=25 then cosT= 53/59.48=.89a

T'=- (4/53)cos^2(T) = (4/53)(.891)^2=about .06 radians per second

rate of change of the angle = about .06 rad/s = about 3.4 degrees per second

rate of change = .06 rad/sec or 3.4 degrees per second

present angle is tan^-1(27/53) = about 26.99584 = about 27 degrees

the angle one second later will be tan^-1((27+4)/53) = tan^-1(31/53) = about 30.3 degrees

the angle increased about 30.3-27 = 3.3 degrees in one second

rate of change =about 3.3 degrees per second

given a small margin of error due to rounding approximations

try a smaller change in time such as 1/8 second

then height changes from 27 to 27.5

angle = tan^-1(27.5/53) = 27.42

the angle increased from about 27 to 27.42 which is a .42 increase in 1/8 second

= .42x8 = 3.4 degree increase per second

angle's rate of change = 3.4 degrees per second

derivative of arctanu = 1/(1+u^2)

u=27/53

(arctanu)' = 1/(1+(27/53)^2) = about .794

= about .8 times 4 = 3.2 degrees per second rate of change

go with 3.4 degrees per second rate of angle's change