A gallon rises at the rate of 8 feet per second from a point on the ground 12 feet from an observer. To 2 decimal places in radians per second, find the rate of change of the angle of elevation when the ball on is 9 feet above the ground. Type your answer in the space bellow. If your answer is a number less than 1, place a leading "0" before the decimal point (ex:0.35)

Question

A gallon rises at the rate of 8 feet per second from a point on the ground 12 feet from an observer. To 2 decimal places in radians per second, find the rate of change of the angle of elevation when the ball on is 9 feet above the ground. Type your answer in the space bellow. If your answer is a number less than 1, place a leading "0" before the decimal point (ex:0.35)

Answer 1

Make your sketch. (What is rising, a gallon, a ball or a balloon ????)

let the height of "whatever is rising" be y ft
let the angle of elevation be Ø
tanØ = y/12
y = 12 tanØ
dy/dt = 12sec^2 Ø dØ/dt

in your sketch, let y = 9
then h^2 = 12^2 + 9^2 = 225
h = 15
then cosØ = 12/15 = 4/5
secØ = 5/4
sec^2 Ø = 25/16

now you got all your values to put into
dy/dt = 12sec^2 Ø dØ/dt

Answer 2

Yes I'm sorry, i meant to type "balloon". I'm still a little confused, can you give me the exact number that is the answer so I can better understand please?

Answer 3

Giving you the "exact" number, will not help you understand it, the solution I gave you

will help you understand it.

Answer 4

ok but I need to finish with the exercise with the answer but I don't understand. Can you finish with my question please?

Answer 5

To find the rate of change of the angle of elevation, we need to calculate the derivative of the angle with respect to time. Given that the gallon rises at a rate of 8 feet per second, we can use the formula for the angle of elevation:

tan(angle) = (height of the gallon) / (distance from the observer)

At a height of 9 feet, the distance from the observer remains constant at 12 feet. So we can write:

tan(angle) = 9 / 12

To find the derivative, we can apply the derivative of the inverse tangent function to both sides of the equation:

d/dt(tan(angle)) = d/dt(9 / 12)

Using the chain rule, we can rewrite this as:

sec^2(angle) * d(angle)/dt = 0

Since sec^2(angle) is never zero, we can divide both sides by sec^2(angle) to solve for d(angle)/dt:

d(angle)/dt = 0 / (sec^2(angle))

The derivative of the angle of elevation with respect to time is therefore equal to zero, since sec^2(angle) is nonzero for all angles.

So, the rate of change of the angle of elevation when the gallon is 9 feet above the ground is 0 radians per second.