Question

PLEASE HELP!! --> A balloon rises at a rate of 4 meters per second from a point on the ground 40 meters from an observer. What is the rate of change of the angle of elevation (in rad/sec) when the balloon is 40 meters above the ground?

Answer 1

Step 1: draw a diagram.

If you done that, tell us what an expression for angle of elevation, call it A in terms of 40 meters and the balloon's height, h. And what is height h as a function of time t?

Answer 2

I don't understand

Answer 3

When the balloon is h meters off the ground, what is the trigonometric expression for the angle of elevation. It involves something like sin or cos or tan or cot or sec or cosec.

Answer 4

sin

Answer 5

The horizontal distance of 40 m and the vertical height of h, form two sides of right-angled triangle.

The observer is sitting at the end of the horizontal 40 m looking up at the balloon. The angle at which she's looking is the angle of elevation, call it A.

What is the trig relationship between that angle, h and 40?

Answer 6

That is, she's looking along the diagonal towards the balloon.

Answer 7

opp/adj

Answer 8

I'm still waiting for an equation in which A appears.

Answer 9

tan A = h/40

That's the equation.

Answer 10

Now, the height h = 4t, because it's rising at a constant velocity. Hence

tan A = 4t/40 = t/10

or

A = arctan(t/10).

The question now asks you for dA/dt when the balloon is 40 meters off the ground.

So you need to
a) calculate dA/dt in general
b) find the value of t for which h = 40 m
c) evaluate dA/dt for that value of t.

Answer 11

ok i got .24