Question

A balloon, 50 feet from an observer is rising at 20 ft/sec. At 5 sec after lift off,

a)how fast is the distance between the observer and the balloon changing?

b)How fast is the angle of elevation changing.

I just need to know what formals to use. I know the first part requires tan = x/50, but I'm not sure what to do next, and what to do for B.

Answer 1

as you say,
tanθ = x/50
so, sec^2 θ dθ/dt = 1/50 dx/dt

You know x at t=5, and you know dx/dt=20, so you can find θ.

The distance z is given by

z^2 = 50^2 + x^2
z dz/dt = x dx/dt

so, find z at t=5 and plug in the numbers.

Answer 2

Thanks for your help, but sorry, I'm still kinda confused.

I'm not sure how to find x. Do I need it to solve for z.

z^2 = 50 ^2 + x^2

Sorry if this is a silly question

Answer 3

well, of course. (it is kind of a silly question) You know after 5 seconds the balloon has gone up 5*50 = 250 feet.

Now you know x and you can get z.