# Thread: Related rates question..here's my work

1. ## Related rates question..here's my work

A balloon rises at the rate of 8 feet per second from a point on the ground 60 feet from an observer. Find the rate of change of the angle of elevation when the balloon is 25 feet above the ground.

My work:

picture: http://i41.tinypic.com/14j0vuv.jpg

tan Θ=y/x
tan Θ=25/60
Θ = 22.78 degrees

derivative of tanΘ=y/x:
sec˛Θ(dΘ/dt) = [ (x)(dy/dt) - (y)(dx/dt) / x˛ ]
1.176(dΘ/dt) = [ 60(8) - (25)(19.2) / 60˛ ]

^^I got the dx/dt number, 19.2 by setting up this proportion:
60/x = 25/8

But then I realized that when I solve this:
60(8) - (25)(19.2), I get 0...what am I doing wrong??

**EDIT: Sorry, you're right, I meant dx/dt and dy/dt

Thanks!

2. In your picture, it looks like you're using dy/dx for the change in height, but the height y is not changing with respect to the change in the ground-distance x; it is changing with respect to the time t. So it should be "dy/dt = 8".

Also, since the balloon is rising directly vertically, then the ground-distance x is not changing; dx/dt = 0.

You have tan(@) = y/x. (I'm using "@" to stand for the angle "theta".) Differentiating with respect to time t, I believe you should get:

. . . . .sec^2(@) (d@/dt) = [(dy/dt)x - y(dx/dt)] / x^2

Since the secant is, for right triangles such as this one, equal to "(hypotenuse) over (adjacent)", use the Pythagorean Theorem to find the value of the hypotenuse when y = 25 and x = 60.

Then plug in the known values for sec(@), dy/dt, dx/dt, x, and y. Solve for d@/dt.