need urgent help with calculus optimization !!!!!!!!!?

Question

A weather balloon is rising vertically at a rate of 2 ft/sec. An observer is situated 300 ft from a point on the ground directly below the balloon. At what rate is the distance between the balloon and the observer changing when the altitude of the balloon is 500 ft?

Answer 1

this is not an optimization problem, but a related rates problem. you are given the rate of change of the height of the balloon and asked for the rate of change of the distance.

you find the relationship of these rates by first finding and equation of the variables height and distance. then, just differentiate this equation (with respect to time) to get the relation of their rates. Since you are given one rate (balloon) you can solve for the other.

Call the distance D and the height of the balloon h, it is pretty easy to see that
D^2 = h^2 +300^2.
Now, implicitly differentiate wrt t and you get
2D dD/dt = 2h dh/dt

you want to solve for dD/dt when h = 500. At this height,
D = sqrt(500^2 + 300^2)
and dh/dt is always 2 ft/sec

So, dD/dt = 2 * 500 * 2 / (2 * sqrt(500^2 + 300^2))

Answer 2

Using the height of the balloon as the vertical side of the triangle (we'll call it b) and the distance of the observer from the ground below the balloon is the horizontal side of the triangle (we'll call it a), we can find the hypotenouse (c) of the triangle and it's derivative. First, c^2 = a^2 + b^2. c^2 = (300)^2 + (500)^2, and so c = 100(rad34). Taking the derivative of the equation,
2c(c') = 2a + 2b(b'), we plug in our values of a, b, and c, as well as (b'), which is 2 ft/sec. (Note that a does not change, it remains at 300 ft, so there is no a', but if that doesn't work and give the right answer, then set a' as 0 and just solve 2c(c') = 2b(b')). Then you solve for c' . Hope that helps!

Answer 3

If y is the height of the balloon and z is the distance between the observer and the balloon, then by the Pythagorean Theorem:

300² + y² = z²

When y = 500, z = √(300² + 500²) = 100√34

Also, using implicit differentiation of 300² + y² = z²:

2yy' = 2zz' so yy' = zz'.

y' is 2ft/sec, y is 500, z is 100√34, so

z' = 500(2)/100√34 = 5/√34 = 5√34/34

I think.