If a hot air balloon pilot inflates the balloon at an average wind speed of about 5 miles per hour with an altitude of 35,000 feet, how fast will the radius of the balloon increase as it goes up higher and higher performing certain tricks? Using the differentiation formula applicable for this situation, estimate the speed of inflation and the speed of landing. Will this be able to help the pilot still perform other tricks before making a colorful landing? Give at least two comprehensible reasons and explain your answer.

"Differentiation, in mathematics, process of finding the derivative, or rate of change, of a function. In contrast to the abstract nature of the theory behind it, the practical technique of differentiation can be carried out by purely algebraic manipulations, using three basic derivatives, four rules of operation, and a knowledge of how to manipulate functions.

The three basic derivatives (D) are: (1) for algebraic functions, D(xn) = nxn − 1, in which n is any real number; (2) for trigonometric functions, D(sin x) = cos x and D(cos x) = −sin x; and (3) for exponential functions, D(ex) = ex."

Let the length of the rope = r

Height of the balloon = h

we are given dr/dt = 15ft/sec

at any time t we know:

[tex]r = 125 + 15*t[/tex]

By the Pythagoras theorem we know:

[tex]h^2 + 125^2 = r^2[/tex]

We have to find dh/dt at time t = 20 sec

[tex]d/dh(h^2 + 125^2) = d/dt(r^2)[/tex]

By chain rule:

[tex]d/dh(h^2 + 125^2)*dh/dt\\ = d/dr(r^2)*dr/dt\\2h*dh/dt = 2r* dr/dt\\h*dh/dt = r*dr/dt\\dh/dt = (r/h)*dr/dt[/tex]

[tex]= [(125 + 15*20)/(425^2 - 125^2)^1/2]*15[/tex]--------------------- at time t = 2

=15.69 ft/sec

Hence dh/dt = 15.69 ft/sec

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