Calculus Rate of Change Word Problem?
Please help me with this problem. I believe I'm doing most of it right but I got stuck at one part. The question is:
A hot-air ballon rises vertically as a rope attached to the base of the ballon is released at a rate of change of 5 ft/ sec. The pulley that releases the rope is 25 feet from the platform where the passengers board. At what rate is the ballon rising when 800 ft of rope has been payed out?
I set the equation up with a triangle where x=25 ft, y= the height of the ballon, and r=the rate of change/ 800ft
Lets say the base is constant i.e. x is 25’ and the y is constantly changing, plus, we know the hypotenuse (lets call this ‘h’) is changing at a rate of 5’/sec. Let's also make a note that when the balloon is resting on ground it would still have a rope of 25’ tied to it from the pulley. At this point x is 25’ and y is 0 and also h is same as 25’.
Lets look at the right triangle
(h+25)^2 = y^2 + x^2 .... since the initial condition says that h is same as x i.e. 25'
(h+25)^2 = y^2 + 25^2
y^2 = (h+25)^2 – 625
y = sqrt((h+25)^2 – 625)… sqrt is square root
Now rate of change of y is dy/dt. so lets differentiate
dy/dt = [1/(2((h+25)^2-625)^0.5)](2(h+25)) * dh/dt
Now simply substitute 800’ for h and 5’/sec for dh/dt
dy/dt = 5.00229 ft/sec
this seems odd but makes sense when u plug in different numbers for h as below
0 infinite
0.1 56.069
1 18.203
2 13.237
5 9.045
10 7.144
100 5.103
300 5.014
500 5.0056
800 5.00229
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