I see what I did, I took the derivative instead of the integral. I need to use the arctan formula instead.
60(1/(1 + t^2)) when you take the integral with the arctan formula
formula:
integral(du/(a^2 + u^2)) = 1/a(arctan(u/a) + C)
so
integral 60(1/1(arctan(t/1)) from 0 to 1...
Ahh, that does make sense that I don't integrate the amount of air that escapes the balloon within a minute but rather the rate at which the air escapes the balloon per minute. If I do that I should integrate 60/(1 + t^2) from 0 to 1. Once done I get 30 as an answer.
Now that I think about it couldn't I find where the slope is undefined by finding where the first derivative is undefined?
If I have
dy/dx = (-3x^2 - x)/(x-2y)
could I set the denominator = 0 and solve?
The curve defined by x3 +xy - y2 = 10 has a vertical tangent line when x = ?
To find when the tangent line is vertical, could I find when the slope is undefined for the original function? The way I previously tried this problem was by taking the first derivative and then finding when x = 0...
Ok, I see that 30t is the rate at which the air escapes the balloon. To find how much air leaves the balloon in a minute, I must take the integral from 0 to 1 of 30t.
Once completed, I got 15 cubic feet escaped the balloon in one minute. Is that even possible? If the air escapes at 30cubic...
Alright thanks, I think I have a further understanding of this now. I did actually when I first started the problem, plug 10 and 10.02 into get the change in volume. Of course that wasn't an available answer and it wasn't the "approximate" answer. But after taking the derivative of the sphere...
Air is escaping from a balloon at a rate of R(t) = 60/(1+t2) cubic feet per minute, where t is measured in minutes. How much air, in cubic feet, escapes during the first minute?
dr/dt = 60/(1 + t2)
if I plug the time in (1) for (t) I should get 60/2 or 30
This seems a tad bit too easy...