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Topic: Calculus related rates hot air balloon problem? (Read 9564 times) |
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iatkrox
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Calculus related rates hot air balloon problem?
« on: Oct 17th, 2012, 1:13am » |
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A hot air balloon rising vertically is tracked by an observer located 4 miles from the lift-off point. At a certain moment, the between the observer's line-of-sight and the horizontal is pi/4 , and it is changing at a rate of 0.1 rad/min. How fast is the balloon rising at this moment?
I'm stuck on this one, any help would be great. Thanks.
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rmsgrey
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Re: Calculus related rates hot air balloon problem
« Reply #1 on: Oct 17th, 2012, 4:35am » |
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The balloon is 4 miles away horizontally, and pi/4 radians above the horizontal, so is 4 miles up.
The tangential component of its velocity is 4 miles * sin(pi/4) * 0.1 rad/min = 0.4 sqrt(2) miles/min.
The actual velocity is at pi/4 radians to the tangential component, so the actual velocity is 0.4*sqrt(2) * cos(pi/4) miles/min = 0.8 miles per minute.
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iatkrox
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Re: Calculus related rates hot air balloon problem
« Reply #2 on: Oct 18th, 2012, 6:44am » |
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Are you sure with the answer ? 
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Grimbal
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Re: Calculus related rates hot air balloon problem
« Reply #3 on: Oct 18th, 2012, 7:04am » |
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Looks correct to me.
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rmsgrey
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Re: Calculus related rates hot air balloon problem
« Reply #4 on: Oct 19th, 2012, 5:12am » |
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I would draw a picture, but it's easy enough to draw your own:
The launch site, the balloon's current position, and the balloon's position in time dt from now are all on the same vertical line (in ascending order). Add in the observer on a horizontal line with the launch site, draw the lines of sight from the observer to current balloon and future balloon, and the perpendicular to the observer - future balloon line through the current balloon.
You can label the horizontal line and the lower angle at the observer easily enough, and the upper angle at the observer is dt times the rate of change of angle. For small enough dt, the middle angle at the current balloon is near enough another right-angle, From there, you should be able to label every angle and length either absolutely or in terms of dt - the only thing you might not know is that, for a small enough angle (in radians), dT, you can replace sin(dT) with dT.
Once you have the distance between current balloon and future balloon in terms of dt, just divide by dt to get the average speed over that time.
As dt approaches 0, the errors of approximation above disappear (the "close enough" right angle and using dT rather than sin dT) and you get an exact speed - if you want to make it more rigorous, you could track the errors and up with something like:
dh = 4*sin(pi/4)*sin(dT)*cos(pi/4+dT)
= 4*sin(pi/4)*(dT+O(dT3))*(cos(pi/4)*(1-O(dT2))-sin(pi/4)(dT+O(dT3)))
= 4*sin(pi/4)*dT*cos(pi/4) - O(dT2)*sin(pi/4) + O(dT3)
= 8dT - O(dT2)*sin(pi/4)
dT = 0.1dt so dh/dt = 0.8 - O(dt) -> 0.8 as dt -> 0
The only use of anything approaching calculus is in the use of limits to justify giving 0.8 as the exact answer rather than as an arbitrarily good approximation, and, possibly, indirectly in the expansions for trig expressions:
cos(a+b) = cos(a)cos(b)-sin(a)sin(b)
cos(dT) = 1-dT2/2+O(dT4) and
sin(dT) = dT - dT3/6+O(dT5)
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