The height (in miles) of an airplane descending from the sky is given by the function f(t)= -t^(3/2). Find the distance travelled by the plane while descending, if it takes 10 minutes to land.
Time is typically measured in seconds, so you should first try to turn those 10 minutes into seconds.
Then you should just be able to substitute that value into the formula given. Replace all instances of \(t\) with the value you get.
okay I got 14696.94 That doesn't look right at all.
I did 600^(3/2)
They're not asking height. They're asking distance which makes this an integral problem if I'm reading correctly.
yes thats correct osirisis
i was thinking the same as osirisis, that's why i deleted my suggestion :P
anyways i hope you get to solve it, gtg
Technically it should be \(-600^{\frac{3}{2}}\). It doesn't matter much though, the answer is correct. This will find the distance because at t=0, f(t)=0.
That's not a general rule, just that \(-0^{\frac{3}{2}}\) is 0
but that number i mentioned before isn't one of the possibilities. So there must be something missing.
What are the possibilities?
-600^(3/2)=-14696.94
29.54 miles 33.46 miles 36.74 miles 40.47 miles
In that case, perhaps using minutes would have been better, if quite unusual. \(-10^{\frac{3}{2}}=-31.62 \: miles\) , which is much closer.
that still doesn't deduce which answer i should choose.
I'm still trying to visualize an integral approach to this. Only thing I have so far is that the distance traveled is the sum of \[\Delta h \Delta t\].
That kind of makes sense..
Sorry I'm at a loss on this one. The graph of the function is already delta h I believe as it is a steadily decreasing value. My gut says to integrate but I have no clue on limits: \[distance=\int\limits_{?}^{?}-t^{\frac{3}{2}}dt=[-\frac{2}{5}t^{\frac{5}{2}}]_{?}^{?}\] Issue is it doesn't give a suitable answer no matter what limits are plugged in. You may want to check again when some of the other calculus folks log in later.
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