Can someone help with #13,14
1 answer:
13) The graph is a negative parabola. The flare shoots up and gravity pushes it back down. The flare hits the water when the height equals zero. To put it in simple words, <u>you are looking for are the x-intercepts.</u>
h = -16t² + 104t + 56
0 = -16t² + 104t + 56
0 = -8(2t² - 13t - 7)
0 = 2t² - 13t - 7
0 = 2t² + t - 14t - 7
0 = 2t( t + 1) - 7(t + 1)
0 = (2t - 7)(t + 1)
0 = 2t - 7 0 = t + 1
t = = 3.5 t = -1
Since time cannot be negative, disregard t = -1
Answer: t = 3.5 seconds
14) same as #13
h = -16t² + 64t
0 = -16t² + 64t
0 = -16( t² - 4)
0 = t² - 4
0 = (t - 2) (t + 2)
0 = t - 2 0 = t + 2
t = 2 t = -2 (disregard t = -2)
Answer: t = 2
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.
Now,
, and 
. The intersection is thus parameterized by the vector-valued function
where
. The arc length is computed with the integral
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Complete the square to get
So in the integral, you can substitute
to get
Next substitute
, so that the integral becomes
This is a fairly standard integral (it even has its own Wiki page, if you're not familiar with the derivation):
So the arc length is
The arc length can be computed with a line integral, but first we'll need a parameterization for
Now,
where
Some rewriting:
Complete the square to get
So in the integral, you can substitute
Next substitute
This is a fairly standard integral (it even has its own Wiki page, if you're not familiar with the derivation):
So the arc length is