11. +-5.88 points LarCalcET6 3.7.033 My Notes Ask Your Teacher As a spherical ra
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11. +-5.88 points LarCalcET6 3.7.033 My Notes Ask Your Teacher As a spherical raindrop falls, it reaches a layer of dry air and begins to evaporate at a rate that is proportional to its surface area (S = 4r2). Show that the radius r of the raindrop decreases at a constant rate. (Let V represent the volume of the raindrop, t represent time, and use k for any necessary constant.) Because the evaporation rate is proportional to the -Selec' dt dt ds dt dV 4 dt However, because the volume of the spherical raindrop is V = 4r, we have However, because the volume of the spherical raindrop is Vwe have dV dt dt Therefore, the radius of the raindrop decreases at a constant rate dt Need Help? L: Radlt-| |Talk to a Tutor Submit Answer Save PrExplanation / Answer
If something decreases at constant rate, then its derivative is constant and negative
Raindrop evaporates (i.e. loses volume) at a rate that is proportional to its surface area:
dV/dt = c (4r²), where c is a positive constant
V = 4/3 r³
Now differentiate with respect to time(t):
dV/dt = 4/3 (3r²) dr/dt
c (4r²) = 4r² dr/dt
dr/dt = c (4r²) / (4r²)
dr/dt = c
Since c is positive constant, then dr/dt = negative constant.
Therefore radius is decreasing at constant rate.
Let x represent the length of the base of the triangle in the figure. Then
cot =x/4;
differentiating with respect to t yields
-csc2 d/dt = (1/4) dx/dt
we know dx/dt = -10 miles per minute because the jet is traveling 600 miles per hour. Thus
d/dt = -(1/4) sin2 dx/dt = (5/2) sin2
(i) When = 45, sin = 1/sqrt2 and so
d/dt = (5/2) sin2(45) = (5/2)x1/2 = 5/4 radian per minute.
(ii) When = 60, sin60 = sqrt 3/2 and so
d/dt = (5/2) sin2(60) = (5/2)x(3/4) = 15/8 radian per minute.
(iii) When = 80 so
d/dt = (5/2) sin2(80) = 2.42 radian per minute.
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