15) A rock is dropped from a height of 576 ft. Write the position function assum

Question

15) A rock is dropped from a height of 576 ft. Write the position function assuming that the rock is dropped on earth and is acted upon by gravity b) How long does it take for the rock to hit the ground? c) What is the average velocity of the rock for the first 3 seconds? d) What is the instantaneous velocity of the rock at time t 3 seconds? e) What is the acceleration at time t 3 seconds? 16) Suppose a liquid is to be cleared of sediment by pouring it through a cone-shaped filter. The height of the cone is 16 in. and the radius at the base is 4 in. If the liquid is flowing out of the cone at a rate of 2 in per minute how fast is the depth of the liquid decreasing when the level is 8 in. deep? 17) Find the extreme values of fix) 6x 3-3x3 on [-1, 1]

Explanation / Answer

y = h0 + v0*t - 1/2gt^2

y = 576 + 0 - 1/2*32*t^2

y = 576 - 16t^2 ----> ANS

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b)
When it hits grnd, y = 0

0 = 576 -16t^2
t^2 = 576/16
t^2 = 36
t = 6

So, 6 seconds to hit grnd

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c)
Avg vel :
s(0) = 576
s(3) = 576 - 16(3)^2 = 432

aVG VEL = (432 - 576) / (3 - 0) = -48 FT/S

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d)
Inst vel at t = 3 :
y = 576 - 16t^2

Deriivng :
v = -32t

-32*3

= -96 ft/s ----> ANS

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e)
a = dv/dt

a = -32 ft/s^2 ---> ANS

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16)
h = 16 and r = 4
So, r/h = 4/16
r/h = 1/4
So, r = h/4

V = 1/3 * pi * r^2 * h

V = 1/3 * pi * h^2/16 * h

V = pi*h^3 / 48

Deriving :
dV/dt = pi/48 * 3h^2 * dh/dt

-2 = pi/16 * 8^2 * dh/dt

-32 = 64pi * dh/dt

dh/dt = -1/(2pi) in/min

So,
the ht is decreasing at
1/(2pi) inches per min ----> ANS

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17)
Deriving :
f' = 8x^(1/3) - x^(-2/3) = 0

8x^(1/3) = 1/x^(2/3)

x = 1/8 which does lie in [-1,1]

Now, when we plug in x = 1/8,
we gte f = -9/8

Now, is this a max or min?
Lets find f''

f'' = 8/3x^(-2/3) + 2/3x^(-5/3)

Plug in x = 1/8, we get :
f'' = 32 > 0

So, this is a LOCAL MIN

So, local min
at (1/8 , -9/8) -----> ANS

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