Rate of change

1.
(a)
V = the volume of the spherical balloon
r = the radius of the spherical balloon

V = (4/3)πr³
dV/dt = (4/3)π(3r²)(dr/dt)
dV/dt = 4πr²)(dr/dt)

When dV/dt = 80 cm³/min and r = 10 cm :
80 = 4π(10)²(dr/dt)
Hence, dr/dt = 0.2/π cm/min

(b)
A = the surface area of the spherical balloon

A = 4πr²
dA/dt = 4π(2r)(dr/dt)
dA/dt = 8πr(dr/dt)

When r = 10 cm, and dr/dt = 0.2/π cm/min
dA/dt = 8π(10)(0.2/π)
dA/dt = 16 cm²/min


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2.
h = the distance between the top of the ladder and the ground
b = the distance between the foot of the ladder and the ground

b² + h² = 13² (Pythagorean theorem)
(d/dt)( b² + h²) = (d/dt)13²
2b(db/dt) + 2h(dh/dt) = 0
b(db/dt) + h(dh/dt) = 0

When b = 5m, db/dt = 0.4 m/s, and h = √(13² - 5²) = 12
5(0.4) + 12(dh/dt) = 0
dh/dt = -1/6 m/s


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3.
x = the horizontal distance between the boat and the lighthouse

tan y = 20/x
(d/dx)(tan y) = (d/dx)(20/x)
(sec²y)(dy/dt) = -(20/x²)(dx/dt)

When dx/dt = -4m/s, x = 50, and sec²y = (20² + 50²)/50² = 29/25
(29/25)(dy/dt) = -(20/50²)(-4)
dy/dt = 4/145 rad/s

2012-05-27 23:54:19 補充：
1.
Since volume is decreasing, dr/dt and dV/dt are thus negative, i.e.
......
......
When dV/dt = -80 cm³/min and r = 10 cm :
-80 = 4π(10)²(dr/dt)
Hence, dr/dt = -0.2/π cm/min ...... (answer)

comment on micatkie's solution to Q1a and Q1b:
negative signs are required as the radius and surface area are decreasing
ie, the solution should be : Q1a) -0.2/π cm/min Q1b) -16 cm²/min