The volume of a cube is increasing at the rate of 4 cubic cm/minute.
How fast is the surface area increasing when the side is 3 cm?
Method 1: non calculus, non rigorous
When s=3, A=54 and V=27
1/1000 min later:
V=27.004,
s = V^(1/3) = 3.00014814
A = 6s^2 = 54.0053332
So A is increasing about 5.333 cm^2/min
Method 2: calculus
dV(t)/dt = 4 cm^3/min integrate
V(t) = 4t + C cm^3 where C is a constant
s(t) = (4t)^(1/3) cm
A(t) = 6 * 4^(2/3) * t^(2/3) cm^2 will differentiate this eqn.
wlog let V(0) = 0, ie let constant=0
Boundary condition is when t = 27/4 min,
V(27/4) = 27 cm^3,
x(27/4) = 3 cm
A(27/4) = 54 cm^2
dA(t)/dt = (2/3)*6*4^(2/3) * t^(-1/3)
dA(27/4)/dt = 4^(5/3) / (27/4)^(1/3)
= 4^(5/3) * 4^(1/3) / 3
= 4^2 / 3
= 16/3 = 5.3333333 cm^2/min
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Posted by Larry
on 2024-10-29 08:40:08 |
2 methods