F.4 app. of differention

1. Volume, V = (4/3)pr^3 where p = pi
so dV/dt = 4pr^2 (dr/dt).
Now dV/dt = 80, r = 10, so dr/dt = 80/400p = 1/5p cm/min.
Surface area , S = 4pr^2
so dS/dt = 8pr(dr/dt) = 80p(1/5p) = 16 cm^2/min.

2. Let side of rectangle be 2x and 2y, so area, A = 4xy.
A^2 = 16x^2y^2
Sub x^2/16 + y^2/9 = 1 into it, we get
A^2 = 144x^2[1 - x^2/16] = 144x^2 - 9x^4
Differentiate both sides,
2A dA/dx = 288x - 36x^3
Put dA/dx = 0, since A is not zero, 288x - 36x^3 = 0
36x(8 - x^2) = 0
x = 0 (rej.) or x = - sqrt 8 (rej.) or x = sqrt 8 = 2 sqrt 2.
When x = 2 sqrt 2, sub into the curve equation,
8/16 + y^2/9 = 1
y^2/9 = 1/2
y = sqrt (9/2) = 3/(sqrt 2)
So max. area A = 4(2 sqrt 2)(3/ sqrt 2) = 24.
Remark : Please check that this is a max. by calculating d^2A/dx^2 yourself.

我想確定一下, rate of change of volume of spherical ballon 是80cm^3/min 還是
80πcm^3/min?

有沒有 π 的?