A Maths問題

1
Let h be the height of water
w be the width of the water level

tanβ=10/60=1/6

w/2=htanβ

w=h/3

Let V be the volume of water

V=1/2wh(100)=50h^2/3

Differentiate both sides with respect to t

dV/dt=(50/3)(2h)dh/dt=(100h/3)dh/dt

Given dV/dt=200,when h=10

200=(100*10/3)dh/dt

dh/dt=0.6

So the depth of the water rise 0.6 cms^-1 when the depth is 10cm


2
Let r, h and s be the radius, height and slant height of cone.

Their relationship can be expressed as:

cos 30 = h/s

OR (√3/2) = h/s

OR h = (√3/2) s


tan 30 = r/h

OR 1/√3 = r/h

OR r = h/√3


sin 30 = r/s

OR 1/2 = r/s

OR s = 2r


As mentioned in question, if h = 4cm, s = 8/√3 cm and r = 4/√3

Volume of cone V = 1/3 π r^2 h = 1/9 π h^3

Differentiating both sides w.r.t. time t, we get:

dV/dt = 1/9 π 3h^2 dh/dt

dV/dt = 1/3 π h^2 dh/dt

8π = 1/3 π 4^2 dh/dt

dh/dt = 1.5

Thus, the rate of change of depth is 1.5 cm/sec

Then, the rate of change of radius is 1.5/√3 cm/sec

Then, the rate of change of slant height is 3/√3 cm/sec

Part a

Curved surface area A = π r s = 2π r^2

We try to differenciate both sides w.r.t time t, we get:

dA/dt = 2π(2r) dr/dt = 4πr dr/dt =4π (4/√3) (1.5/√3) = 8π

Thus, the rate of increase of the area of the wet surface is 8π cm^2/sec.

Part b

From the analysis part, we has already get:

dV/dt = 1/3 π h^2 dh/dt

(8π - dW/dt)= 1/3 π 4^2 (-1) = -16/3 π

dW/dt = 40/3 π where dW/dt is the rate of water flowing out from funnel.

Thus, the rate of water flows out from funnel is 40/3 π cm^3/sec.