# P3 DifferentiationWatch

This discussion is closed.
#1
edges of a cube are length of xcm.
given that vol of cube is being increased at pcm^3s^-1 where p is a constant.
calculate in terms of p in cm^2 s^-1, rate at which the surface area of cube increase when x = 5..

thanks
0
13 years ago
#2
(Original post by Vishpatel)
edges of a cube are length of xcm.
given that vol of cube is being increased at pcm^3s^-1 where p is a constant.
calculate in terms of p in cm^2 s^-1, rate at which the surface area of cube increase when x = 5..

thanks
You know dV/dt=p
You want to find dS/dt at x=5.
To do this you need to use the chain rule:
dS/dt=dV/dt.dS/dV.
So you'll need an expression connecting S and V.
The surface area of the cube is 6x^2 while the volume is x^3.
dS/dx=12x, dV/dx=3x^2
dS/dV=dS/dx.dx/dV=12x/(3x^2)=4/x
So dS/dt= (p)(4/x), x=5 so dS/dt=4p/5 and so the surface area is increasing at the rate of (4/5)p cm^2 s^-1.
0
13 years ago
#3
volume = (x^3)
surface area = 6(x^2)

dv/dt = p
ds/dt = ?

chain rule says dv/ds would be useful as...

dv/ds = (dv/dt)(dt/ds)

dv/dx = 3(x^2)
ds/dx = 12x

dv/ds = (dv/dx)(dx/ds)
dv/ds = (dv/dt)(dt/ds)

(dv/dt)(dt/ds) = (dv/dx)(dx/ds)
(ds/dt) = (ds/dx)(dx/dv)(dv/dt)
ds/dt = [12x][p]/[3(x^2)]
ds/dt = 4p/x

x = 5 => ds/dt = 4p/5 (cm^2)(s^-1)
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