Related Rates Question (1 Viewer)

[_ShiFTy_]

A rubber washer is being compressed. At a certain time, the outer diatmeter is found to be 3cm, the inner diameter is 1cm, the thickness of the washer is decreasing at a rate of 1/4 cm/min and the outer diameter is increasing at a rate of 1/2 cm/min. If the volume remains pi cm^3, at what rate is the inner diameter changing...

V = pi/4 (D^2 - d^2).T

dT/dt = -1/4
dD/dt = 1/2
dV/dt = 1?

 

[_ShiFTy_]

Answer is 7/2...anyone else?

 

[Mountain.Dew]

im not too sure if 7/2 is the right answer...putting it into context, 7/2 does seem an unlikely answer. i got 1/2 instead.

here is my working...

please discuss if u feel i have done something fishy...

 

[alcalder]

Can't be. The answer has to be negative. The inner diameter would be decreasing. Surely??

EDIT: Mountain.Dew, problem with what you have done is the rate of change of diameter is given, not radius.

So, using diameter, I still don't get 7/2 but -1/2.

If one thinks creatively about 7/2 it is conceivable that in -1/2 the negative sign was very close to the answer and was transcribed as 7/2.