Related Rates (1 Viewer)

nimrod_dookie

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Can anyone please provide assistance with the following related rates questions?

1. For the given mass of gas in a cylinder pv^1.5=400 where p is the pressure in Nm^-2 and v is the volume in cubic metres. If the pressure increases at 3Nm^-2 per minute find the rate at which the volume is changing at the instant when the pressure is 50 Nm-2.

2. Wheat runs from a hole in a silo at a constant rate and forms a conical heap whpse base radius is treble the height. If after 1 minute, the height of the heap is 20 cm, find the rate at which the height is rising at this instant.

Any help would be greatly appreciated thankyou.:)
 

Riviet

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1. Make v the subject of pv^1.5=400 (p.v3/2 or [pv]3/2?)

Then you can differentiate with respect to p to find dv/dp.

We are given dp/dt = 3,

Use the chain rule:

dv/dt = dv/dp x dp/dt

Substitute dv/dp, dp/dt and p=50 to find dv/dt.
 

ianc

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Question 2:


I am going to treat the falling wheat as a volume.
  • Volume of the formed cone = 1/3 [FONT=&quot]π r[/FONT]<sup>[FONT=&quot]2[/FONT]</sup>[FONT=&quot] h
    [/FONT]Since we know that r = 3h:
    V = 1/3 [FONT=&quot]π (3h)[/FONT]<sup>[FONT=&quot]2[/FONT]</sup>[FONT=&quot] h
    [/FONT]= [FONT=&quot]3πh[/FONT]<sup>[FONT=&quot]3 [/FONT]</sup>
  • Rate of falling wheat (the amount of wheat fallen = volume of cone)
    Let dv/dt = x
    Integrating:
    v=xt + c --> (c=0 because when t=0, v=0)
    v=xt
  • Hence xt = [FONT=&quot]3πh[/FONT]<sup>[FONT=&quot]3
    [/FONT]</sup>x = [FONT=&quot](3πh[/FONT]<sup>[FONT=&quot]3[/FONT]</sup>) / t
  • After 1 minute, h=20 cm
    Therefore x= 75398.2 [FONT=&quot]cm[/FONT]<sup>[FONT=&quot]3[/FONT]</sup>/minute
    Therefore dv/dt = 75398.2
  • v= [FONT=&quot]3πh[/FONT]<sup>[FONT=&quot]3
    [/FONT]</sup>dv/dh = [FONT=&quot]9πh[/FONT]<sup>[FONT=&quot]2
    [/FONT]</sup>
Therefore dh/dt = dv/dt * dh/dv
= 75398.2 / ([FONT=&quot]9πh[/FONT]<sup>[FONT=&quot]2 [/FONT]</sup>)

when h=20cm:
dh/dt = 6.67 cm/minute
 

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