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Rate of change help plz (1 Viewer)

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1. A vessel is of such a shape that when the depth of water is x cm the volume is V cm^3, where V = 30x+x^3. Water is pured into the empty vessel at the rate of 25cm^3/s. At what rate is the level of the water rising when the depth is 5cm?

Thanks in Advance
 

~caramello~

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we need to find dx/dt (rate the water level is changing)

dx/dt = dV/dt x dx/dV
we know dV/dt = 25 cm^3/s and that
V = 30x + x^3
dV/dx = 30 + 3x^2
dx/dV = 1/(30+3x^2)

therefore dx/dt = 25 x 1/(30+3x^2)
where x = 5
therefore dx/dt = 0.238
 
Joined
Mar 29, 2009
Messages
48
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Male
HSC
2009
we need to find dx/dt (rate the water level is changing)



dx/dt = dV/dt x dx/dV

we know dV/dt = 25 cm^3/s and that

V = 30x + x^3

dV/dx = 30 + 3x^2

dx/dV = 1/(30+3x^2)



therefore dx/dt = 25 x 1/(30+3x^2)

where x = 5

therefore dx/dt = 0.238
Thanks caramello
this was from fitzpatrick book rite?? i think i remember it xD
lol it's from Jones & Couchman book
Probably different numbers then i guess?
 

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