finding rate of depth decrease of an inverted cone
A funnel has a circular top of diameter 20cm and a height of 30cm. When the depth of the liquid in the funnel is 12cm, the liquid is dropping at a rate of 0.2cm²/s. At what rate is the depth of the liquid in the funnel decreasing at this moment
this was found in a chapter using the "chain rule" aka "composite function rule" or "function of a function rule" which states:
if y = f(F(x)), and u = F(x) so that y=f(u), then dy/dx = dy/du*du/dx
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We'll start by recalling the formula that gives the volume of the cone:
V = `pi` `r^(3)` h/3
In this case, w'ell consider V as being the volume of the liquid remained in the funnel, h is the depth of remaining liquid and r is the radius of the circle which represents the base of the new cone formed by the remaining liquid.
Therefore, if h = 12, dV/dt = -0.2
Since in the formula that gives the volume of the cone, we have two variables, we'll express the radius in terms of the height. For this reason, we'll use similar triangles:
r/h = (20/2)/30 = 1/3
r = h/3
We'll re-write the formula that gives the volume of cone, in terms of h.
V = `pi` `h^(3)` /27
We'll differentiate with respect to t:
dV/dt = (3`pi``h^(2)` /27)(dh/dt)
dV/dt = (`pi` `h^(2)` /9)(dh/dt)
But dV/dt = -0.2
-0.2 = (` `144`pi` /9)(dh/dt)
-0.2 = (16`pi`) (dh/dt)
(dh/dt) = -0.2/16`pi`
(dh/dt) `~~` -0.003 cm/s
Therefore, the depth of the liquid in the funnel is decreasing at a rate of 0.003 cm/s.
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