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Q. A funnel (cone shaped) has a circular top of diameter 20 cm and a height of 30 cm. When the depth of liquid in the funnel is 15 cm, the liquid is dripping from the funnel at a rate of 0.2 cm^3 s^-1. At what rate is the depth of the liquid in the funnel decreasing at this instant?

brut

Q. A funnel (cone shaped) has a circular top of diameter 20 cm and a height of 30 cm. When the depth of liquid in the funnel is 15 cm, the liquid is dripping from the funnel at a rate of 0.2 cm^3 s^-1. At what rate is the depth of the liquid in the funnel decreasing at this instant?

Let:

V = Volume of water

h = Depth

t = Time

r = Radius of cone =

dh/dt = dh/dv x dv/dt [Chain rule]

But V = (1/3)πr²h

Thus dV/dh = (1/3)πr²

And so dh/DV = 3 ÷ πr²

Work out the radius when h = 15 based on h=30 when 2r = 20:

I think I'm right in saying that the diameter will halve, hence r = 5 at this point

Therefore, using dh/dV = 3 ÷ 25π and dv/dt = 0.2 (from data given):

dh/dt = (3 ÷ 25π x 0.2

dh/dt = 3 ÷ 125π

I hope that's right, anyway

dV/dt = -0.2 (given in the question)

V=h/3 * πr² (volume of liquid in the cone)

Ratio r/h is constant (steepness of the cone). Initially r=10 and h = 30 therefore r = h/3.

Substitute back into V.

V=(h/3)³π

Differentiating:

dV/dh=(h/3)²π

We want dh/dt, which is dh/dV * dV/dt (chain rule)

So, at dV/dt = -0.2 and h=15:

dh/dt = (-0.2)/((15/3)²π )

=-1/125π

Hope that helps, let me know if you need any of it clearing up.

Edit: Beaten to it, this forum's quick. JohnSPals, it looks like you differentiated with respect to h treating r as a constant. I don't think you can do that because r is variable with h.

V=h/3 * πr² (volume of liquid in the cone)

Ratio r/h is constant (steepness of the cone). Initially r=10 and h = 30 therefore r = h/3.

Substitute back into V.

V=(h/3)³π

Differentiating:

dV/dh=(h/3)²π

We want dh/dt, which is dh/dV * dV/dt (chain rule)

So, at dV/dt = -0.2 and h=15:

dh/dt = (-0.2)/((15/3)²π )

=-1/125π

Hope that helps, let me know if you need any of it clearing up.

Edit: Beaten to it, this forum's quick. JohnSPals, it looks like you differentiated with respect to h treating r as a constant. I don't think you can do that because r is variable with h.

Robert602

dV/dt = -0.2 (given in the question)

V=h/3 * πr² (volume of liquid in the cone)

Ratio r/h is constant (steepness of the cone). Initially r=10 and h = 30 therefore r = h/3.

Substitute back into V.

V=(h/3)³π

Differentiating:

dV/dh=(h/3)²π

We want dh/dt, which is dh/dV * dV/dt (chain rule)

So, at dV/dt = -0.2 and h=15:

dh/dt = (-0.2)/((15/3)²π )

=-1/125π

Hope that helps, let me know if you need any of it clearing up.

Edit: Beaten to it, this forum's quick. JohnSPals, it looks like you differentiated with respect to h treating r as a constant. I don't think you can do that because r is variable with h.

V=h/3 * πr² (volume of liquid in the cone)

Ratio r/h is constant (steepness of the cone). Initially r=10 and h = 30 therefore r = h/3.

Substitute back into V.

V=(h/3)³π

Differentiating:

dV/dh=(h/3)²π

We want dh/dt, which is dh/dV * dV/dt (chain rule)

So, at dV/dt = -0.2 and h=15:

dh/dt = (-0.2)/((15/3)²π )

=-1/125π

Hope that helps, let me know if you need any of it clearing up.

Edit: Beaten to it, this forum's quick. JohnSPals, it looks like you differentiated with respect to h treating r as a constant. I don't think you can do that because r is variable with h.

Thanks, the answer was 0.0025 cm s^-1.

Robert602

Edit: Beaten to it, this forum's quick. JohnSPals, it looks like you differentiated with respect to h treating r as a constant. I don't think you can do that because r is variable with h.

Ahh. Fair enough, that seems like a good point

I think I took it at that single point. The question said "at that specific point" so I treat the radius as being constant.

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