Yes, is correct. Saying that "the plastic flows at 300 litres/min, which is "volume per time" means that . Knowing "the volume of the mold at any height, h" means that you can find dV/dh which is 1/(dh/dV).
So you know .
Solve that for h(t).
An antique anchor is to be preserved in a plastic mold. The mold is 3 m high and is to be placed on top of a stand of height 1 m. Inside the mold, the anchor is to be placed on top of a marble box. The figure below shows the longitudinal cross-section of this arrangement. The parabolic arcs of the mold are given by the equation .
(Apologies for the crude drawing and wording -- I left my book at school so I'm doing this from memory.)
From previous questions I have:
- the volume of the mold, and
- the volume of the mold at any height, h, above the ground.
1) Plastic flows into the mold at 300 litres/min. Find the height of the anchor and hence the height of the marble block if the top of the anchor is covered after exactly 10 minutes.
I was thinking something along the lines of .
2) If the plastic now flows at 329 litres/min, express h in terms of t.
If you could point me in the right direction, that would be great. Thanks.
Yes, is correct. Saying that "the plastic flows at 300 litres/min, which is "volume per time" means that . Knowing "the volume of the mold at any height, h" means that you can find dV/dh which is 1/(dh/dV).
So you know .
Solve that for h(t).
So I found h(t), substituted t = 10 and got h = 1.188 m, which should be the height of the anchor if it's covered after 10 minutes. But it makes no sense -- if the top of the anchor is 1.188 m above the ground, then it's only 18.8 cm tall inside the 3 m high mold.