# Thread: Diff. Equations #2

1. ## Diff. Equations #2

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 $\displaystyle y = \frac{4}{7x^2} - \frac{3}{7}$.

(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 $\displaystyle \frac{dh}{dt} = \frac{dV}{dt}\cdot\frac{dh}{dV}$.

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.

2. ## Re: Diff. Equations #2

Yes, $\displaystyle \frac{dh}{dt}= \frac{dV}{dt}\cdot\frac{dh}{dV}$ is correct. Saying that "the plastic flows at 300 litres/min, which is "volume per time" means that $\displaystyle \frac{dV}{dt}= 300$. 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 $\displaystyle \frac{dh}{dt}= \frac{\frac{dV}{dt}}{\frac{dV}{dh}}= \frac{300}{\frac{dV}{dh}}$.

Solve that for h(t).

3. ## Re: Diff. Equations #2

For the volume at any height, h, I got (in integral form for brevity's sake)$\displaystyle \int_1^h \frac{4}{7y-3}dy$. Does this seem right?

4. ## Re: Diff. Equations #2

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.

5. ## Re: Diff. Equations #2

Someone suggested using h+1 as the upper limit, but that gave an even smaller height. I'm out of ideas.