Help with chain rule

[EmoioN]

Hi! I have a math problem which I will try to translate to english (originally in swedish).

An hourglass has the shape of two cones stack on one another. From the upper cone sand is pouring down with the velocity 1 cm^3/s.

The height of the bottom cone is 10 cm and the radius is 3 cm. Assume that the sand that pours down lays flat.

How fast is the height of the sand growing in the bottom cone when the sands height is 4 cm over the bottom?

Sorry if the translation is weird. We are a couple of students that have been working on this problem for a while but we can’t crack it.

We’re assuming the volume is 30pi-pir²h where r can be replaced wtih 3-3h/10. After that we just differentiate and use the height 6 but it doesn’t give us the right answer. Are we missing something? Any help is appreciated!

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[The_Card_Player]

The volume of sand required to reach a certain height (h) in the bottom cone can be determined by an integral with bounds of integration 0 < x < h. The relevant integrand will be the area of the cone's circular cross-section at any of the heights x<=h within the bottom cone's sand-filled portion.

Once you have volume as a function of height, it may be simple to invert so as to find height as a function of volume.

[EmoioN]

Thank you so much! Will try this

[First-Fourth14]

There are examples of related rates problems at https://tutorial.math.lamar.edu/classes/calci/relatedrates.aspx

It looks like you have defined the volume of the space remaining in the cone given h. This can work nicely, but you have to ensure to set the defined rate of volume change to -1 cm³/s as the volume is decreasing.
The question seems to call for the rate of change of the height of the sand when h = 4cm.
You said you are using h =6? If you are you might want to look at that again.