Derivative as an Instantaneous rate of change

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Homework Help Overview

The discussion revolves around understanding the derivative as an instantaneous rate of change, specifically in the context of finding the volume of a cone as a function of depth. Participants are exploring how to formulate the volume equation and differentiate it.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants are attempting to establish the volume function V(d) based on the geometry of a cone and are discussing the implications of differentiating this function. Questions about starting the equation and the relationship between volume and depth are raised.

Discussion Status

Some participants have provided insights into the volume formula and its dependence on depth, while others express confusion about the initial setup and the implications of the volume's mathematical representation. There is an ongoing exploration of how to correctly formulate and differentiate the volume function.

Contextual Notes

Participants are navigating the challenge of applying geometric formulas to a variable context, with some uncertainty about the correct relationships between the dimensions involved.

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I am having trouble with this problem. Any help how to start it would be appreciated?

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Mathematically, what is asked is [tex]V'(d)[/tex] (prime denotes differentiation), where V(d) is the volume as a function of the depth d. So find V(d) first.
 
Yes, mathematically, the result should be a constant, but there's no serious physical sense here.
 
Thats the thing i am having trouble with, starting the equation? I know I need the Volume for a cylinder equation but how do I use it to figure it out?
 
The volume of a cone is 1/3 b*h, where b is the area of the base (pi r^2) and h is the height.
 
So would the new equation be 1/3 4 - d * 4 - d (pi r^2)?
 
The equation is V = 1/3 * d^3 * Pi . Since d can change, it is a variable, which makes the equation a function, so you can write V(d) = 1/3 * d^3 * Pi , because the volume of the fluid V depends of it's depth d.
 
Okay, Thanks a lot i didn't get the d^3.
 
The d^3 is implied by radius = height.
 

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