Troubleshooting a Simple Problem: Tips & Tricks

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Discussion Overview

The discussion revolves around a problem related to the volume of an expanding cube, specifically focusing on the differentiation of the volume formula and the application of given data to calculate rates of change. The scope includes mathematical reasoning and technical explanation.

Discussion Character

  • Technical explanation, Mathematical reasoning, Homework-related

Main Points Raised

  • One participant presents the formula for the volume of a cube, \( V = s^3 \), and differentiates it with respect to time to find \( \frac{dV}{dt} = 3s^2\frac{ds}{dt} \).
  • Another participant calculates \( \frac{dV}{dt} \) for specific values of \( s \) (2 cm and 10 cm) and provides the results as \( 72 \text{ cm}^3/\text{s} \) and \( 1800 \text{ cm}^3/\text{s} \), respectively.
  • A subsequent post reiterates the calculations and suggests additional clarity for exam purposes by explicitly stating the differentiation steps and substituting values.
  • One participant expresses understanding of the problem and indicates a willingness to continue discussing further calculations.

Areas of Agreement / Disagreement

Participants appear to agree on the calculations presented, but there is no explicit consensus on the overall correctness of the approach or the problem's context, as one participant is still seeking confirmation of their understanding.

Contextual Notes

There may be limitations related to the assumptions made about the values of \( s \) and \( \frac{ds}{dt} \), as well as the dependence on the clarity of the problem statement, which is not fully detailed in the discussion.

karush
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I know this is a simple problem but new at it. answer not in book so hope correct.
 
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Re: expanding cube

You have the formula for volume:

$$V=s^3$$

Differentiating with respect to time $t$, we find:

$$\frac{dV}{dt}=3s^2\frac{ds}{dt}$$

Now plug in the given data...and keep in mind your units...you should get $$\frac{\text{cm}^3}{\text{s}}$$
 
Re: expanding cube

$$3\cdot 2^2 \cdot 6 = 72 \text { cm}^3\text{/ sec}$$

and

$$3\cdot 10^2 \cdot 6 = 1800 \text { cm}^3\text{/ sec}$$
 
Re: expanding cube

karush said:
$$3\cdot 2^2 \cdot 6 = 72 \text { cm}^3\text{/ sec}$$

and

$$3\cdot 10^2 \cdot 6 = 1800 \text { cm}^3\text{/ sec}$$

Looks good. If you wish to be absolutely clear on an exam, I would write (in addition to showing your differentiation with respect to $t$ to obtain the formula):

a) $$\left. \frac{dV}{dt} \right|_{s=2\text{ cm}}=3\left(2\text{ cm} \right)^2\left(6\,\frac{\text{cm}}{\text{s}} \right)=72\,\frac{\text{cm}^3}{\text{s}}$$

b) $$\left. \frac{dV}{dt} \right|_{s=10\text{ cm}}=3\left(10\text{ cm} \right)^2\left(6\,\frac{\text{cm}}{\text{s}} \right)=1800\,\frac{\text{cm}^3}{\text{s}}$$
 
Re: expanding cube

makes sense
I will post some more to see how close I am
 

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