Derivation Application in Differential Calculus, verification question problem.

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Discussion Overview

The discussion revolves around a verification of a solution to a problem involving implicit differentiation in the context of a cylindrical tank draining water. Participants are examining the rate of change of the height of water in the tank as it drains at a specified volume rate.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant presents a solution using the formula for the volume of a cylinder, suggesting that the height change rate, dh/dt, is 1/2 pi ft/min.
  • Another participant points out the need to account for the negative sign due to the volume decreasing, leading to the equation -200 = 100π(dh/dt).
  • A later reply reiterates the importance of the negative sign and proposes that the correct calculation leads to -1/2 pi ft/min.
  • Another participant emphasizes the need to check units associated with the calculations, suggesting that the final answer's units should be clarified.
  • One participant acknowledges a typo in their previous message, indicating the answer should be -2 pi ft^3/min but questions the placement of values in the implicit differentiation process.

Areas of Agreement / Disagreement

Participants express differing views on the correct interpretation of the calculations and the final answer. There is no consensus on the correct value of dh/dt, and the discussion remains unresolved regarding the final answer and the implications of unit consistency.

Contextual Notes

There are unresolved issues regarding the assumptions made in the calculations, particularly concerning the treatment of the negative sign and the units of the final answer. The discussion highlights the importance of careful attention to detail in implicit differentiation.

jamescv31
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Greetings everyone in MHB. :)

Well I've just created a thread to just verify if my answer is correct. On a simple problem that using implicit differentiation.

A cylindrical tank of radius 10 ft is having drained with water at the rate of 200 f^3/ min. How fast is the height of water changed?

Find dh/ dt

My solution goes here:

The formula used is V= (\pi) (r^2) (h)

then substitute the values

200 ft^3/min =pi (100ft) (dh/dt)

therefore my answer is 1/2 pi ft^3/min

Can anyone check this if its a right solution made?

Thanks
 
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As you correctly surmised, we have:

$$\d{V}{t}=\pi r^2\d{h}{t}$$

The cylindrical tank is being drained, so it's volume is decreasing! $-200 \frac{f^3}{min}$, you have to remember the negative.

Plugging what we know:

$$-200=100\pi \d{h}{t}$$

How can we isolate for $\d{h}{t}$?
 
Rido12 said:
As you correctly surmised, we have:

$$\d{V}{t}=\pi r^2\d{h}{t}$$

The cylindrical tank is being drained, so it's volume is decreasing! $-200 \frac{f^3}{min}$, you have to remember the negative.

Plugging what we know:

$$-200=100\pi \d{h}{t}$$

How can we isolate for $\d{h}{t}$?

by using implicit difference where the -200ft/min is the numerator and the 100 ft is th denominator hence the answer is -1/2 pi ft ^ 3min.
 
Last edited:
jamescv31 said:
by using implicit difference where the -200ft/min is the numerator and the 100 ft is th denominator hence the answer is -1/2 pi ft ^ 3min.

$$\frac{-200}{100\pi}=\text{what again?}.$$

Also check your units. The $100\pi$ isn't just $100\pi$. It has some units attached to it, which means your final answer might have different units. What units would make sense?
 
its -2 pi ft^3min a typo error. But the implicit difference where the placing of values for numerator and denominator is correct right?
 

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