A related rates problem (shouldn't be complicated)

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Homework Help Overview

The problem involves an inverted conical container with specific dimensions and a rate of water flowing out. Participants are tasked with determining how fast the depth of the water is decreasing at a given height.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the relationship between the radius and height of the cone, using similar triangles to express radius in terms of height. There is an exploration of the implications of treating the radius as a constant versus a variable.

Discussion Status

Some participants have attempted to derive the rate of change of height based on the volume formula, while others have pointed out potential errors in assuming radius is constant. There is a recognition of differing results compared to a reference solution, prompting further questioning of the setup and assumptions.

Contextual Notes

Participants are working under the constraints of a related rates problem, with specific values provided for the dimensions of the cone and the rate of water flow. There is an ongoing discussion about the correct application of the relationships between the variables involved.

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Homework Statement


An inverted conical container has a diameter of 42 in and a depth of 15 in. If water is flowing out of the vertex at a rate of 35 [tex]\pi[/tex] in^3 / sec, how fast is the depth of the water dropping when the height is 5 in?


Homework Equations


v= (1/3) [tex]\pi[/tex] r^2 * h


The Attempt at a Solution


What I know:
D=42 in
R=21 in
h=15 in
(dV/dT)=35 [tex]\pi[/tex] in^3/min
(dH/dT)=? When h=5.

(r1/h1)=(r2/h2) ... (21/15)= (r2/5) ...r2=7

plugging in 7 for r, I get...
v= (1/3) [tex]\pi[/tex] (49) * h

Taking the derivative in terms of T, I get...
(dV/dT)=(49 [tex]\pi[/tex] /3) (dH/dT)
35 [tex]\pi[/tex] = (49 [tex]\pi[/tex] / 3) (dH/dT)



I get (dH/dT) to = 15/7 in/sec, but the back of my book says 5/7 in/sec. Is it just supposed to be the ratio of height to radius? Thanks in advance.
 
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Did I go wrong in plugging 7 in for r before I took the derivative? But if I took the derivative I'd have (dr/dt) in the equation, which I wouldn't be able to solve for.
 

Homework Statement


An inverted conical container has a diameter of 42 in and a depth of 15 in. If water is flowing out of the vertex at a rate of 35 [tex]\pi[/tex] in^3 / sec, how fast is the depth of the water dropping when the height is 5 in?


Homework Equations


v= (1/3) [tex]\pi[/tex] r^2 * h


The Attempt at a Solution


What I know:
D=42 in
R=21 in
h=15 in
(dV/dT)=35 [tex]\pi[/tex] in^3/min
(dH/dT)=? When h=5.

(r1/h1)=(r2/h2) ... (21/15)= (r2/5) ...r2=7

plugging in 7 for r, I get...[/v= (1/3) [tex]\pi[/tex] (49) * h

Taking the derivative in terms of T, I get...

(dV/dT)=(49 [tex]\pi[/tex] /3) (dH/dT)
Here is your error- you are treating r as if it were a constant. Since (r/h)= (21/7), r= (21/7)h and [itex]V= (1/3)\pi r^2 h= (1/3)\pi(21/7)^2 h^3[/itex]. differentiate both sides of that with respect to h.

35 [tex]\pi[/tex] = (49 [tex]\pi[/tex] / 3) (dH/dT)



I get (dH/dT) to = 15/7 in/sec, but the back of my book says 5/7 in/sec. Is it just supposed to be the ratio of height to radius? Thanks in advance.
 
Thank you very much. I felt iffy when I used r as a constant..
:)
 

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