How do related rates apply to circles, rectangles, and spheres?

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Homework Help Overview

The discussion revolves around related rates in the context of geometric shapes, specifically circles, rectangles, and spheres. Participants explore how changes in dimensions affect other related measurements, such as circumference, diagonal lengths, and volume.

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  • Mixed

Approaches and Questions Raised

  • Participants attempt to apply the concept of related rates to various geometric scenarios, questioning the implications of given rates of change and how to set up equations correctly. Some express confusion regarding the interpretation of specific problems, particularly concerning the second question about the rope around the Earth.

Discussion Status

There is ongoing clarification regarding the setup of equations for the rectangle problem, with some participants suggesting that additional data may be needed for precise calculations. The discussion includes verification of approaches, with some participants confirming the correctness of certain calculations while others express uncertainty.

Contextual Notes

Participants note potential issues with the clarity of the second question and the sufficiency of data for the rectangle problem. There is also a mention of the importance of including units in answers.

courtrigrad
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Hello all

Just had questions on related rates:

1. The radius of a circle is growing by [tex]\frac{dr}{dt} = 7[/tex]. How fast is the circumference growing? Ok so [tex]C = 2\pi r[/tex] and [tex]\frac{dC}{dr} = 2\pi \frac{dr}{dt} = 2\pi(7) = 14\pi[/tex]

2. #1 has some amazing implications. Suppose you want to put a rope around the Earth that any 7-footer can walk under. If the distance is 24,000 miles, what is the additional length of rope? Do I just put [tex]C = 24,000[/tex]? I am not sure if I understand what it is asking.

3. The sides of a rectangle increase in such a way that [tex]\frac{dz}{dt} = 3\frac{dy}{dt}[/tex] where z is the diagonal. At the instant when [tex]x = 4 y = 3[/tex] what is the value of [tex]\frac{dx}{dt}[/tex]? So [tex]x^2 + y^2 = z^2[/tex]. [tex]2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 2[/tex] So then do I just substitute in the given values do get [tex]\frac{dx}{dt}[/tex]? How would I use the fact that [tex]\frac{dx}{dt} = 3\frac{dy}{dt}[/tex]?

4. Air is being pumped into a spherical balloon at the rate of [tex]5.5[/tex] cubic inches per minute. Find the rate of change of the radius when the radius is 4 inches. Ok so I know that [tex]V = \frac{4}{3}\pi r^3[/tex]. So [tex]\frac{dV}{dt} = 5.5[/tex] So [tex]5.5 = 4\pi (4)^{2} \frac{dr}{dt}[/tex]. I get [tex]\frac{5.5}{64\pi}[/tex] Is this correct?


Thanks :smile:
 
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First question is right. I don't understand the second question. Have you copied it word for word?

For the third question the equation should be
[tex]2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 2z\frac{dz}{dt}[/tex]

Your question is confusing... Is:
[tex]\frac{dx}{dt} = \frac{dz}{dt} = 3\frac{dy}{dt}[/tex]

In one part you've written dx/dt=3dy/dt... another part you've written dz/dt=3dy/dt.
 
Your last question is solved correctly,though you should have added the unit...

As for the second,the way i see it...You don't need too much data...

Daniel.
 
I copied it word for word. I am sorry. For #3 it should be: [tex]\frac{dz}{dt} = 3\frac{dy}{dt}[/tex]
 
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courtrigrad said:
I copied it word for word. I am sorry. For #3 it should be: [tex]\frac{dz}{dt} = 3\frac{dy}{dt}[/tex]

That's ok.

For the second question, I guess that it's asking for the length of rope a 7ft can walk under around Earth - circumference of earth.

For the third question... I believe there's insufficient data to get an exact value for dx/dt. Double check the question, to see if there's more data.
 
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for #3 we have The sides of a rectangle increase such that [tex]\frac{dz}{dt} = 1[/tex] and [tex]\frac{dx}{dt} = 3\frac{dy}{dt}[/tex]. Find [tex]\frac{dx}{dt}[/tex] when [tex]x = 4, \ y = 3[/tex]

I get: [tex]2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 2z\frac{dz}{dt}[/tex]. Substituting [tex]x,y,z[/tex] we get [tex]8\frac{dx}{dt} + 6\frac{dy}{dt} = 10[/tex] Substituting in [tex]\frac{dx}{dt} = 3\frac{dy}{dt}[/tex] I got [tex]\frac{dx}{dt} = 1[/tex] Is this correct?


Whoops
 
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courtrigrad said:
I get: [tex]2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 2z\frac{dz}{dt}[/tex]. Substituting [tex]x,y,z[/tex] we get [tex]8\frac{dx}{dt} + 6\frac{dy}{dt} = 10[/tex] Substituting in [tex]\frac{dx}{dt} = 3\frac{dy}{dt}[/tex] I got [tex]\frac{dx}{dt} = 1[/tex] Is this correct?


Whoops

Looks good. :smile:
 

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