Related Rates Solutions

  • #1
1,235
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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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Answers and Replies

  • #2
learningphysics
Homework Helper
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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.
 
  • #3
dextercioby
Science Advisor
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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.
 
  • #4
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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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  • #5
learningphysics
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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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  • #6
1,235
1
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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  • #7
learningphysics
Homework Helper
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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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