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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

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# Homework Help: Related Rates Solutions

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