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 [quote="mith"]It's asking where the numbers are the same, ignoring the dimensions.[/quote]
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Posted: Fri Mar 26, 2004 8:09 pm    Post subject: 1

Quote:
a. Find drV/dt when r=10 cm.
Dr. Borodog
Posted: Fri Mar 26, 2004 2:41 am    Post subject: 0

Thanks mith. I should have gotten that from the original statement: ". . . increasing at the same numerical rate . . ."

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You will respect my philosophai.

[This message has been edited by Dr. Borodog (edited 03-25-2004 09:41 PM).]
mith
Posted: Fri Mar 26, 2004 2:38 am    Post subject: -1

It's asking where the numbers are the same, ignoring the dimensions.
Dr. Borodog
Posted: Fri Mar 26, 2004 2:24 am    Post subject: -2

Although, dV/dt can never equal dr/dt, as they are dimensionally different, so I don't know what they want there.

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You will respect my philosophai.
Dr. Borodog
Posted: Fri Mar 26, 2004 2:19 am    Post subject: -3

dV/dt = 4pr2 dr/dt
dA/dt = 2pr dr/dt

That should get you started.

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You will respect my philosophai.
Specter
Posted: Fri Mar 26, 2004 1:35 am    Post subject: -4

I was looking through an old Calculus textbook and came upon this R.O.C. problem. I don't think it's too difficult, but I really have no idea how to solve it.

quote:

The radius (r) of a sphere is increasing at a constant rate 0.04 cm/sec.

Find:
a. At the time when the radius is 10 centimeters, what is the rate of increase of its volume?
b. At the time when the volume is 36pi cm3, what is the rate of increase of the area of a cross section through the center of the sphere?
c. At the time when the volume and the radius are increasing at the same numerical rate, what is the radius?

I think I know how to define some of the changing variables as derivatives.
Based on what I remember, I think I can say...
dr/dt = 0.04
a. Find dr/dt when r=10 cm.
b. Find dA/dt when V=36 pi.
c. When dV/dt = dr/dt, find r.
(right?)

Vsphere=4/3*pi*r3
Acircle=pi*r2

That is about as far into Calculus that I can remember.