then set dV(A)/dt(A) = dV(B)/dt(B) and solve for x
9x^2 = 6x^2 + 12x + 6
3x^2 - 12x -6 = 0
x^2 -4x - 2 = 0
using the quadratic formula I come up with
2 + sqrt(6).
ruling out the negative solution.
so the side lenght of (A) = 2 + sqrt(6)
and Side lenght of (B) = 3 + sqrt(6)
Now, my second question. If i were to ask, what time will the rates of volume be the same, assuming start times were the same, would i need to add another piece of information, like one of the original starting volumes to deduce this?
Originally posted by joe shmo I haven't done these types of problems yet, so the math I have done on it so far may be incorrect, but i'll let you be the judge of that.
This is just for fun
Ok, here it is broken into two parts
start with a cube(A) of side length x, nested inside a cube(B) of side lenght (x+1), at what side lengths will their rates of change in volume be equal
...[text shortened]... to add another piece of information, like one of the original starting volumes to deduce this?
if my notation doesnt make sence, I'll be happy to clarify what I thought it meant..🙂
Originally posted by wolfgang59 I cant follow this. If the rates of change of the sides are constant then the rates of change of the volumes will be constant. Namely 8 and 27. They are never going to be equal!
The volume (V) is equal to x^3 so
DV/dx = 3x^2
DV/dt = DV/dx * dx/dt
= (3x^2) * 3
I'm unclear as to the realtionship between the two cubes - ...[text shortened]... constantly 1 unit larger than the smaller?
ie does y = x+1 (y is side length of larger cube)
sorry, I think I meant to ask, when are their rates of change in volume with respect to time equal
And yes, the side of the larger cube is (x+1)
so the two volumes are V= x^3 & V = (x+1)^3
the cubes are growing at different rates relative to each other, so the larger cube isnt always 1 unit larger
given enough time V= x^3 should become greater than V = (x+1)^3 right?
Originally posted by wolfgang59 How can x^3 be greater than (x+1)^3 ????
Your problem I think is that you are using x as a variable as well as using it as a discrete value.
yeah, your right about that..all logic escapes me somtimes😳. haha
This may seem like a strange question to ask but, I have a hunch that you are right about me using a discrete value. Can you explain to me, if possible, what you think the connection I'm making is, and its flaw?
Its probably best to use examples to answer this, if you are willing to try it.😕