Related rates and cube with a sphere inside of it

  • #1
I was pondering a question today. If you have a cube with a sphere inside of it, and the sphere is growing at 2 m/s. The cube itself is expanding at 1 m/s. If the cube is 5 x 5 x 5, and the sphere has a radius of 2. Is it possible to calculate when the sphere and the cube will be expanding at the same rate? I also assumed that eventually the sphere will overcome the cube.

V=4/3*pi*r^2
dv=4/3*pi*2r dr/dt

V=x^3
dv=3x^2
 

Answers and Replies

  • #2
arildno
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First, you are interested in:
When does the volume expansion rate of the sphere equal the volume expansion rate of the cube?

Okay, let's start:

Sphere:
The radius r, as a function of time, is given by r(t)=2+2t
Thus, its volume V, as a function of time, is given by: [itex]V(t)=\frac{4\pi}{3}(2+2t)^{3}[/itex]
Cube:
The cube's side, as a function of time, is given by s(t)=5+t.

Thus, its volume v, as a function of time is [itex]v(t)=(5+t)^{3}[/itex]

We wish to find t so that:
[tex]\frac{dV}{dt}=\frac{dv}{dt}[/tex]

This is essentially the setup.

However, with the measures you've given, even at t=0, the volumetric expansion rate of the sphere is greater than that of the cube.
 
  • #3
HallsofIvy
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I was pondering a question today. If you have a cube with a sphere inside of it, and the sphere is growing at 2 m/s. The cube itself is expanding at 1 m/s. If the cube is 5 x 5 x 5, and the sphere has a radius of 2. Is it possible to calculate when the sphere and the cube will be expanding at the same rate? I also assumed that eventually the sphere will overcome the cube.

V=4/3*pi*r^2
dv=4/3*pi*2r dr/dt

V=x^3
dv=3x^2
First clarify the problem. You say the sphere is "growing at 2 m/s". That's a "rate of change of distance". Do you mean the radius is increasing at 2 m/s? Or the diameter? You say the cube is "expanding at 1 m/s". Do you mean that the length of an edge is increasing at 1 m/s? Since you are clearly using rate of change of a length here, I am not happy with the use of the same terminology, "expanding at the same rate", to refer to (apparently) rate of increase of the volume.

Second be careful of your formulas. The volume of a sphere is given by "[ITEX](4/3)\pi r^3[/itex]", not "[itex]r^2[/itex]". For the sphere, [itex]dV/dt= 4\pi r^2 dr/dt[/itex] not what you have. Assuming that "growing at 2 m/s" means the radius is increasing at that rate, you know that dr/dt= 2 (If it is diameter that is increasing at that rate, then dD/dt= 2 dr/dt= 2 so dr/dt= 1). For the cube, yes, V= x2 so dV/dx= 3x2 dx/dt and, assuming that "expanding at 1 m/s" means the edge length is changing at that rate, dx/dt= 1. The "cube and sphere will be increasing at the same rate" when those are the same: [itex]4\pi r^2 dr/dt= 3x^2 dx/dt[/itex] which, assuming what I said above, is [itex]8\pi r^2= 3x^2[/itex]. that gives you a relation between r and x. You didn't say anything about how large the cube and sphere are initially so I don't know what actual values x and r might have.
 
  • #4
Well let's go with the radius is expanding at 2 m/s.

I do give the original dimensions of the cube and the sphere (well I just gave the r as 2).
 
  • #5
HallsofIvy
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Again, what is the formula for volume of a sphere?
 

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