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I've encountered a problem I cannot seem to be able to solve. 1 = the problem 2 = my solution

_____1 A ball has the volume of 3.0 cm^3. The volume decreases with time t (in months), the change per unit of time is proportional to the area of the ball. After 1 month, the volume is 2.0 cm^3.

What is the volume of the ball after 4 months?

_____2 https://i.stack.imgur.com/r6xiZ.png

My guess is that I am doing something wrong at the step I've marked in a rectangle.

My question to you is: am I completely in the wrong direction and if so, how would I solve it?

Arcthor
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  • The title is a bit comical. I thought you were asking for a differential equation which models a high school. Don't change the title. It's pretty amusing :P – Cameron Williams Mar 28 '15 at 23:51
  • I can't read the image. We want $\frac{dV}{dt}=-kV^{2/3}$. So write $V^{-2/3}\frac{dV}{dt}=-k,dt$ and integrate. – André Nicolas Mar 28 '15 at 23:56

1 Answers1

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you need these: $$\frac{dV}{dt} = -k_1A, V = \frac 43 \pi r^3, A = 4\pi r^2$$

you can get rid of $r$ and find a relation between $V$ and $A$ so that we can get $$\frac{dV}{dt} = -3kV^{2/3} $$ for some constant $k.$

separating the variables and integrating we get $$\int_3^2 \frac {dV}{ 3V^{2/3}} = -k\int_0^1dt \to k = 3^{1/3} - 2^{1/3}$$ we need $V$ at time $t = 4$ so $$ 3^{1/3} - V(4)^{1/3} = 4k \to V(4) = \left(3^{1/3} -4(3^{1/3} - 2^{1/3})\right)^3 = \left(4 \times 2^{1/3} - 3 \times 3^{1/3}\right)^3=0.362 \, cm^3$$

abel
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