# Connected rates of change question

1. Nov 23, 2008

### thomas49th

1. The problem statement, all variables and given/known data
At time t seconds the surface area of a cube is A cm² and the volume is V cm³. The surface area of the cube is expanding at a constant rate 2 cm²/s
Show that
$$\frac{dV}{dt} = \frac{1}{2}V^{\frac{1}{3}}$$

2. Relevant equations
This is a rates of change question and therefore I will need to dervie differential equations from the information presented.
I can say V = x³
A = 6x²
Where x is the length of a side

3. The attempt at a solution

$$\frac{dA}{dT} = A + 2$$
$$\frac{dA}{dT} = x^{2} + 2$$

Am I going in the right direction?

Thanks :)

2. Nov 23, 2008

### Pere Callahan

No.

The rate of change of A being constantly two means
$$\frac{dA}{dt}=2$$.
You wrote down a formula for V(x) and for A(x). Combine these two to get a formular V(A) for the volume in terms of the area. Then ther ate of change for the volume would be
$$\frac{dV}{dt}=\frac{dV}{dA}\frac{dA}{dt}=2\frac{dV}{dA}$$.
Then in this expression, express the area A again in terms of the volume V, and you will be done.

3. Nov 23, 2008

### thomas49th

ahh cool. how do i got about combining these formulae? Should I use x as the parameter (like in parametrics)

Thanks :)

4. Nov 23, 2008

### Pere Callahan

Yes you know A(x). Make x the subject to get x(A). Plug it into V(x) and viola you have V(A).

5. Nov 23, 2008

### thomas49th

x = sqrt(a)
v = a^{3/2}

however when i diff that $$\frac{dV}{dA} = \frac{3}{2} V^{1}{2}$$

that is not the correct answer

Where am I messing up?

Thanks :)

6. Nov 23, 2008

### Staff: Mentor

You are asked to show that $$\frac{dV}{dt} = \frac{1}{2} V^\frac{1}{3}$$
You stopped with $$\frac{dV}{dA}$$

How can you get $$\frac{dV}{dt}$$ from $$\frac{dV}{dA}$$?

7. Nov 23, 2008

### thomas49th

multiply by dA/dt which is 2?
which gives me 3V ^ {3/2}

that isn't the right answer though?

Thanks :)

8. Nov 23, 2008

### Pere Callahan

A=6x2, you missed that 6, didn't you?

Moreover, your final exponent of V is wrong. You have $V\sim A^{3/2}$, so
$$\frac{dV}{dA}\sim A^{1/2}\sim V^{1/3}$$

You should be able to get the factos right.

9. Nov 24, 2008

### thomas49th

right im getting this:
yeh i missed a = 6x²

dv/da = 3/2 * 6 & A ^(1/2)
= 9A^(1/2)

now multiply this by da/dt giving us 18A^(1/2)

now remeber a = 6 x 3root(V) ^ 2
= 6V^{2/3}
and a is to the power of a half

so

6^(1/2) x 6V^(1/3)
now im almost there but I cant get the 1/2 at the front of V

interestingly 6^2 = 36 and 18/36 is 1/2 so im guessing it's somthing round there?

Thanks :)

10. Nov 24, 2008

### Pere Callahan

No, what are you doing there. A=6x2, so
$$x=\left(\frac{A}{6}\right)^{1/2}$$
Plug this into V=x3, and you get
$$V=\left(\frac{A}{6}\right)^{3/2}$$
Differentiating gives
$$\frac{dV}{dA}=\frac{3}{2}\frac{1}{6}\left(\frac{A}{6}\right)^{1/2}$$
Together with dA/dt=2, this means
$$\frac{dV}{dt}=\frac{1}{2}\left(\frac{A}{6}\right)^{1/2}$$
Now,
$$\left(\frac{A}{6}\right)^{1/2}=x=V^{1/3}$$
which is the correct result