# Rates of change - confused (again!)

1. Jun 4, 2004

### crookesm

Imagine a cube that 'grows' by five cubic inches per week. How fast is its surface area increasing when the length of one of its sides is seven inches?

I know that the derivative of volume (V) with respect to time (t) is 5, e.g:

$$\frac{dV}{dt} = 5$$

To calculate the surface area of a cube from a given volume I would use:

$$S=6(\sqrt[3]{V})^2$$

Therefore, would I need to calculate the derivate of S with respect to V in order to reach my goal of deriving S with respect to t?

Or am I going totally in the wrong direction?

2. Jun 4, 2004

### arildno

Differentiate S with respect to t, by using the chain rule.
Use your knowledge of V and dV/dt to calculate how fast the surface area changes.

3. Jun 9, 2004

### crookesm

Thanks Arildno,

1. I know the cube is growing at $$5in^2$$ per week.

2. This can be represented as $$5x$$ (where x is the week number)

3. I need to figure out the week where the length is 7 inches. Solving $$\sqrt[3]{5x} = 7$$, I find that the week is 68.6

4. Therefore, I need to find the cube surface area's rate of change at week 68.6 - represented as: $$6(\sqrt[3]{5x})^2$$

Plugging this into my calculator returns a rate of $$2.857in^2$$ per week. This is the correct answer I was looking for! Hurrah!

However, I have been trying to calculate the derivative manually without much success. I have been using the chain rule:

I am taking $$f(x)=6x^2$$ and $$g(x)=\sqrt[3]{5x}$$. I understand I need to take the derivate of f(x) and g(x) as follows - $$f'(g(x))g'(x)$$. I can calculate $$f'(x)=12(\sqrt[3]{5x})$$ but I cannot get $$g'(x)$$. I'm totally stumped when trying to derive this. Any hints...?

Last edited: Jun 9, 2004
4. Jun 9, 2004

### arildno

Write g(x) like this:
$$g(x)=\sqrt[3]{5x}=5^{\frac{1}{3}}x^{\frac{1}{3}}$$

Does that help?

Last edited: Jun 9, 2004
5. Jun 9, 2004

### crookesm

Sorry about that - I kept screwing up my Latex formatting. I have edited my post it should be correct now..

6. Jun 9, 2004

### arildno

I've posted a hint, I don't know if you find it sufficient, though..

7. Jun 9, 2004

### jdavel

crookesm,

I think the notation you're using is getting you confused. Stick with V for volume and S for surface area. You're trying to find dS/dt. According to the chain rule:

dS/dt = dS/dV*dV/dt

You have equations for S(V) and V(t), so you can find those deriviatives.

Does that help?

8. Jun 9, 2004

### Gokul43201

Staff Emeritus
Here's another way,

$$\frac {dV} {dt} = \frac {d} {dt} (a^3) = 3a^2 \frac {da} {dt} = 5$$
$$=> \frac {da} {dt} = 5/(3*7*7)$$

And
$$\frac {dS} {dt} =\frac {d} {dt} (6a^2) = 12a \frac {da} {dt} = \frac {12*7*5} {3*7*7}=20/7=2.857$$