Rates associated calculus - derivative

1. Jun 7, 2015

masterchiefo

1. The problem statement, all variables and given/known data
Growth is observed for a cubic crystal. Initially the height of the cube is 1 cm . the surface of the cube increases at a rate of 6 cm2 / hour.

Question: Calculate dh/dt

2. Relevant equations
ds/dt = ds/dh * dh/dt

3. The attempt at a solution
ds/dt = 6 cm2 / hour
ds/dh = h*h = 2*h

The answer I should get: 1/2h

I do not understand how to get to that answer thank you.

2. Jun 7, 2015

Staff: Mentor

You're trying to do too much all at once here. h * h ≠ 2h

What is the equation of S in terms of h?
What is the rate of change of S with respect to h?
If you want this to be interpreted as what you intended (but not as what you wrote), you need parentheses: 6/(2h).
That answer is not correct, at least with regard to what you wrote, which is the same as (1/2)h. If you want to indicate that 2h is in the denominator, put parentheses around it, like this: 1/(2h)
Or write it using LaTeX, like so $\frac {1} {2h}$
What I wrote is this: # #\frac {1} {2h}# #, without the extra spaces.
See above.

3. Jun 7, 2015

masterchiefo

Surface in terms of h = height * height = h2
d/dh = 2*h

Not really sure I understand how different I should do this...

4. Jun 7, 2015

Staff: Mentor

Earlier you wrote "ds/dh = h*h = 2*h"
This is incorrect, since 1) dS/dh ≠ h*h and 2) h*h ≠ 2h. It appeared that you were munging together two concepts into one mush that didn't make much sense.
My aim was to get you to write two equations: one for S in terms of h, and the other for $\frac{dS}{dt}$

In this thread you wrote "d/dh = 2*h", which is incorrect for a technical reason. d/dh is not the derivative - it is the differentiation operator - something that you apply to a function to get a derivative. What you probably meant was dS/dh = 2h. (Note that for surface area, S is usually used. Lower case s is usually used for distance, as in s = (1/2)gt2. I have no idea what s stands for, but it's used a lot for this purpose in many physics books.)
In your relevant equation, you have dS/dt = dS/dh * dh/dt
Use this equation and the quantities you know to solve for dh/dt.

5. Jun 7, 2015

masterchiefo

dS/dh = 6*h2 =12h
dS/dt = 6cm2/hour

6/ (12h) = 1/(2h)

6. Jun 7, 2015

Staff: Mentor

You're doing it again - writing things that aren't true.
dS/dh ≠ 6h2 and 6h2 ≠ 12h

7. Jun 8, 2015

vela

Staff Emeritus
I always saw the use of $s$ as a math book thing. :) According to Wikipedia, the reason $s$ is used is because the Latin word for length is spatium.