The discussion focuses on calculating the rate of change of height (dh/dt) for a cubic crystal with an initial height of 1 cm and a surface area increase rate of 6 cm²/hour. The relevant equation used is ds/dt = ds/dh * dh/dt, where ds/dt is given as 6 cm²/hour. The correct interpretation of the surface area in terms of height is S = h², leading to the derivative dS/dh = 2h. The final calculation confirms that dh/dt equals 1/(2h) when properly expressed with parentheses to clarify the denominator.
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You're trying to do too much all at once here. h * h ≠ 2hmasterchiefo said:Homework Statement
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
Homework Equations
ds/dt = ds/dh * dh/dt
The Attempt at a Solution
ds/dt = 6 cm2 / hour
ds/dh = h*h = 2*h
If you want this to be interpreted as what you intended (but not as what you wrote), you need parentheses: 6/(2h).masterchiefo said:My answer: 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)masterchiefo said:The answer I should get: 1/2h
See above.masterchiefo said:I do not understand how to get to that answer thank you.
Surface in terms of h = height * height = h2Mark44 said: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.
Earlier you wrote "ds/dh = h*h = 2*h"masterchiefo said:Surface in terms of h = height * height = h2
d/dh = 2*h
Not really sure I understand how different I should do this...
dS/dh = 6*h2 =12hMark44 said: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.
You're doing it again - writing things that aren't true.masterchiefo said:dS/dh = 6*h2 =12h
masterchiefo said:dS/dt = 6cm2/hour
6/ (12h) = 1/(2h)
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.Mark44 said:(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.)