The discussion revolves around a problem related to rates of change in calculus, specifically concerning the growth of a cubic crystal. The original poster presents a scenario where the height of the cube is initially 1 cm, and the surface area increases at a rate of 6 cm²/hour, leading to a question about calculating dh/dt.
The conversation includes attempts to clarify the equations involved, with some participants providing guidance on how to express the relationships correctly. There is an ongoing exploration of the definitions and relationships between the variables, but no consensus has been reached on the correct approach or solution.
There are indications of confusion regarding the notation and the application of derivatives, with participants highlighting potential misunderstandings in the original poster's attempts. The discussion also touches on the conventions used in mathematical notation, particularly in relation to surface area and height.
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.)