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Rates associated calculus - derivative

  1. Jun 7, 2015 #1
    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

    My answer: 6/2h


    The answer I should get: 1/2h

    I do not understand how to get to that answer thank you.
     
  2. jcsd
  3. Jun 7, 2015 #2

    Mark44

    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.
     
  4. Jun 7, 2015 #3
    Surface in terms of h = height * height = h2
    d/dh = 2*h

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


    the answer is 1/(2h).
     
  5. Jun 7, 2015 #4

    Mark44

    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.
     
  6. Jun 7, 2015 #5
    dS/dh = 6*h2 =12h
    dS/dt = 6cm2/hour

    6/ (12h) = 1/(2h)
     
  7. Jun 7, 2015 #6

    Mark44

    Staff: Mentor

    You're doing it again - writing things that aren't true.
    dS/dh ≠ 6h2 and 6h2 ≠ 12h
     
  8. Jun 8, 2015 #7

    vela

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    Staff Emeritus
    Science Advisor
    Homework Helper
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    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.
     
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