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Rates of change - confused (again!)

  1. Jun 4, 2004 #1
    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:

    [tex] \frac{dV}{dt} = 5[/tex]

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

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

    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. jcsd
  3. Jun 4, 2004 #2

    arildno

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    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.
     
  4. Jun 9, 2004 #3
    Thanks Arildno,

    I have been thinking about this more clearly...

    1. I know the cube is growing at [tex]5in^2[/tex] per week.

    2. This can be represented as [tex]5x[/tex] (where x is the week number)

    3. I need to figure out the week where the length is 7 inches. Solving [tex] \sqrt[3]{5x} = 7 [/tex], 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: [tex] 6(\sqrt[3]{5x})^2 [/tex]

    Plugging this into my calculator returns a rate of [tex]2.857in^2[/tex] 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 [tex]f(x)=6x^2[/tex] and [tex]g(x)=\sqrt[3]{5x}[/tex]. I understand I need to take the derivate of f(x) and g(x) as follows - [tex]f'(g(x))g'(x)[/tex]. I can calculate [tex]f'(x)=12(\sqrt[3]{5x})[/tex] but I cannot get [tex]g'(x)[/tex]. I'm totally stumped when trying to derive this. Any hints...?
     
    Last edited: Jun 9, 2004
  5. Jun 9, 2004 #4

    arildno

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

    Does that help?
     
    Last edited: Jun 9, 2004
  6. Jun 9, 2004 #5
    Sorry about that - I kept screwing up my Latex formatting. I have edited my post it should be correct now..
     
  7. Jun 9, 2004 #6

    arildno

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    Your not alone in this..
    I've posted a hint, I don't know if you find it sufficient, though..
     
  8. Jun 9, 2004 #7
    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?
     
  9. Jun 9, 2004 #8

    Gokul43201

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    Here's another way,

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

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