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liizzyliizz

  • 4 years ago

The volume of a cube is increasing at a a rate of 300 in^3/min. at the instant when the edge is 20 inches at what rate is the edge changing.

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  1. anonymous
    • 4 years ago
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    edge is \[x = \sqrt[3]{V}\] so \[x'(t)=\frac{1}{3\sqrt[3]{V^2}}\times V'(t)\]

  2. anonymous
    • 4 years ago
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    you are told that \[V'(t)=300\] so you get \[x'(t)=\frac{300}{3\sqrt[3]{V^2}}=\frac{100}{\sqrt[3]{V^2}}\] so all you need to finish is to find out what V is when x is 20

  3. liizzyliizz
    • 4 years ago
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    Ahh, so for future problems similar to this how would one know to go about this.

  4. liizzyliizz
    • 4 years ago
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    I still blame my algebra skills, but that is just me lol

  5. lgbasallote
    • 4 years ago
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    dV/dT = 300 when e = 20 volume of a cube = e^3 dV/dT = 3e^2(de/dt) 300 = 3e^2 (de/dt) de/dt = 300/3e^2 when e =20 de/dt = 300/3(20)^2 de/dt = 1/4 in/min is the answer I came up with.

  6. liizzyliizz
    • 4 years ago
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    ^ that made a lot more sense lol

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