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Math Help - Use differentials to approximate?

  1. #1
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    Use differentials to approximate?

    Use differentials to approximate the value of 7.8^3??

    Should be easy, but canīt find the correct answer. How do you do this... help....
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  2. #2
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    Hello, math34!

    Use differentials to approximate the value of 7.8^3

    Evidently, we have the function: . f(x) \:=\:x^3
    Then: . df \:=\:3x^2dx

    We know that: . f(8) \,= \,8^3 \,= \,512

    Let x = 8,\;dx = -0.2

    Then: . dy \:=\:3(8^2)(-0.2)\:=\:-38.4

    Hence: . 7.8^3\:\approx\:512 - 38.4 \:=\:\boxed{473.6}

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    How good is this approximation?

    The actual value is: . 7.8^3\:=\:474.552

    Our answer is short by only: . 474.552 - 473.6 \:=\:0.952

    As a percent: . \frac{0.952}{474.552}\:=\:0.002066103 \:\approx\:0.2\%

    We were off by only two-tenths of a percent (2 parts in 1000) . . . not bad!

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  3. #3
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    Quote Originally Posted by math34 View Post
    Use differentials to approximate the value of 7.8^3??

    Should be easy, but can´t find the correct answer. How do you do this... help....
    F(a+e)~=F(a)+eF'(a)

    Put F(x)=x^3, F'(x)=3x^2, and a=8 and e=-0.2, then:

    F(7.8)=F(8)-0.2*3*8^2=512 - 0.6*64=473.6,

    which compares with an exact value of 474.552.

    RonL
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  4. #4
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    How did you get -0.2 for dx?

    Thanks
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  5. #5
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    How did you get -0.2 for dx? ,,

    xdfgsdfg
    Quote Originally Posted by Soroban View Post
    Hello, math34!


    Evidently, we have the function: . f(x) \:=\:x^3
    Then: . df \:=\:3x^2dx

    We know that: . f(8) \,= \,8^3 \,= \,512

    Let x = 8,\;dx = -0.2

    Then: . dy \:=\:3(8^2)(-0.2)\:=\:-38.4

    Hence: . 7.8^3\:\approx\:512 - 38.4 \:=\:\boxed{473.6}

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    How good is this approximation?

    The actual value is: . 7.8^3\:=\:474.552

    Our answer is short by only: . 474.552 - 473.6 \:=\:0.952

    As a percent: . \frac{0.952}{474.552}\:=\:0.002066103 \:\approx\:0.2\%

    We were off by only two-tenths of a percent (2 parts in 1000) . . . not bad!

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  6. #6
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    Because that was the decrease in the value. You went from 8 to 7.8 thus the change was -.2
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