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Math Help - Approximation

  1. #1
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    Approximation

    Hi, I have some problems with approximation.I'll be glad if you could help.We just learnt basics.I can't solve this kind of questions.

    App.(2.9)

    2x^4 + 5x - 7x ^2 +8x -3 X= 1.03

    How accurately must the edge of a cube be measured so that the volume will be correct to within 3% ?


    Thanks in advance. I have an exam on Saturday ! Your help will be much appreciated.
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  2. #2
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    Quote Originally Posted by truevein View Post
    Hi, I have some problems with approximation.I'll be glad if you could help.We just learnt basics.I can't solve this kind of questions.

    App.(2.9)

    2x^4 + 5x - 7x ^2 +8x -3 X= 1.03

    How accurately must the edge of a cube be measured so that the volume will be correct to within 3% ?


    Thanks in advance. I have an exam on Saturday ! Your help will be much appreciated.
    Hopefully, the "basics" included using the derivative to find the tangent line approximation to a graph: if y= f(x) then the tangent line at x= a is y= f'(a)(x- a)+ F(a).

    For the first problem, find the tangent line to y= x^3 at x= 3. The derivative of x^3 is 3x^2 and at x= 3, that is 27. Also 3^3 is 27. The tangent line is y= 27(x- 3)+ 27. Now let x= 2.9

    Do the same with y= 2x^4 + 5x - 7x ^2 +8x -3. Find the tangent line to the graph at x= 1 and let x= 1.03.


    If a side of a cube is "s", its volume is V= s^3. The derivative is \frac{dV}{ds} = 3s^2 and, in terms of differentials dV= 3s^2 ds. You can think of "dV" and "ds" as the errors in V and s. Saying that you want "the volume will be correct to within 3%" means that you want \frac{dV}{V}= \frac{3s^2}{s^3}ds= \frac{3}{s}ds less than .03. "How accurately must the edge of a cube be measured", as a percentage, is \frac{ds}{s}.
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  3. #3
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    Thanks a lot! I could solve the first two myself with the help of f(x) = f'(xo)(x-xo) +f(xo). I didn't memorise it, btw, it is the slope formula But the third question was really bugging me. Thanks for your time and effort again.
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