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dmcampos

  • 7 months ago

18.01SC Single Variable Calculus: Problem Set 1. Question 1F-7c: The result I got was (-3mgr) / ( (1+r^2)^(7/4) ) but the solutions say it is (-3mgr) / ( (1+r^2)^(5/2) ) Can anyone explain to me what I did wrong? I used both the chain rule and the quotient rule, and after doing it, I got this: [ - mg * ( (3/2) * (1 + r^2)^(1/2) * 2r ) ] / ( (1 + r^2)^(9/4) )

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  1. phi
    • 7 months ago
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    the power rule might be the easiest way: \[ d \ x^n = n x ^{n-1} dx \] \[ \frac{d}{dr} \frac{m g}{\left( 1+r^2\right)^{\frac{3}{2}} } = \frac{d}{dr} m g \left( 1+r^2\right)^{-\frac{3}{2}} \] m and g are constants, so we can move them outside the derivative \[ m g \frac{d}{dr}\left( 1+r^2\right)^{-\frac{3}{2}} \\ = - \frac{3}{2}m g\left( 1+r^2\right)^{-\frac{3}{2}- 1} \frac{d}{dr}\left( 1+r^2\right)\\ = - \frac{3}{2}m g\left( 1+r^2\right)^{-\frac{5}{2}} 2 r\\ = \frac{-3mgr}{\left( 1+r^2\right)^{\frac{5}{2}}} \]

  2. phi
    • 7 months ago
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    You can use the quotient rule (though normally when the numerator is a constant you would not bother) your answer is close: \[ \frac{-mg \cdot \frac{3}{2} (1+r^2)^{\frac{1}{2}} \cdot 2 r}{(1 + r^2)^{\frac{9}{4}}} \] The "bug" is that the square of the denominator is \[ \left( (1+r^2)^\frac{3}{2} \right)^2= (1+r^2)^{\frac{3}{2}\cdot 2}= (1+r^2)^3\] if we use the correct denominator we have \[ \frac{-mg \cdot \frac{3}{\cancel{2}} (1+r^2)^{\frac{1}{2}} \cdot \cancel{2} r}{(1 + r^2)^{3}} \\ = \frac{-3mg r}{(1 + r^2)^{3-\frac{1}{2}}} \\ = \frac{-3mg r}{(1 + r^2)^\frac{5}{2}} \] which is the correct result

  3. dmcampos
    • 7 months ago
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    Thanks a lot, your answer was perfect, and I figured out my mistake. (and I'm sorry I didn't give many details, I was just going update the post and you had already answered) It was really simple: I assumed \[((1+r^{2})^{\frac{ 3 }{ 2 }})^{2} = (1+r^{2}) ^{\frac{ 9 }{ 4 }}\]instead of \[(1+r^{2})^{3}\]which is obviously the correct way.

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