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Math Help - Approximate change in area

  1. #1
    Member McScruffy's Avatar
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    Approximate change in area

    Given the problem:

    Two sides of a triangle measure 9 in. and 6 in. The measure of the inclusive angle changes from 35^\circ to 32^\circ. Use differentials to approximate the change in the area of the triangle.

    I used A={\frac {1} {2}} ab \sin {C}, where a and b are the lengths of the sides and C is the included angle, for my area function. I am just wondering if anyone has another approach for the area equation that would work well with this problem.

    Thanks
    Last edited by McScruffy; July 8th 2009 at 03:53 PM.
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  2. #2
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    Hello, McScruffy!


    Your area formula is the best one to use.

    Don't forget to convert all angles to radians.

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  3. #3
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by Soroban View Post

    Don't forget to convert all angles to radians.
    Could you eplain why?
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  4. #4
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    Quote Originally Posted by VonNemo19 View Post
    Could you eplain why?
    Formulae such as \frac{d}{d \theta} \cos \theta = - \sin \theta etc. are only valid when \theta is measured in radians.
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  5. #5
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by mr fantastic View Post
    Formulae such as \frac{d}{d \theta} \cos \theta = - \sin \theta etc. are only valid when \theta is measured in radians.
    I did not know that. I'm trying really hard to understand why. Don't tell me, give me a hint.
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  6. #6
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    Quote Originally Posted by VonNemo19 View Post
    I did not know that. I'm trying really hard to understand why. Don't tell me, give me a hint.
    Start from the first principles defination. The various trig limits (that you've probably taken for granted as standard forms) are found using radian measure. Google will turn up relevant pages. And there's an old thread somwehere that discusses the derivation of \lim_{\theta \rightarrow 0} \frac{\sin \theta}{\theta} = 1 using a geometric approach ....
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  7. #7
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by mr fantastic View Post
    Start from the first principles defination. The various trig limits (that you've probably taken for granted as standard forms) are found using radian measure. Google will turn up relevant pages. And there's an old thread somwehere that discusses the derivation of \lim_{\theta \rightarrow 0} \frac{\sin \theta}{\theta} = 1 using a geometric approach ....
    Say no more. I'm very familiar with the proof. I now can see why the use radians was so covienient. It is because the area sector of the portion of the unit circle is squeezed between the two triangles as the angle tends to zero. Right? And using radians, the area sector is simply \theta. Is this right or wrong?
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  8. #8
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    Quote Originally Posted by VonNemo19 View Post
    Say no more. I'm very familiar with the proof. I now can see why the use radians was so covienient. It is because the area sector of the portion of the unit circle is squeezed between the two triangles as the angle tends to zero. Right? And using radians, the area sector is simply \theta. Is this right or wrong?
    Looks OK.
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