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Math Help - Find The maximum error (differentials in Calc3)

  1. #1
    No one in Particular VonNemo19's Avatar
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    Find The maximum error (differentials in Calc3)

    OK, here's the problem:

    The angle of elevation of the top of a hill is 21^{\circ}\pm30'' and the distance to the top of the hill is (500\pm.3)\text{ m} . Find the maximum error in determining the height of the hill.

    So, I let t be the distance to the top, \theta be the angle of elevation, and h be the height of the hill, then

    h(t,\theta)=t\sin{\theta}

    Now, the total differential is dh=\frac{\partial{h}}{\partial{t}}dt+\frac{\partia  l{h}}{\partial{\theta}}d\theta

    Now what do I do?
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  2. #2
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    Re: Find The maximum error (differentials in Calc3)

    Hello, VonNemo19!

    The angle of elevation of the top of a hill is: 21^{\circ}\pm30''
    and the distance to the top of the hill is: (500\pm 0.3)\text{ m}.
    Find the maximum error in determining the height of the hill.

    So, I let t be the distance to the top, \theta be the angle of elevation,
    and h be the height of the hill, then: . h(t,\theta)\:=\:t\sin{\theta}

    Now, the total differential is: . dh \:=\:\frac{\partial h}{\partial t}dt+\frac{\partial h}{\partial \theta}d\theta
    Now what do I do?

    Do some math . . . Find dh ... then substitute the given values.

    We have: . h \:=\:t\sin\theta \quad\Rightarrow\quad \begin{Bmatrix}\dfrac{\partial h}{\partial t} &=& \sin\theta \\ \\[-3mm] \dfrac{\partial h}{\partial\theta} &=& t\cos\theta \end{Bmatrix}

    Hence: . dh \;=\;(\sin\theta)dt + (t\cos\theta)d\theta


    We are given: . \begin{Bmatrix}\theta &=& 0.3665 && d\theta &=& 0.0087 \\ t &=& 500 && dt &=& 0.3\end{Bmatrix} . (Angles are in radians.)

    Therefore: . dh \;=\;(\sin0.3665)(0.3)(500\cos0.3665)(0.0087) \;=\;4.16860972

    The maximum error is: . dh \;\approx\;4.17\text{ m}

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  3. #3
    No one in Particular VonNemo19's Avatar
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    Re: Find The maximum error (differentials in Calc3)

    Quote Originally Posted by Soroban View Post
    Hello, VonNemo19!


    Do some math . . . Find dh ... then substitute the given values.

    We have: . h \:=\:t\sin\theta \quad\Rightarrow\quad \begin{Bmatrix}\dfrac{\partial h}{\partial t} &=& \sin\theta \\ \\[-3mm] \dfrac{\partial h}{\partial\theta} &=& t\cos\theta \end{Bmatrix}

    Hence: . dh \;=\;(\sin\theta)dt + (t\cos\theta)d\theta


    We are given: . \begin{Bmatrix}\theta &=& 0.3665 && d\theta &=& 0.0087 \\ t &=& 500 && dt &=& 0.3\end{Bmatrix} . (Angles are in radians.)

    Therefore: . dh \;=\;(\sin0.3665)(0.3)(500\cos0.3665)(0.0087) \;=\;4.16860972

    The maximum error is: . dh \;\approx\;4.17\text{ m}

    OK, so good. This means that I was on the right track.


    So, the total differential is the same as the max error?
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