Find The maximum error (differentials in Calc3)

**VonNemo19** · June 15th 2011, 12:22 PM

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?

**Soroban** · June 15th 2011, 01:45 PM

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: . $dh \;\approx\;4.17\text{ m}$

**VonNemo19** · June 16th 2011, 10:18 AM

Originally Posted by Soroban

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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