# Thread: Find The maximum error (differentials in Calc3)

1. ## Find The maximum error (differentials in Calc3)

OK, here's the problem:

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

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

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

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

Now what do I do?

2. ## Re: Find The maximum error (differentials in Calc3)

Hello, VonNemo19!

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

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

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

Do some math . . . Find $\displaystyle dh$ ... then substitute the given values.

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

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

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

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

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

3. ## Re: Find The maximum error (differentials in Calc3)

Originally Posted by Soroban
Hello, VonNemo19!

Do some math . . . Find $\displaystyle dh$ ... then substitute the given values.

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

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

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

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

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

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

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