Where should the exponent be written in numbers with units of measurement?

Question

If you are to calculate the hypotenuse of a triangle, the formula is:

$h = \sqrt{x^2 + y^2}$

If you don't have any units for the numbers, replacing x and y is pretty straightforward: $h = \sqrt{4^2 + 6^2}$

But what if the numbers are in meters?
$h = \sqrt{4^2m + 6^2m}$ (wrong, would become $\sqrt{52m}$)
$h = \sqrt{4m^2 + 6m^2}$ (wrong, would become $\sqrt{10m^2}$)
$h = \sqrt{(4m)^2 + (6m)^2}$ (correct, would become $\sqrt{52m^2}$)

Or should I just ignore the unit of measurement in these cases?

Answer 1

Suppose you have been given $x$ and $y$ in metres, and you'd like to know the quantity, $z=\sqrt{x^2+y^2}$. Then, as you have predicted this quatity will be in metres.

Two things have been involved:

Homogeneity of Dimension

Two quantities of different dimensions cannot be added. This is one of the axioms of numerical physics.

Example: It is clear that adding $5$ metres to $3$ seconds does not give a physically meaningful quantity that can be interpreted in real life.

Certain functions only take values in dimensionless quantities

For instance, $\sin (\sqrt{x^2+y^2})$ would not make sense even if $x$ and $y$ have same dimensions. This is a bit subtler, but this is what it is!

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• Coming to your question, the first quantity you tell us in dimension of $m^{1/2}$ which is against your guess!

• The second quantity is dimensionally fine while numerically this is not what you want.

• The third quantity is fine in all ways.

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My suggestion:

First manipulate the numbers and then the units separately. This is a good practice in Numerical Physics. The other answers have done it all at one go. But, I don't prefer it that way!

Answer 2

The last one: $$\begin{align} h&=\sqrt{(4\text{ m})^2+(6\text{ m})^2} \\ &=\sqrt{16\text{ m}^2+36\text{ m}^2} \\ &=\sqrt{52\text{ m}^2} \\ &=2\sqrt{13}\text{ m} \end{align}$$

Answer 3

The last one is correct. The general idea is that you treat the unit as if it were another number:

$$\begin{align*}h &= \sqrt{(4\mathrm{m})^2+(6\mathrm{m})^2} \\ &= \sqrt{16\mathrm{m}^2 + 36\mathrm{m}^2} \\ &= \sqrt{52 \mathrm{m}^2} \\ &= \sqrt{52}\sqrt{\mathrm{m}^2} \\ &= \sqrt{52}\mathrm{m}\end{align*}$$