Check Your Understanding Is the equation v = a t dimensionally consistent? One further point thin needs to be mentioned is the effect of the operations of calculus on dimensions. We have seen that dimensions obey the rules of algebra, just like units, but what happens when we take the derivative of one physical quantity with respect to another or integrate a physical quantity over another? The derivative of a function is just the slope of the line tangent to its graph and slopes are ratios, so for physical quantities v and t , we hive that the dimension of the derivative of v with respect to us just the ratio of the dimension of v over that of t : [ d v d t ] = [ v t ] . Similarly, since integrals are just sums of products, the dimension of the integral of v with respect to t is simply the dimension of v times the dimension of t : [ ∫ v d t ] = [ v ] ⋅ [ t ] . By the same reasoning, analogous rules hold for the units of physical quantities derived from other quantities by integration or differentiation.
Check Your Understanding Is the equation v = a t dimensionally consistent? One further point thin needs to be mentioned is the effect of the operations of calculus on dimensions. We have seen that dimensions obey the rules of algebra, just like units, but what happens when we take the derivative of one physical quantity with respect to another or integrate a physical quantity over another? The derivative of a function is just the slope of the line tangent to its graph and slopes are ratios, so for physical quantities v and t , we hive that the dimension of the derivative of v with respect to us just the ratio of the dimension of v over that of t : [ d v d t ] = [ v t ] . Similarly, since integrals are just sums of products, the dimension of the integral of v with respect to t is simply the dimension of v times the dimension of t : [ ∫ v d t ] = [ v ] ⋅ [ t ] . By the same reasoning, analogous rules hold for the units of physical quantities derived from other quantities by integration or differentiation.
Check Your Understanding Is the equation
v
=
a
t
dimensionally consistent?
One further point thin needs to be mentioned is the effect of the operations of calculus on dimensions. We have seen that dimensions obey the rules of algebra, just like units, but what happens when we take the derivative of one physical quantity with respect to another or integrate a physical quantity over another? The derivative of a function is just the slope of the line tangent to its graph and slopes are ratios, so for physical quantities
v
and
t
, we hive that the dimension of the derivative of
v
with respect to us just the ratio of the dimension of v over that of
t
:
[
d
v
d
t
]
=
[
v
t
]
.
Similarly, since integrals are just sums of products, the dimension of the integral of
v
with respect to
t
is simply the dimension of
v
times the dimension of
t
:
[
∫
v
d
t
]
=
[
v
]
⋅
[
t
]
.
By the same reasoning, analogous rules hold for the units of physical quantities derived from other quantities by integration or differentiation.
In the following problem I am asked to calculate a volume respecting the rules of significant figures:
v=(4/3)pi (R^3)Where R=4.5cmHow should I do the calculation taking into account the significant figures?
Should I do the direct calculations and at the end use as the number of significant figures the number that has fewer figures after the decimal point?or parse R separately first? applying the same ruleFirst of all, Thanks.
The purpose of this problem is to show the entire concept of dimensional consistency can be
summarized but the old saying "You can't add apples and oranges." It you have studied power series
expansions in a calculus course, you know the standard mathematical funstions such as trigonometric
functions, logarithms, and exponential function can be expressed as infinite sums of the form where
the an are dimensionless constants for all n = 0, 1, 2, ... and x is the argument of the function. (If you
have not studied power series in calculus yet, just trust us.) Use this fact to explain why the
requirement that all terms in an equation have the same dimensions is sufficient as a definition of
dimensional consistency. That is, it actually implies the arguments of standard mathematical funstions
must be dimensional consistency. That is, it actually implies the arguments of standard mathematical
functions must be dimensionless, so it is not really necessary to make this latter condition a…
So we take a pile of paper, we count the number of sheets, and obtain 226. Then, we measure how tall the pile is with a ruler: 24.0 mm.
As discussed in section 1, we could have a long argument about how much exactly should the uncertainty of this measurement be. Let’s settle to ± 0.5 mm
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