Video: Using Dimensional Analysis to Find the Dimensions of Constants in a Fifth Degree Polynomial

Consider the equation 𝑠 = 𝑠₀ + 𝑣₀𝑑 + π‘Žβ‚€π‘‘Β²/2 + 𝑗₀𝑑³/6 + 𝑆₀𝑑⁴/24 + 𝑐𝑑⁡/120, where 𝑠 is a length and 𝑑 is a time. What is the dimension of 𝑠₀? What is the dimension of 𝑣₀? What is the dimension of π‘Žβ‚€? What is the dimension of 𝑗₀? What is the dimension of 𝑆₀? And what is the dimension of 𝑐?

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Video Transcript

Consider the equation 𝑠 equals 𝑠 zero plus 𝑣 zero 𝑑 plus π‘Ž zero 𝑑 squared divided by two plus 𝑗 zero 𝑑 cubed divided by six plus capital 𝑆 zero times 𝑑 to the fourth divided by 24 plus 𝑐 times 𝑑 to the fifth divided by 120, where 𝑠 is a length and 𝑑 is a time. What is the dimension of 𝑠 zero? What is the dimension of 𝑣 zero? What is the dimension of π‘Ž zero? What is the dimension of 𝑗 zero? What is the dimension of capital 𝑆 zero? And what is the dimension of 𝑐?

This is a question in dimensional analysis, where we want to solve for the dimension of each unknown term in the equation for 𝑠. We can symbolise the dimension of a variable by writing β€œdim” and then in parentheses that variable. For example, we’re told the dimension of 𝑠 is a length, which we’ll abbreviate capital 𝐿, and that the dimension of 𝑑 is time, which we’ll abbreviate capital 𝑇.

Considering this equation, we can now solve for the dimensions of each of the variables in it. Looking at 𝑠 zero where it appears in the equation, we see that it’s combined with no other terms but stands on its own, so the dimension of 𝑠 zero must be the same as the dimension of 𝑠. That means the dimension of 𝑠 zero is length. Next, we look at 𝑣 zero. 𝑣 zero we see is multiplied by 𝑑, which has units of time, so the dimension of 𝑣 zero multiplied by capital 𝑇 must be equal to the dimension of 𝑠 length. Therefore, 𝑣 zero has dimensions of length per time.

One way to write that is length times inverse time. When we consider π‘Ž zero, we see that that value is multiplied by 𝑑 squared, so the dimension of π‘Ž zero when multiplied by 𝑑 squared must yield the dimensions of 𝑠, length, therefore π‘Ž zero has dimensions of length per time squared.

Another way to write that is length times time to the negative two. Looking next at 𝑗 zero, that value is multiplied by 𝑑 cubed since 𝑗 zero times 𝑑 cubed must have units of length. In order for it to match with the units of 𝑠, the dimension of 𝑗 zero is length per time cubed or length times time to the negative three. Next, we look at capital 𝑆 sub zero.

Capital 𝑆 sub zero multiplied by 𝑑 to the fourth has units of length, so the dimensions of 𝑠 zero equals length per time to the fourth power or length times time to the negative fourth. Finally, we look at 𝑐. This value multiplied by 𝑑 to the fifth has the same units as 𝑠, the units of length, therefore the dimensions of 𝑐 are length per 𝑑 to the fifth, which can be equivalently written length times time to the negative five.

These are the various dimensions of the variables written in this equation.

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