Figuring out the Constant of Variation from a Dataset

[Beren]

Hey all,
Say I have a table that shows the relationship between the surface area of a sphere (A), and its radius (r), but I don't know the formula.

Radius (cm) [ 2 ] 7 ] 15 ] 20 ] 50 ]
Surface area (cm^2) [ 50.24 ] 615.44 ] 2826 ] 5024 ] 31 400 ]

How would I (a) figure out the constant of variation (not quite sure what they mean by this)
(b) Express this relationship as an equation using pi = 3.14?

Thanks in advance.

 

[stapel]

Say I have a table that shows the relationship between the surface area of a sphere (A), and its radius (r), but I don't know the formula.

Radius (cm) [ 2 ] 7 ] 15 ] 20 ] 50 ]
Surface area (cm^2) [ 50.24 ] 615.44 ] 2826 ] 5024 ] 31 400 ]

How would I (a) figure out the constant of variation (not quite sure what they mean by this)

To learn what "variation" equations are, and how their constants work, try here.

As for how you are supposed to find this particular relationship, we would need to know what techniques or tools have been provided to you by your class and textbook. The exercise provides only a table of values. You are not told what sort of "relationship" you seek (in this case, a direct-variation with the square of the radius), so there must be some other method you're supposed to use to "find" the formula for the surface area A of a sphere in terms of its radius r.

What method(s) are you supposed to use? "Technology" for finding the regression equation, or something else? Thank you!

 

[Beren]

uhm, not sure, I was curious as to whether it is possible to figure out the formula from the dataset, I ended up just searching up the formula . That was all the information given in the question however, I wasn't told anything else... Anyway thanks for the link.

 

[Beren]

To learn what "variation" equations are, and how their constants work, try here.

We would need to know what techniques or tools have been provided to you by your class and textbook. The exercise provides only a table of values. You are not told what sort of "relationship" you seek (in this case, a direct-variation with the square of the radius), so there must be some other method you're supposed to use to "find" the formula for the surface area A of a sphere in terms of its radius r.
What method(s) are you supposed to use? "Technology" for finding the regression equation, or something else? Thank you!

I wasn't given any other information by the actual question, since the exercise was about direct proportions I presume that was the method I should use to figure it out. I ended up just directly plugging in the formula for surface area. Anyway thanks for the link.

 

[Ishuda]

Hey all,
Say I have a table that shows the relationship between the surface area of a sphere (A), and its radius (r), but I don't know the formula.

Radius (cm) [ 2 ] 7 ] 15 ] 20 ] 50 ]
Surface area (cm^2) [ 50.24 ] 615.44 ] 2826 ] 5024 ] 31 400 ]

How would I (a) figure out the constant of variation (not quite sure what they mean by this)
(b) Express this relationship as an equation using pi = 3.14?

Thanks in advance.

What I would do: First call the radius x and the surface area y so that I would not be tempted to use what I know about the surface area of a sphere. Next, from (b), I would assume the constant of variation is related to 3.14, so I would divide y (to get y₁) by 3.14 just to see if I get something 'nice'. I do:
(x,y₁): (2, 16), (7, 196), (15, 900), (20, 1600), (50, 10000)

If I plot that, I get something that looks like a parabola which goes through (0,0) suggesting a form of y₁=a x + b x². Since (a) indicates this is a constant of variation problem [that is only one constant], because of the way the plot looks, it is very likely that a is zero so lets divide by x² to get y₂
(x,y₂): (2, 4), (7, 4), (15, 4), (20, 4), (50, 4)

So
y₂ = 4 = y1 / x² = (y / 3.14) / x²
or
y= 4 * 3.14 x²
or
A = 4 \(\displaystyle \pi\) r²

EDIT: You could also divide y₁ by x and plot that to notice that it was a straight line going through (0,0) with a slope of 4 to arrive at the same conclusion.