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What is the name of this type of related equation?

1

$begingroup$

So I was playing with this idea of taking any equation ($f(x)$) and be able to have a parametric equation so that both equations slopes are parallel. I was able to generate the said parametric equation.

$x(t)=t+cos(tan^-1(f'(t))-fracpi2)$

$y(t)=f(t)+sin(tan^-1(f'(t))-fracpi2)$

Then I thought how could I calculate the area between these two curves given 2 lines of length 1 that pass through the corresponding points $(a,f(a))$ and $(x(a),y(a))$. And the other 2 points $(b,f(b))$ and $(x(b),y(b))$. Here is a picture from Desmos to help illustrate what I mean. Picture The shaded region would be what I am trying to find. In this case $f(x)=x^2$, $a=0$ and $b=1$. I found the equation for the area as follows.

$A=frac12int_a^b(sqrt1+f'(t)^2+sqrtx'(t)^2+y'(t)^2)dt$

With the case used in the picture, $A=2.03251721644$. Now I didn't to a prof yet but I did do some preliminary tests to see if this works. And the equation passes the tests.

Here are my questions. What is this called? Is this something that is already known about in the math community? And if so, please point me to where I can read more about this topic. Thank you!

-Follow up question. Is my area equation correct? From what I've gathered it is but I would like a second opinion.

calculus trigonometry algebraic-geometry parametric

share|cite|improve this question

edited Mar 26 at 18:26

GeoCalc

asked Mar 26 at 16:03

GeoCalcGeoCalc

162

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Presumably, you've already seen this?
  $endgroup$
  – J. M. is a poor mathematician
  Mar 26 at 16:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you, I had not seen that @J.M.isnotamathematician
  $endgroup$
  – GeoCalc
  Mar 26 at 16:43
  

  
  
  
  

add a comment | 

1

$begingroup$

So I was playing with this idea of taking any equation ($f(x)$) and be able to have a parametric equation so that both equations slopes are parallel. I was able to generate the said parametric equation.

$x(t)=t+cos(tan^-1(f'(t))-fracpi2)$

$y(t)=f(t)+sin(tan^-1(f'(t))-fracpi2)$

Then I thought how could I calculate the area between these two curves given 2 lines of length 1 that pass through the corresponding points $(a,f(a))$ and $(x(a),y(a))$. And the other 2 points $(b,f(b))$ and $(x(b),y(b))$. Here is a picture from Desmos to help illustrate what I mean. Picture The shaded region would be what I am trying to find. In this case $f(x)=x^2$, $a=0$ and $b=1$. I found the equation for the area as follows.

$A=frac12int_a^b(sqrt1+f'(t)^2+sqrtx'(t)^2+y'(t)^2)dt$

With the case used in the picture, $A=2.03251721644$. Now I didn't to a prof yet but I did do some preliminary tests to see if this works. And the equation passes the tests.

Here are my questions. What is this called? Is this something that is already known about in the math community? And if so, please point me to where I can read more about this topic. Thank you!

-Follow up question. Is my area equation correct? From what I've gathered it is but I would like a second opinion.

calculus trigonometry algebraic-geometry parametric

share|cite|improve this question

edited Mar 26 at 18:26

GeoCalc

asked Mar 26 at 16:03

GeoCalcGeoCalc

162

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Presumably, you've already seen this?
  $endgroup$
  – J. M. is a poor mathematician
  Mar 26 at 16:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you, I had not seen that @J.M.isnotamathematician
  $endgroup$
  – GeoCalc
  Mar 26 at 16:43
  

  
  
  
  

add a comment | 

$begingroup$

So I was playing with this idea of taking any equation ($f(x)$) and be able to have a parametric equation so that both equations slopes are parallel. I was able to generate the said parametric equation.

$x(t)=t+cos(tan^-1(f'(t))-fracpi2)$

$y(t)=f(t)+sin(tan^-1(f'(t))-fracpi2)$

Then I thought how could I calculate the area between these two curves given 2 lines of length 1 that pass through the corresponding points $(a,f(a))$ and $(x(a),y(a))$. And the other 2 points $(b,f(b))$ and $(x(b),y(b))$. Here is a picture from Desmos to help illustrate what I mean. Picture The shaded region would be what I am trying to find. In this case $f(x)=x^2$, $a=0$ and $b=1$. I found the equation for the area as follows.

$A=frac12int_a^b(sqrt1+f'(t)^2+sqrtx'(t)^2+y'(t)^2)dt$

With the case used in the picture, $A=2.03251721644$. Now I didn't to a prof yet but I did do some preliminary tests to see if this works. And the equation passes the tests.

Here are my questions. What is this called? Is this something that is already known about in the math community? And if so, please point me to where I can read more about this topic. Thank you!

-Follow up question. Is my area equation correct? From what I've gathered it is but I would like a second opinion.

calculus trigonometry algebraic-geometry parametric

share|cite|improve this question

edited Mar 26 at 18:26

GeoCalc

asked Mar 26 at 16:03

GeoCalcGeoCalc

162

$endgroup$

share|cite|improve this question

edited Mar 26 at 18:26

GeoCalc

asked Mar 26 at 16:03

GeoCalcGeoCalc

162

share|cite|improve this question

edited Mar 26 at 18:26

GeoCalc

asked Mar 26 at 16:03

GeoCalcGeoCalc

162

asked Mar 26 at 16:03

GeoCalcGeoCalc

162

asked Mar 26 at 16:03

GeoCalcGeoCalc

162