topic Re: Tangent Line extrapolation from a known point on cuve in Discussions

Tangent Line extrapolation from a known point on cuve

dr_d — Sun, 11 Jun 2023 11:21:00 GMT

Greetings,

Hope someone can help me resolve this calculation. I have hundreds of curves each being different. I calculated a critical point on each of the curves, from this point, I need to extrapolate perhaps using a tangent line to the horizontal axis and have that value recorded.

Regards,

Re: Tangent Line extrapolation from a known point on cuve

Mark_Bailey — Thu, 13 Jan 2022 19:59:19 GMT

What kind of curves? What is their equation? What are the values for the parameters in the model equation? Can you use the derivative of the curve to calculate the tangent to the critical point? And is 'perhaps' a statement of your confidence in this approach?

Re: Tangent Line extrapolation from a known point on cuve

dr_d — Thu, 13 Jan 2022 20:16:12 GMT

The curve will be similar to the picture below, the point where the blue line originates is the critical point for the tangent line to be extrapolated to horizontal axis.

Let me know if this makes sense

Thanks

Re: Tangent Line extrapolation from a known point on cuve

Mark_Bailey — Fri, 14 Jan 2022 15:56:08 GMT

It looks like you are hunting for the linear portion of the curve and then estimating the x-intercept.

Re: Tangent Line extrapolation from a known point on cuve

dr_d — Fri, 14 Jan 2022 17:41:15 GMT

Yes, I have hundreds of these curves and all being different would be nice to automate those x-intercepts. I can probably calculate the linear slope using a point above and a point below in the array, then plot that line an record the value for the x-intercept...I'm just having a hard time conceptualizing that in JMP, I was able to do it VB but if JMP can do it will be much easier for the next set of evaluations.

Thanks

Re: Tangent Line extrapolation from a known point on cuve

txnelson — Fri, 14 Jan 2022 18:10:46 GMT

Using the JSL function called Linear Regression(), it should be easy to move the X and Y data that is right of the critical point, and to retrieve the intercept. Look into the Scripting Index for Linear Regression.

Re: Tangent Line extrapolation from a known point on cuve

burakbagdatli — Fri, 14 Jan 2022 23:53:28 GMT

Take a look at the table script I put into a data table I made up for this problem.

You probably don't need the interactivity but it should give you an idea for solving the intercept based on a critical point selection.

The main solution in in the graph script:

selected_rows = Current Data Table() << Get Selected Rows();
If( N Items( selected_rows ) > 0,
	Current Data Table() << Clear Row States();
	critical_row = selected_rows[1, 0][1];
	Current Data Table() << Select Rows( critical_row );
	Current Data Table() << Colors( "Red" );
	Current Data Table() << Clear Select();
);
Try(
	x_prev = :X[critical_row - 1];
	x_crit = :X[critical_row];
	y_prev = :Y[critical_row - 1];
	y_crit = :Y[critical_row];
	slope = (y_crit - y_prev) / (x_crit - x_prev);
	x_intercept = x_crit - y_crit / slope;
	Y Function( slope * (a - x_intercept), a );
	Marker( Marker State( 2 ), {x_intercept, 0} );
	Text( {x_intercept, 0}, Eval Insert( "  intercept = ^x_intercept^" ) );
);

Re: Tangent Line extrapolation from a known point on cuve

dr_d — Mon, 17 Jan 2022 16:21:13 GMT

thanks, I look into this further

Re: Tangent Line extrapolation from a known point on cuve

dr_d — Mon, 17 Jan 2022 16:22:48 GMT

This is really elegant, it worked great for a single curve, I just need to figure out how to implement to multiple curves.

Regards,