Geometry Geometry Math Forum

October 25th, 2019, 05:51 AM   #1
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Could someone tell me about the relation in the attached image.
One in the top follows the linear relation and extended length can be obtained through multiplying the height*angle.

What would be relation for bottom one?
Attached Images Q.jpg (7.4 KB, 4 views) October 25th, 2019, 06:47 AM #2 Global Moderator   Joined: Dec 2006 Posts: 21,124 Thanks: 2332 Can you define the curve in some way? Thanks from topsquark and HazyGeoGeek October 25th, 2019, 08:49 AM #3 Newbie   Joined: Oct 2019 From: France Posts: 4 Thanks: 0 Concave It can be of conic, but it has to be a concave shape. Thank you for your response.. October 25th, 2019, 12:27 PM #4 Senior Member   Joined: Jun 2019 From: USA Posts: 386 Thanks: 211 Is this a design problem? As in, you are supposed to choose a concave shape, and then calculate the length? A parabola, a hyperbola, a circular arc, and an elliptical arc are all conics that can be concave and can fit there. Once you have the shape (whether it be given, or whether you design it yourself), just write the equation as y = f(x). x is the vertical distance, and y is the horizontal distance. Then the ? is just f(h). For example, the surface of the top shape is described by the equation y = ax. Substitute x = h, and y(h) = ah. Thanks from topsquark October 28th, 2019, 05:04 AM   #5
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Quote:
 Originally Posted by skipjack Can you define the curve in some way?
How to find the points for conic curve? You can find the explanation in the picture attached.
Attached Images qq.jpg (22.0 KB, 2 views) October 28th, 2019, 05:59 AM   #6
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 Originally Posted by HazyGeoGeek . . . height*angle.
It looks like height*tan(angle) to me, where tan is the usual trigonometric tangent function. October 28th, 2019, 06:01 AM   #7
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Question Explained

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 Originally Posted by DarnItJimImAnEngineer Is this a design problem? As in, you are supposed to choose a concave shape, and then calculate the length? A parabola, a hyperbola, a circular arc, and an elliptical arc are all conics that can be concave and can fit there. Once you have the shape (whether it be given, or whether you design it yourself), just write the equation as y = f(x). x is the vertical distance, and y is the horizontal distance. Then the ? is just f(h). For example, the surface of the top shape is described by the equation y = ax. Substitute x = h, and y(h) = ah.
Yes This is a design problem. And I have to create a concave curve. Could you plaese specify how to obtain the concave curve points. When I tried with the parabolic equation it doesn't work.
Attached Images qq.jpg (22.0 KB, 1 views) October 28th, 2019, 06:06 AM #8 Senior Member   Joined: Jun 2019 From: USA Posts: 386 Thanks: 211 What do you mean by, "it didn't work?" Your simple choices for conics are: $y = \sqrt{r^2-x^2}$, where $r\geq h_5$ $y = \sqrt{a^2-\frac{a^2}{b^2}x^2}$, where $b \geq h_5$ $y = ax^2$, where $a > 0$ $y = \sqrt{\frac{a^2}{b^2}x^2+a^2}$, for any $a, b$ Last edited by DarnItJimImAnEngineer; October 28th, 2019 at 06:56 AM. Reason: Fixed hyperbolic equation for consistency Tags nonlinear, relation Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post uniquegel Algebra 4 September 8th, 2014 05:18 PM RGNIT Calculus 0 March 21st, 2014 12:21 AM tools Linear Algebra 1 September 21st, 2012 01:38 PM Regnes Algebra 3 July 13th, 2011 05:56 PM gaussrelatz Algebra 0 December 31st, 1969 04:00 PM

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