# Non-Linear Relation

#### HazyGeoGeek

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?

#### skipjack

Forum Staff
Can you define the curve in some way?

• 2 people

#### HazyGeoGeek

Concave

It can be of conic, but it has to be a concave shape.

#### 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.

• 1 person

#### HazyGeoGeek

Question Explained

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.

#### skipjack

Forum Staff
. . . height*angle.
It looks like height*tan(angle) to me, where tan is the usual trigonometric tangent function.

#### HazyGeoGeek

Question Explained

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.

#### DarnItJimImAnEngineer

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 $a, b$

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