Area between Triangle and Rectangle

[orchidthief]

Homework Statement

Consider a large triangle, the tip is located at the origo x=0, it is sloped at an angle θ and -θ relative to the x-axis, relative to the x-axis, its dimensions in x can be considered infinite.

A large stripe/strip/band is placed on top of the triangle but perpendicular to the x-axis so to speak, it has a width b and it's center is located at x=0 and it can be considered infinite in the y direction. (So one edge of the stripe at the start is at x=b/2 and the other at x=-b/2.)

This large stripe can be shifted in the x direction to gradually cover a larger and larger trapezoidal area of the triangle.

The exercise is to find a generalised expression for the area shared by the triangle and the stripe as a function of x.2. The attempt at a solution

What I deduced was that this was a piecewise solution, while x<b/2 the area they share is:

A(x)= tanθ(x+b/2)^2 (a basic area of a triangle, that gives it an area when x=0)

And when x>b/2 it

A(x)=1/2*b*(q+p) (the area of a trapezoid)

Where q=2tanθ(x-b/2) and p=2tanθ(x+b/2) giving me a final expression in this part of:

A(x)=2bx*tanθ

Now my question is, the actual wording of the problem makes it sound like it should be possible to find one elegant expression covering the x>0 regardless of the zone, but I'm just completely lost in how one such might be found. Any help would be nice.

 

[LCKurtz]

Homework Statement

Consider a large triangle, the tip is located at x=0, it is sloped at an angle θ, it's dimensions in x can be considered infinite.

A large rectangle is placed on top of the triangle but perpendicular to the x-axis so to speak, it has a width b and it's center is located at x=0 and it can be considered infinite in the y direction. (So one side of the rectangle at the start is at x=b/2 and the other at x=-b/2.)

This large rectangle can be shifted in the x direction to gradually cover a larger and larger trapezoidal area of the triangle.

The exercise is to find a generalised expression for the area shared by the triangle and the rectangle as a function of x.

So one side of your triangle is a line through the origin at angle θ. What is a second side, the x axis? If it's a triangle, what is the third side? And your rectangle, what it its height or are you saying it is a vertical strip. I'm having trouble picturing the trapezoid to which you refer. Even so, once you explain that, I think almost certainly you will have a multiple piece formula for your answer.

 

[orchidthief]

Yeah the tip of the triangle is at origin. The axis down the middle of the triangle is the x-axis, and then one side of the triangle running at an angle θ and one at -θ. Can see how that would be confusing. It can be considered infinite in the x direction.

Yes, the rectangle is a stripe that is infinite in the y direction, and has width b in the x direction, centered around the y-axis at the start, but then the whole thing can be shifted in the x direction. At the start the tip and the center of the stripe are both at x=0, so the area they share is just a small triangle, but as the stripe is moved in the x direction, eventually the area they share goes from being a triangle to a trapezoid. (When the stripe has moved b/2 in the x direction and the triangle starts emerging on the other side of the stripe.)

Btw, I realize this is in the precalc forum, since it seems like a fairly straightforward geometrical problem, but if calculus can be helpful then that's fair game as well.

 

[LCKurtz]

OK. With that clarification, I get the same answers you have. Good work. The answer is inherently a two piece formula and the only way to express it as a "single" formula would be to introduce an artifice like a unit step function into the expression.

 

[orchidthief]

Yeah, didn't really have any doubts about the correctness of the answers I got, more about consolidating it into a single expression. How would one go about trying to introduce such a unit step artifice? Heavisides step function I take it? Some hints and pointers would be useful, only having to deal with one expression would certainly simplify the rest of the homework set.

 

[LCKurtz]

Yeah, didn't really have any doubts about the correctness of the answers I got, more about consolidating it into a single expression. How would one go about trying to introduce such a unit step artifice? Heavisides step function I take it? Some hints and pointers would be useful, only having to deal with one expression would certainly simplify the rest of the homework set.

Let's call the Heaviside unit step function u(x). If you want a function F(x) to equal f(x) for x < a and g(x) for x > a you could write:

F(x) = f(x) + u(x-a)(g(x) - f(x)).

When x < a the second term is 0 and when x > a the second term subtracts out the f and adds in the g. You can add a second similar term if it changes formula again for some other value of x greater than a.

 

[orchidthief]

Let's call the Heaviside unit step function u(x). If you want a function F(x) to equal f(x) for x < a and g(x) for x > a you could write:

F(x) = f(x) + u(x-a)(g(x) - f(x)).

When x < a the second term is 0 and when x > a the second term subtracts out the f and adds in the g. You can add a second similar term if it changes formula again for some other value of x greater than a.

That makes good sense. Thanks a bunch for the help.