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Area Of An Isosceles Triangle As A Function Of Theta

  1. Aug 12, 2013 #1
    Would this would the proper function, as described in the title of this thread, [itex]A = 1/2 x^2 \cos \frac{\theta}{2}[/itex]?

    And suppose that the side x and the angle were changing with time, would the derivative, with respect to time, be [itex]\frac{dA}{dt} = x \cos \frac{\theta}{2} - 1/4 x^2 \sin \frac{\theta}{2}[/itex]?

    Did I properly derive these?
     
  2. jcsd
  3. Aug 12, 2013 #2

    LCKurtz

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    You could use some parentheses in those expressions. A much easier formula to work with is$$A =\frac 1 2 x^2\sin\theta$$
     
  4. Aug 13, 2013 #3

    LCKurtz

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    I neglected to answer that. The answer is no.
     
  5. Aug 13, 2013 #4

    HallsofIvy

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    You really have left a lot to us to interpret. I assume that you mean that "[itex]\theta[/itex]" is the "vertex" angle- the one opposite the "odd" side. Further, I assume that you are taking "x" to be the length of the two congruent sides. I can't be certain because you do not say any of that.

    If that is what you intended, the dropping a perpendicular from the vertex gives two right triangles with angle [itex]\theta/2[/itex] and hypotenuse of length x. The "opposite side", half the base of the original triangle, has length [itex]x sin(\theta/2)[/itex] and the "near side", the altitude of the original triangle, is [itex]x cos(\theta/2)[/itex]. The whole base of the triangle is [itex]2x sin(\theta/2)[/itex] and so the area is [tex]A= (1/2)(2 x sin(\theta/2))(x cos(\theta/2))= x^2 sin(\theta/2) cos(\theta/2)[/itex].

    In any case, your result, [itex]dA/dt= xcos(\theta/2)- (1/4)x^2 sin(\theta/2)[/itex] cannot possibly be correct because there is no derivative with respect to time. If A is a function of x and y and both are functions of t, then [itex]dA/dt= (\partial A/\partial x)(dx/dt)+ (\partial A/\partial y)(dy/dt)[/itex]. You have only "[itex]\partial A/\partial x+ \partial A/\partial y[/itex]".
     
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