Area Of An Isosceles Triangle As A Function Of Theta

[Bashyboy]

Would this would the proper function, as described in the title of this thread, A = 1/2 x^2 \cos \frac{\theta}{2}?

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

Did I properly derive these?

 

[LCKurtz]

Would this would the proper function, as described in the title of this thread, A = 1/2 x^2 \cos \frac{\theta}{2}?

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

Did I properly derive these?

You could use some parentheses in those expressions. A much easier formula to work with is$$A =\frac 1 2 x^2\sin\theta$$

 

[LCKurtz]

Would this would the proper function, as described in the title of this thread, A = 1/2 x^2 \cos \frac{\theta}{2}?

I neglected to answer that. The answer is no.

 

[HallsofIvy]

You really have left a lot to us to interpret. I assume that you mean that "\theta" 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 \theta/2 and hypotenuse of length x. The "opposite side", half the base of the original triangle, has length x sin(\theta/2) and the "near side", the altitude of the original triangle, is x cos(\theta/2). The whole base of the triangle is 2x sin(\theta/2) and so the area is A= (1/2)(2 x sin(\theta/2))(x cos(\theta/2))= x^2 sin(\theta/2) cos(\theta/2)[/itex].<br /> <br /> In any case, your result, dA/dt= xcos(\theta/2)- (1/4)x^2 sin(\theta/2) 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 dA/dt= (\partial A/\partial x)(dx/dt)+ (\partial A/\partial y)(dy/dt). You have only &quot;\partial A/\partial x+ \partial A/\partial y&quot;.