# Homework Help: Area Of An Isosceles Triangle As A Function Of Theta

1. Aug 12, 2013

### 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?

2. Aug 12, 2013

### LCKurtz

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

3. Aug 13, 2013

### LCKurtz

4. Aug 13, 2013

### 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 [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, $dA/dt= xcos(\theta/2)- (1/4)x^2 sin(\theta/2)$ 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 "$\partial A/\partial x+ \partial A/\partial y$".