I am stuck on the following problem;
Let a cone with height $h$ and base area $B$ have the density $\rho (x) = \rho_{0} \frac{3x^{2} + 2xh}{h^2}, 0 \leq x \leq h$ Where $x$ is the distance from the cone apex and $\rho_{0}$ is a real constant.
(a) Show that the relation between cone radius $y$ and distance from cone apex $x$ is given by: $y = (\frac{B}{\pi h^{2}})^{2}x$
I would really appreciate a pointer on where to get started. Is the information regarding density relevant here? I know how to use it later on where i'm asked to find the cone's mass.
Thanks for any help you can give!
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$\begingroup$The density is not relevant, since at this point it's pure geometry. I assume we're talking about a right cone, otherwise this doesn't make any sense.
To prove the equation given in the question, look at the cone from the side - suddenly it becomes an isoceles triangle. What's the base of the triangle? What's the height? Now drop down a perpendicular from the top of the triangle to the base, which it will bisect. Pick a point on that perpendicular, and mark it as being distance $x$ from the apex. Draw the perpendicular at that point. The base of the right triangle formed by that perpendicular is of length $y$. Now do you have any similar triangles? Can you relate $y$ and $x$ to the base and height of the larger triangle, and hence to $B$ and $h$?
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