I am doing question 6 of a practice calculus exam, namely:
A container in the shape of a right circular cone with vertex angle a
right angle is partially filled with water. a) Suppose water is added
at the rate of 3 cu.cm./sec. How fast is the water level rising when
the height h = 2cm.?
My answer was ${dh\over dt} = {3\over 4\pi} \cdot {c^2}$, where $c$ is a fixed constant equal to the ratio of the height to the radius.
The given solution, however, is ${dh\over dt} = {3\over 4\pi}$. The solution uses the assumption that "$r=h$ since this is a right circular cone." Is this a reasonable assumption? I have checked definition of right circular cone and cannot find anything about this.
Thanks,
Josh
Best Answer
Let me amplify Roddy's answer.
A right circular cone has a circular base; let's assume that it's at the origin of 3-space, and that the base lies in the $xy$-plane
The cone is "right" so that its vertex lies along the perpendicular to the base-plane, through the base-center. So let's go ahead and say that the vertex is at the point $(0,0,h)$, just to give the thing a coordinate. ("Right" here says nothing about "right angle" -- it means "upright rather than slonched over to one side", to use the technical term :) )
Looking solely at the $xz$-plane, i.e., a slice through the cone, we see a cross section that looks like this shape: $\wedge$, with the angle at the vertex (location $(0,0,h)$, remember) being a right angle (because the problem says "with vertex angle a right angle"). The location of the bottom vertex of the right leg is then $(r, 0, 0)$ (because $y = 0$ in the $xz$-plane, and because the whole bottom of the cone is in the $z = 0$ plane).
If we drop a vertical line from the top of the wedge to the $z = 0$ line in our picture, the half-angle at the top is 45 degrees. That makes the ratio of the legs of the triangle (which are $r$ and $h$ in length) be $\tan 45 = 1$. Hence $\frac{r}{h} = 1$, so $r = h$.