|
||||||||||||
|
||||||||||||
| ||||||||||||
Hi Joel, Here is my diagram of your inverted bowl. The cross section in the XY-plane is a parabola opening downwards so its equation is \[y = a x^2 + b\] where $a$ and $b$ are real numbers. I have made the height of the bowl $H$ cm and the radius of the case $R$ cm. Hence in the XY-plane $(0, H)$ and $(R, 0)$ are on the parabola and hence $b = H$ and $a = - \large \frac{H}{R^2}$ and thus the equation of the parabola is \[y = - \frac{H}{R^2} x^2 + H.\] Now suppose you slice the bowl by a plane which is $h$ cm above the base to obtain a circle of radius $r$ cm. The point $(r, h)$ in the XY-plane is on the parabola and hence \[h= - \frac{H}{R^2} r^2 + H.\] Solving for $r$ yields \[r = \sqrt{\frac{H-h}{H}} R.\] Using $r$ you can calculate the circumference of the red circle in the diagram above. Penny | ||||||||||||
|
||||||||||||
Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences. |