Now we {{CMD_317}}find equations{{CMD_318}} that relate all the variables. Notice that water in the container assumes the shape of the container. In this example, the shape is a cone. Therefore, we use the formula for the volume of a cone {{MATH_319}} Let's draw a cross section of the cone. {{CMD_320}} {{ENV_321}} {{CMD_322}} Notice a big right triangle and a smaller right triangle inside the big one. These two right triangles are similar, because their corresponding angles are equal. Since the ratios of corresponding sides in similar triangles are equal, we get {{MATH_323}} At this point we could {{CMD_324}}differentiate{{CMD_325}} both sides of each equation, but we can take a simpler approach by combining these two equations and expressing {{MATH_326}} in terms of {{MATH_327}}.