Next, we {{CMD_31}}draw a picture{{CMD_32}}. Note that our pizza dough is shaped as a cylinder. {{CMD_33}} {{ENV_34}} {{CMD_35}} Next we need to {{CMD_36}}find equations{{CMD_37}}. Recall the formula for the volume of the cylinder, {{MATH_38}} and note that {{MATH_39}}, and write {{MATH_40}} which immediately simplifies to {{MATH_41}} Since both {{MATH_42}} and {{MATH_43}} are functions of time, we now {{CMD_44}}differentiate{{CMD_45}} both sides of the equation using implicit differentiation, {{MATH_46}} Now we'll {{CMD_47}}evaluate{{CMD_48}} all the quantities at the instant when {{MATH_49}}. But, first, we have to compute {{MATH_50}} at that moment. Since {{MATH_51}} it follows that {{MATH_52}}, when {{MATH_53}}. Now, we are ready to {{CMD_54}}solve{{CMD_55}}. {{MATH_56}} {{MATH_57}} Therefore, when {{MATH_58}}, {{MATH_59}} Hence, the thickness of the dough is changing at a rate of {{MATH_60}} cm/min when {{MATH_61}} cm. {{CMD_62}} {{CMD_63}} In our next example we consider a melting snowball. {{CMD_64}} {{ENV_65}} {{CMD_66}} {{CMD_67}} Consider a melting snowball.