dP dt = − × dt . Substituting x dt , we get dP dt = − × = − ≈− rupee/week. That is the price is changing, in fact decreasing at the rate of ` . per week.
Example . Salt is poured from a conveyer belt at a rate of cubic metre per minute forming a conical pile with a circular base whose height and diameter of base are always equal. How fast is the height of the pile increasing when the pile is metre high? Let h and r be the height and the base radius.
Therefore h = . Let V be the volume of the salt cone. V = r h h dV dt ; min m / Hence, dV dt = p h dh dt Therefore, dh dt = dV dt h ⋅π That is, dh dt = × × = p m / min. Example .
(Two variable related rate problem) A road running north to south crosses a road going east to west at the point P . Car A is driving north along the first road, and car B is driving east along the second road. At a particular time car A is kilometres to the north of P and traveling at km/hr, while car B is kilometres to the east of P and traveling at km/hr. How fast is the distance between the two cars changing?
Fig. . h - - Let a t ( ) be the distance of car A north of P at time t , and b t ( ) the distance of car B east of P at time t , and let c t ( ) be the distance from car A to car B at time t . By the Pythagorean Theorem, c t a t b t Taking derivatives, we get c t c t a t a t