**Unformatted text preview: **volume are all increasing as water is poured into the container. This means that we actually have three things varying with time: the water level h (the height of the cone of water), the radius r of the circular top surface of water (the base radius of the cone of water), and the volume of water V . The volume of a cone is given by V = πr 2 h/ 3. We know dV/dt , and we want dh/dt . At ﬁrst something seems to be wrong: we have a third variable r whose rate we don’t know. But the dimensions of the cone of water must have the same proportions as those of the container. That is, because of similar triangles, r/h = 10 / 30 so r = h/ 3. Now we can eliminate r from the problem entirely: V = π ( h/ 3) 2 h/ 3 = πh 3 / 27. We take the derivative of both sides and plug in h = 4 and dV/dt = 10, obtaining 10 = (3 π · 4 2 / 27)( dh/dt ). Thus, dh/dt = 90 / (16 π ) cm/sec....

View
Full Document

- Fall '07
- JonathanRogawski
- Math, Calculus, Derivative, ... ...