4. A tank in the shape of a cone (with vertexpointing down) of height 5 meters and radius 4meters drains through a small hole at its vertexinto a cylindrical tank of height 5 meters andradius 4 meters that lies below it. The water isdraining such that the height of the water withinthe conical tank is falling by meter per minute.A. If V is the volume of the water within theconical tank and H is the height of the waterwithin the conical tank, find an expression forin terms of H.dVdt.B. If v is the volume of the water within thecylindrical tank and ŉ is the height of thewater within the cylindrical tank, find anexpression for de in terms of hand dhdvdtdtC. At what rate does the height of the waterlevel in the cylindrical tank rise when theheight of the water level in the conical tank isexactly 3 m?

Height of cone = 5 meters The radius of the cone = 4 meters Since the water is dripping from the vertex, therefore, the water itself will be in the shape of a cone. And as the water is falling down the radius of the cone at the different levels of water also changes and decreases. That implies that radius is also a function of time. Let the volume of the cone at any time, t = V and the height of the water at time t = H Radius at time t = r. Since the water is draining from the cone, therefore the value of H will keep on decreasing and its rate is 1/2. Therefore, -dHdt = 12 dHdt = -12 m/min The volume of a cone is defined as, 13π(r)^2 h Therefore, at time t, V = 13πr2H Even though the cone-shaped water is changing its shape the angle at the vertex will remain the same. Therefore, half of the angle at the vertex initially = α = half of the angle at the vertex at time t Tan (α) = 45 = rH r = 4H5 Differentiating on both sides, dVdt = 13π r^2 (dHdt) Putting the value of r in terms of H and the value ofdHdt, dVdt = 13π (16 H^225 )(-12) = -875π(H)^2 Here, the negative sign shows that water is decreasing. radius of the cylindrical tank = 4m, height of the tank = 5m the volume of water in the cylindrical tank at time t = v The height of the water in the cylindrical tank at time t = h Water will remain in the shape of the cylinder with a radius equal to the radius of the tank. Therefore, the radius of the cylindrical water = 4m The volume of the cylinder = π radius2 * height v = π 42 * h Differentiating on both sides, with respect to t, dvdt =16 πdhdt m3/min Writing in terms of h and dhdt, dvdt =16 πdhdt(h)0 m3/min Since the water is draining from the conical tank and then filling the cylindrical tank. Therefore, Rate of increase in water in the cylindrical tank = - the rate of decrease in water in the conical tank. dvdt = -dVdt 16πdhdt = 875π(H^2)dhdt = 1150(H^2)Given, H = 3mdhdt = 9150 = 350m/min