Question

Can someone please help with these two questions please?

1. Assume that the radius r of a sphere is expanding at a rate of 60 cm/min. The volume of a sphere is V = 4/3r^3 and its surface area is 4r^2. Determine the rate of change in surface area at t = 2 min, assuming that r = 16 at t = 0.

2. A conical tank has height 3 m and radius 2 m at the top. Water level is rising at a rate of 2.1 m/min when it is 1.8 m from the bottom of the tank. At what rate is water flowing in? (Round your answer to three decimal places.)

Answer 1

Answer:

Step-by-step explanation:

S=4πr²

ds/dt=4π×2r×dr/dt=8πr dr/dt

dr/dt=60 cm/min

ds/dt=8π×60r=480πr

r=16,t=0

at t=2,r=16+32=48 cm

so ds/dt=480π×48=23040 πcm²≈72382.79 cm²

2.

$V=\frac{1}{3} \pi r^2*h\\\frac{r}{h} =\frac{2}{3} \\r=\frac{2}{3} h\\V=\frac{1}{3} \pi( \frac{2}{3} *h)^2h\\V=\frac{4}{27} \pi h^3\\\frac{dV}{dt} =\frac{4}{27}\pi *3h^2\frac{dh}{dt} \\\\when ~h=1.8 ~m,\frac{dh}{dt} =2.1~m/min\\\frac{dV}{dt} =\frac{4}{9}\pi h^2\frac{dh}{dt} =\frac{4}{9} \pi*(1.8)^2*2.1=27.216 \pi \approx 85.502 m^3$