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

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