Whatts the awnser help 1⁄3 + 1⁄6 + 3⁄4
2 answers:
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Answer:
ds/dt=69743.35m^2/min
Step-by-step explanation:
The radius r(t)r(t)r, (, t, )of a sphere is increasing at a rate of 7.57.57, point, 5 meters per minute. At a certain instant t_0t 0 t, start subscript, 0, end subscript, the radius is 555 meters. What is the rate of change of the surface area S(t)S(t)S, (, t, )of the sphere at that instant?
If i could understand the question correctly, the typos notwithstanding
r=radius, 555m
dr/dt=rate of change in radius 5m/min
ds/dt=rate of change in surface area ?
S=area of a sphere 4
ds/dt=ds/dr*dr/dt.................................1
ds/dr=differentiation of the area of a sphere=
ds/dt=*5
ds/dt=69743.35m^2/min
The surface area of the sphere will change by 69733.35 square meters for every minute.