Question

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?

Answer 1

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$\pi r^{2}$

ds/dt=ds/dr*dr/dt.................................1

ds/dr=differentiation of the area of a sphere=$8\pi r$

ds/dt=$8\pi 555$*5

ds/dt=69743.35m^2/min

The surface area of the sphere will change by 69733.35 square meters for every minute.