Question

A cell is sphere. The rate of increase of weight is proportional to the surface area. The weight x(t) at time t will be proportional to the square of radius.

When t= 0, x=a. Determine the weight x(t) of the cell at time t. Let c be proportional constant

Answer 1

x = k*r^2 (the weight is proportional to the square of radius)
x'(t) = k2*4 pi r^2 (the rate of increase of the weight is proportional to the surface area)
x'(t) = k3*x(t), for some constant k3
x(t) = a e^(k3 t)

Answer 2

thank you, i still confuse at second equation. x is differentiate by t, then right hand side is differentiate by r? can you explain more here

Answer 3

x'(t) = k2*4 pi r(t)^2
radius and weight are both functions of time

Answer 4

k3 = 4 pi k2/k

Answer 5

understood. thank you very much

Answer 6

cool :)