Question

TISSUE GROWTH Suppose a particular tissue culture has area AA(tt) at time tt and a

potential maximum area MM. Based on properties of cell division, it is reasonable to assume that

the area AA grows at a rate jointly proportional to �AA(tt) and MM − AA(tt); that is

dddd

dddd = kk�AA(tt) [MM − AA(tt)]

where kk is a positive constant.

a. Let RR(tt) = AA′

(tt) be the rate of tissue growth. Show that RR′

(tt) = 0 when AA(tt) = MM/3.

b. Is the rate of tissue growth greatest or least when AA(tt) = MM/3? [Hint: Use the first

derivative test or second derivative test.]

c. Based on the given information and what you discovered in part (a), what can you say

about the graph of AA(tt)?

Answer 1

$$\dfrac{dA}{dt}=k\sqrt{A(t)}(M-A(t))$$

a.

$$R(t)=A'(t)$$

$$R'(t)=A''(t)=\bigg(k\big(M\sqrt{A(t)}-A(t)\sqrt{A(t)}\big)\bigg)'$$

$$=k(\dfrac{M}{2\sqrt{A}}-\dfrac{3\sqrt{A}}{2})A'$$

$$=\dfrac{k}{2\sqrt{A}}(M-3A)\big(k\sqrt{A}(M-A)\big)$$

$$=\dfrac{k^2}{2}(M-3A)(M-A)$$

$$R'(t)=0=>\dfrac{k^2}{2}(M-3A)(M-A)=0$$

$$A=\dfrac{M}{3}\ or\ A=M$$

b.

If $A<\dfrac{M}{3}, R'>0, R$ increases.

If $\dfrac{M}{3}<A<M, R'<0, R$ decreases.

The rate of tissue growth $R(t)=A'(t)$ is greatest when $A(t)=\dfrac{M}{3}.$

c.

$A''(t)=0,$ when $A=\dfrac{M}{3}.$

If $A<\dfrac{M}{3}, A''>0, A$ is concave up.

If $\dfrac{M}{3}<A<M, A''<0, A$ is concave down.

The graph of $A(t)$ has an inflection point when $A(t)=\dfrac{M}{3}.$