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Let $ A(t) $ be the area of a tissue culture at time $ t $ and let $ M $ be the final area of the tissue when growth is complete. Most cell divisions occur on the periphery is proportional to $ \sqrt {A(t)}. $ So a reasonable model for the growth of tissue is obtained by assuming that the rate of growth of the area is jointly proportional to $ \sqrt {A(t)} $ and $ M - A(t). $(a) Formulate a differential equation and use it to show that the tissue grows fastest when $ A(t) = \frac {1}{3} M. $ (b) Solve the differential equation to find an expression for $ A(t). $ Use a computer algebra system to perform the integration.

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a) $$=\frac{1}{2} k A^{-1 / 2}[k \sqrt{A}(M-A)][M-3 A]=\frac{1}{2} k^{2}(M-A)(M-3 A)$$b) $$C=\frac{\sqrt{M}+\sqrt{A_{0}}}{\sqrt{M}-\sqrt{A_{0}}}$$

Calculus 2 / BC

Chapter 9

Differential Equations

Section 3

Separable Equations

Campbell University

Harvey Mudd College

University of Michigan - Ann Arbor

Lectures

13:37

A differential equation is a mathematical equation for an unknown function of one or several variables that relates the values of the function itself and its derivatives of various orders. An ordinary differential equation (ODE) is a differential equation containing one or more derivatives of a function and their rates of change with respect to the function itself; it can be used to model a wide variety of phenomena. Differential equations can be used to describe many phenomena in physics, including sound, heat, electrostatics, electrodynamics, fluid dynamics, elasticity, quantum mechanics, and general relativity.

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Let $A(t)$ be the area of …

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GU In the model of Exercis…

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Atissue culture grows unti…

All right. So for this problem we need to consider A. And T. B. The area of a tissue culture at time. T. And let m be the final area of the tissue when growth is complete. So most cells on the periphery is proportional of the tissue and the number of cells on the periphery is proportional to the square root of A. Of T. So a reasonable model for the growth of tissue is obtained by assuming that the rate of growth of the area is jointly proportional to the square root of eight of T. And m minus A. Of T. So for a the rate of growth of the area is jointly proportional to the square root of A. F. T. And M minus a. Ft. So that the rate is proportional to the product of these two quantities. So below for for what we need to do is with that formula D. A. Over D. F. T equals square root of a. Um and minus a. K. For that concept we need to differentiate and we would differentiate it using the chain rule. So below I have it using the chain rule. So given that we use the trend rule, this represents a maximum by the first derivative since di of DT the derivative of diva of D. T. Goes from positive one A. Of T equals M over three. Moving to the next question, we have the differential equation. Now from the previous question. So for some constant K we need to find the value of A. That maximizes D. A. Of T D. A. Over D. T. So therefore take the derivative of D. A over D. T. With respect to A. So here we have D A over DT equals we're rid of a and minus a K. So we would take that derivative and then D over D. A. Um in parentheses D. A over D. T. With equal zero within a equals M three. So that is the value of A. That makes the largest growth rate given that we use the differential. And we took um we took the derivative and then now we compare it on the conditional To equal a equals one over square root KC -2.

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