Explanation Given information: The given function is P = L 1 + C e − k t , where P is the total number of people, infected with a virus with time t in weeks. Initially 10 people have the virus and the number of people infected is increasing exponentially, with a continuous growth rate of 1.78. In the long run, about 5000 people will become infected. Calculation: The curve is the steepest when y is maximum. We have: y = − 50 ( 1 + 6 e − 2 t ) 2 ( − 12 e − 2 t ) When the second derivative is zero, we get the critical points. y ″ = − 2 e − 2 t ( 1 + 6 e − 2 t ) 2 − e − 2 t .2

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Chapter 4.7, Problem 15P

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Find the point where the curve y=501+6e−2t for t≥0 is steepest.

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Chapter 4.7, Problem 14P

Chapter 4.7, Problem 16P

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Find the point where the curve y = 50 1 + 6 e − 2 t for t ≥ 0 is steepest.

Solution Summary: The author explains that the curve y=501+6e-2t for ge 0 is steepest, and the function is the sharpest at t=1.242

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Find the point where the curve y=501+6e−2t for t≥0 is steepest.

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Chapter 4 Solutions