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A population rate changes at a rate modeled by the differential equation $ \frac{dy}{dt} = .2y(1000-y), $ where t is measured in years. What are all y values for which y is increasing at a decreasing rate?

An answer online claims that the second derivative of this would be:

$ 200 - .4y $, however, isn't this answer leaving out dy/dt? In other words, should not the second derivative be:

$200 - .4y\frac{dy}{dt} $?

On this site, the solution does not have the last dy/dt: http://www.rtmsd.org/cms/lib9/PA01000204/Centricity/Domain/211/2013%20BC%20Multiple%20Choice%20SOLUTIONS.pdf

Isn't that incorrect?

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    Here is the question you are asked to solve: you know that $\frac{dy}{dt}=g(y)$, determine the set of values $z$ such that the function $g$ is decreasing at $z$. – 2017-02-06
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    And the answer is indeed that the solutions are every $y$ such that $g(y)>0$ and $g'(y)<0$, that is, $y(1000-y)>0$ and $1000-2y<0$, that is, $500 – 2017-02-06

2 Answers 2

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We have $\frac{dy}{dt} = .2y(1000-y)$

$\frac{dy}{dt} = 200y - .2y^2$

1.) If finding derivative with respect to t, then it is -

$\frac{d^2y}{dt^2} = 200 \frac{dy}{dt} - 0.4y\frac{dy}{dt}$

2.) If finding derivative with respect to y, then it is -

$= 200 - 0.4y$

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Yes it seems that the person who posted that answer took the derivative with respect to y and did not use the chain rule to get the derivative with respect to t.

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    Yeah, *and they were right to do so*... – 2017-02-06