Question

Populations of aphids and ladybugs are modeled by the following equations. $$ \frac{dA}{dt} = 2A - 0.01AL $$ $$ \frac{dL}{dt} = -0.5L + 0.0001AL $$ (a) Find the equilibrium solutions. smaller A-value $$(A, L) = (\square)$$ larger A-value $$(A, L) = (\square)$$ (b) Find an expression for $$ \frac{dL}{dA} $$. $$ \frac{dL}{dA} = \square $$

Name: populations of aphids and ladybugs are modeled by the following equations fracdadt 2a 001al fracdldt 05l 00001al a find the equilibrium solutions smaller a value a l square larger a value a 93275
Uploaded: 2022-12-28T15:25:31-08:00
Channel: Lainey Roebuck
Description: populations of aphids and ladybugs are modeled by the following equations fracdadt 2a 001al fracdldt 05l 00001al a find the equilibrium solutions smaller a value a l square larger a value a 93275

          Populations of aphids and ladybugs are modeled by the following equations.
$$ \frac{dA}{dt} = 2A - 0.01AL $$
$$ \frac{dL}{dt} = -0.5L + 0.0001AL $$
(a) Find the equilibrium solutions.
smaller A-value $$(A, L) = (\square)$$
larger A-value $$(A, L) = (\square)$$
(b) Find an expression for $$ \frac{dL}{dA} $$.
$$ \frac{dL}{dA} = \square $$

$populations of aphids and ladybugs are modeled by the following equations fracdadt 2a 001al fracdldt 05l 00001al a find the equilibrium solutions smaller a value a l square larger a value a 93275$

Added by Manuela R.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Lainey Roebuck

12/28/2022

Step 1

From the first equation: $2A - 0.01AL = 0$ $A(2 - 0.01L) = 0$ This gives two possibilities: $A = 0$ or $2 - 0.01L = 0$. From the second equation: $-0.5L + 0.0001AL = 0$ $L(-0.5 + 0.0001A) = 0$ This gives two possibilities: $L = 0$ or $-0.5 + 0.0001A = 0$. Show more…

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Populations of aphids and ladybugs are modeled by the following equations. $$ \frac{dA}{dt} = 2A - 0.01AL $$ $$ \frac{dL}{dt} = -0.5L + 0.0001AL $$ (a) Find the equilibrium solutions. smaller A-value $$(A, L) = (\square)$$ larger A-value $$(A, L) = (\square)$$ (b) Find an expression for $$ \frac{dL}{dA} $$. $$ \frac{dL}{dA} = \square $$