Question

a field is full of crickets. some lizards notice the field of crickets and enter the field to start eating the crickets. if c represents the number of crickets in the field, and l represents the number of lizards in the field, then the number of crickets in the field in terms of the number of lizards in the field is given by c(l)=5700 - 30log₂(l). if the number of lizards in the field is increasing at a rate of 3 lizards per minute, how is the population of crickets changing per minute when there are 15 lizards in the field? at a rate of crickets per minute

Explanation:

Step1: Differentiate the function $C(l)$ with respect to $l$.

The derivative of a constant is 0, and the derivative of $- 30\log_2(l)$ using the formula $\frac{d}{dx}\log_a(x)=\frac{1}{x\ln(a)}$ is $-\frac{30}{l\ln(2)}$. So, $C^\prime(l)=-\frac{30}{l\ln(2)}$.

Step2: Find the rate of change of crickets when $l = 15$.

Substitute $l = 15$ into $C^\prime(l)$. We get $C^\prime(15)=-\frac{30}{15\ln(2)}=-\frac{2}{\ln(2)}\approx - 2.89$. The negative sign indicates that the number of crickets is decreasing. The rate of change of the population of crickets per minute is $\frac{2}{\ln(2)}\approx2.89$ crickets per minute.

Answer:

$\frac{2}{\ln(2)}\approx2.89$ crickets per minute