Question

attempt 1: 10 attempts remaining. a population model is given by (p(t)=\frac{ln(t^{2}+2)}{t^{2}+3t + 7}), (tgeq0). what is the rate of change of the population at (t = 4)? (round your answer to 2 decimal places.) (p(4)=)

Explanation:

Step1: Apply quotient - rule

The quotient - rule states that if $y=\frac{u}{v}$, then $y'=\frac{u'v - uv'}{v^{2}}$. Here, $u = \ln(t^{2}+2)$ and $v=t^{2}+3t + 7$. First, find $u'$ and $v'$.
$u'=\frac{2t}{t^{2}+2}$ (by the chain - rule, since the derivative of $\ln(f(t))$ is $\frac{f'(t)}{f(t)}$ with $f(t)=t^{2}+2$ and $f'(t) = 2t$), and $v'=2t + 3$.

Step2: Calculate $P'(t)$

$P'(t)=\frac{\frac{2t}{t^{2}+2}(t^{2}+3t + 7)-\ln(t^{2}+2)(2t + 3)}{(t^{2}+3t + 7)^{2}}$.

Step3: Substitute $t = 4$

First, calculate the values in the numerator and denominator when $t = 4$.
For the numerator:

• $\frac{2\times4}{4^{2}+2}(4^{2}+3\times4 + 7)-\ln(4^{2}+2)(2\times4 + 3)$
• $\frac{8}{18}(16 + 12+7)-\ln(18)(8 + 3)$
• $\frac{4}{9}(35)-11\ln(18)$
• $\frac{140}{9}-11\ln(18)\approx15.56 - 11\times2.8904\approx15.56-31.7944=-16.2344$.
• The denominator is $(4^{2}+3\times4 + 7)^{2}=(16 + 12 + 7)^{2}=35^{2}=1225$.
• Then $P'(4)=\frac{-16.2344}{1225}\approx - 0.01$.

Answer:

$-0.01$