\( 1 \leftarrow \) The size P of a certain insect population at time t (in days) obeys the function \( \mathrm{P}(\mathrm{t})=600 e^{0.04 t} \) (a) Determine the number of insects at \( \mathrm{t}=0 \) days. (b) What is the growth rate of the insect population? (c) Graph the function using a graphing utility. (d) What is the population after 10 days? (e) When will the insect population reach 720 ? (f) When will the insect population double?

To tackle the questions regarding the given insect population function \( P(t) = 600 e^{0.04 t} \), let's break it down: (a) To determine the number of insects at \( t = 0 \), we substitute \( t \) with 0 in the function: \[ P(0) = 600 e^{0.04 \times 0} = 600 e^{0} = 600 \text{ insects}. \] (b) The growth rate of the insect population is represented by the coefficient of \( t \) in the exponent of the exponential function, which is \( 0.04 \). This means the population grows at a rate of \( 4\% \) per day. (c) To graph the function, you can use any graphing utility (like Desmos or a graphing calculator). Plot \( P(t) = 600 e^{0.04 t} \) and observe how it rises exponentially. (d) To find the population after 10 days, substitute \( t = 10 \) into the function: \[ P(10) = 600 e^{0.04 \times 10} = 600 e^{0.4} \approx 600 \times 1.49182 \approx 894.91. \] Thus, the population after 10 days is approximately \( 895 \) insects. (e) To find when the population reaches 720, set \( P(t) = 720 \): \[ 720 = 600 e^{0.04 t} \] Divide both sides by 600: \[ 1.2 = e^{0.04 t}. \] Taking the natural logarithm of both sides: \[ \ln(1.2) = 0.04 t \implies t = \frac{\ln(1.2)}{0.04} \approx 5.007. \] So, the population will reach 720 insects in about \( 5.01 \) days. (f) To determine when the population will double, set \( P(t) = 1200 \): \[ 1200 = 600 e^{0.04 t} \implies 2 = e^{0.04 t}. \] Taking the natural logarithm: \[ \ln(2) = 0.04 t \implies t = \frac{\ln(2)}{0.04} \approx 17.328. \] Hence, the insect population will double in about \( 17.33 \) days.