( 6 points) The number of bacteria in a culture is growing at a rate of 20e^t/4 per unit of time t. At time t=0, the number of bacteria present was 23000 . Find the number present at t=4

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Step:1 ：The problem is a differential equation problem. The rate of growth of the bacteria is given by (20e^{t/4}). We also know the initial condition, that at t=0, the number of bacteria is 23000. We need to find the number of bacteria at t=4.Step:2 ：To solve this, we can integrate the rate of growth function from 0 to 4, and add the initial number of bacteria.Step:3 ：[int_{0}^{4} 20e^{t/4} dt + 23000]Step:4 ：Calculate the integral to get the number of bacteria at t=4.Step:5 ：[boxed{23137}]

Answer

Step: 1 ：The problem is a differential equation problem. The rate of growth of the bacteria is given by (20e^{t/4}). We also know the initial condition, that at t=0, the number of bacteria is 23000. We need to find the number of bacteria at t=4.Step: 2 ：To solve this, we can integrate the rate of growth function from 0 to 4, and add the initial number of bacteria.Step: 3 ：[int_{0}^{4} 20e^{t/4} dt + 23000]Step: 4 ：Calculate the integral to get the number of bacteria at t=4.Step: 5 ：[boxed{23137}]

( 6 points) The number of bacteria in a culture is growing at a rate of $20e^t/4 per unit of time t. At time t=0, the number of bacteria present was 23000 . Find the number present at t=4

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