Science:Math Exam Resources/Courses/MATH110/April 2019/Question 01 (h)

MATH110 April 2019

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Question 01 (h)
Suppose the number of fish in a pond $t$ months from now is $40e^{0.2t}$ . At what rate (in fish per month) is the population of fish growing at $t=2$ months?

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
How can you translate "rate of the population" into mathematical symbols? In class, you may have discussed how the derivative of a function gives the rate of change at a given time. See the next hint for more detail.

Hint 2
We are given the function $P(t)=40e^{0.2t}$ , which describes the number of fish in a pond at time $t$ . The rate of change of the number of fish in a pond at time $t$ is given by $P^{'}(t)$ . Calculate the derivative of $f$ !

Hint 3
Use the chain rule.

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Solution
Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Let the population function be $P(t)=40e^{0.2t}$ . The derivative is $P'(t)=0.2\cdot 40e^{0.2t}$ fish/month $=8e^{0.2t}$ fish/month. Therefore the rate that the population is growing in 2 months is $P'(2)=8e^{0.4}$ fish/month.