# Video: AP Calculus AB Exam 1 • Section I • Part B • Question 90

Michelle made a \$16000 investment. The Investment is supposed to grow at the rate of 460𝑒^(0.2𝑡) dollars per year, where 𝑡 is measured in years. What is then the approximated amount Michelle will have in the bank after 5 years?

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### Video Transcript

Michelle made a 16000-dollar investment. The Investment is supposed to grow at the rate of 460𝑒 to the 0.2𝑡 dollars per year, where 𝑡 is measured in years. What is then the approximated amount Michelle will have in the bank after five years?

Let’s start by labelling the amount Michelle has in the bank as 𝐴. In fact, we could call this 𝐴 of 𝑡, such that 𝐴 is a function of 𝑡. Therefore, 𝐴 of 𝑡 is the amount in the bank after 𝑡 years. Now, the question tells us that Michelle makes an initial investment of 16000 dollars. This means that at 𝑡 is equal to zero, 𝐴 is equal to 16000 dollars. Alternatively, we could write 𝐴 of nought is equal to 16000 dollars. The next piece of information given in the question, which we can use, is that the investment is supposed to grow at a rate of 460𝑒 to the 0.2𝑡 dollars per year. This is the rate at which the amount grows. So we can say that d𝐴 by d𝑡 is equal to 460𝑒 to the 0.2𝑡.

Now the question is asking us to find how much Michelle will have in the bank after five years. And so, therefore, what we’re trying to find is 𝐴 of five. In order to find 𝐴 of five, we first need to find 𝐴 of 𝑡. We could do this by integrating d𝐴 by d𝑡 with respect to 𝑡. When integrating 460𝑒 to the 0.2𝑡, we first notice that 460 is a constant. And so we can factor it out of the integral. Now, what we need to integrate is 𝑒 to the 0.2𝑡 with respect to 𝑡. Here, we have a function within a function. In order to integrate this, we can use the reverse chain rule.

We find the differential of the function within the function, which is 0.2𝑡. The differential of 0.2𝑡 with respect to 𝑡 is simply 0.2. Therefore, in order to integrate 𝑒 to the 0.2𝑡, we start by integrating the exponential part as usual, which simply gives us each 𝑒 to the 0.2𝑡. And then, we mustn’t forget to divide by the differential of 0.2𝑡, which is 0.2. This is equivalent to multiplying by one over 0.2.

Now that we have completed our indefinite integration, we mustn’t forget to add a constant of integration, which we can call 𝑐. Simplifying our function, we find that 𝐴 of 𝑡 is equal to 2300𝑒 to the 0.2𝑡 plus 𝑐. Now, we can use our initial condition which is that 𝐴 of naught is equal to 16000. We simply substitute 𝐴 is equal to 16000 and 𝑡 is equal to zero into this equation. Now, we can simplify this and we obtain that 𝑐 is equal to 13700.

For our next step, we’ll simply substitute the value of 𝑐 back into 𝐴 of 𝑡. Now, we have found an equation for 𝐴 of 𝑡 which is 𝐴 of 𝑡 is equal to 2300𝑒 to the 0.2𝑡 plus 13700. Since we’re trying to find 𝐴 of five, our final step here will be to substitute 𝑡 is equal to five into 𝐴. This gives us a value of 19952 dollars and five cents. Rounding this to the nearest dollar gives us the amount in Michelle’s account after five years will be 19952 dollars.