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Interest If $\$ 1000$ is invested in an account that pays 5$\%$ compounded continuously, the total amount, $A(t),$ in the account after $t$ years is$$A(t)=1000 e^{0.05}$$a. Find the average rate of change per year of the total amount in the account for the first five years of the investment (from $t=0$ to $t=5$ ).b. Find the average rate of change per year of the total amount in the account for the second five years of the investment (from $t=5$ to $t=10 )$ .c. Estimate the instantaneous rate of change for $t=5$ .

a) 56.8b) 72.94c) 64.88

Calculus 1 / AB

Chapter 11

The Derivative

Section 3

Rates of Change

Derivatives

Differentiation

Applications of the Derivative

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We're told that if we invested $1000 into account, that pays 5% interest that compounds continuously with T being how many years we leave it in. Then we can see how much money we have in the account. By the equation eight TZ until 1000 times ive rates than 0.5 And we want to find the average rate of change off, how much money we have in the account in the 1st 5 years and the 2nd 5 years. And then they want to estimate the instantaneous rate of change for when t is equal to five. So let's go ahead and plug everything into this equation we have up here. So we're going to look at So instead of f, we're going to use a so it's gonna be a five minus a of zero all over five minus zero. Now, if we were to find a five. So first, let's just pluck five and still be 1000 times e raised to the 0.5 times five. So that should be 0.25 and then minus 1000 raised to zero, which would just be one. And then in the denominator, we would have five. So first knows we could divide this five into what we have in the numerous s just gonna be too 100 for each of those. So we're gonna have e raised the 0.25 times 200 then minus 200. So that's going to be something around 50 six point 81 dollars Now for B, it's going to be the same set up. We're just going to use five intent. This times will be a of 10 minus a five all over 10 minus five. So plugging in 10 we would have 1000 e to the 0.5 minus 1000 e toothy 0.25 and then 10 minus five is just gonna be five. So again, we can go ahead. Divide those fives and each so we'd have 200. Reach those. And once we simplify this town, let's see what we could get. So, e rates. Is there a point? Uh, minus. He raise to the 0.25 and we multiply that by 200 then divide not just multiplied by 200. So that's going to be something around 72 point $94 so century will be just found was the average rate of change for each year. So I should probably do per year. So in the 1st 5 years we expected to increase by about $56.81. And in the 2nd 5 years, we expect each year for it to increase by about $72.94. Now for this next part, it says estimate the instantaneous tray of change at five. So they want to find exactly what would be our rate of change for how much money were earning after five years. Well, we could just plug everything into the instantaneous rate change equation and then just keep taking smaller, smaller values of H. But notice how in these last two questions, we found the average rate of change and it was centered around five. So one way we can estimate this is by just adding those two values together and then dividing by two. Because if we found the slope of the rate of change before that and estimated it and then we do it the same thing on the right, we would expect it to be something in the middle of those two values, so begin just right. This has to the average rate of change from 0 to 5, plus the average rate of change from 5 to 10 all over to so spoke everything good. It would be 56.81 plus 72.94 all over, too. And once we add these up, that should give. So that's 129.75 and divide that by two. So after we've around, there should be something around 64 point 88 dollars per year. So after five years, we would expect us to gain about $64.88 for the remainder of that year.

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