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When the Thompson's have their first child, Will, they decide to begin a college savings account for him by depositing $350.00 into an interest-bearing savings account. Will's account is continuously compounded at a fixed rate of 3.4%. Three years later, the Thompson's have a second child, Sarah, and they also decide to begin a college savings account for her by depositing $350.00 into an interest-bearing savings account. Sarah's account is continuously compounded at a higher fixed rate of 5.1%. After the initial deposit into each account, the Thompson's do not withdraw or deposit any money into either account. How many years will it take, since the initial deposit into Sarah's account, for the balance of Sarah's account to be greater than the balance of Will's account?

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znk

Answer:

6 yr

Step-by-step explanation:

The formula for interest compounded continuously is

A = Pe^rt

Let w = the accrued amount in Will's account

and s = the accrued amount in Sarah's account

and t = the years since the initial deposit in Will"s account

We can set the two functions equal and solve for t.

350e^(0.034t) = 350e^{0.051(t - 3)]

     e^(0.034t) = e^{0.051(t - 3)]      Divided each side by 350

           0.034t = 0.051(t - 3)           Took the ln of each side

           0.034t = 0.051t - 0.153      Distributed the 0.051

                    0 = 0.017t - 0.153      Subtracted 0.034t from each side

            0.017t = 0.153                   Added 0.153 to each side

                     t = 0.153/0.017         Divided each side by 0.017

                     t = 9 yr

Sarah's account  started three years after Will's, so the balance in her account will exceed that in Will's account in 6 yr after the initial deposit in her account.

The diagram shows her account (blue line) catching up to Will's in six years.

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