the value of a savings account after t years is given by: v(t)=131(1.15^t) thousand dollars. calculator…

the value of a savings account after t years is given by: v(t)=131(1.15^t) thousand dollars. calculator checkpoint: v(2)=173.2475 a. calculate the value of the savings account after 8 years. (round your answer to three decimal places.) b. calculate the rate of change of the savings account after 8 years. (round your answer to three decimal places.) thousand dollars per year c. write the linearization of v after 8 years. (round all numerical values to three decimal places.) v_l(t)= thousand dollars d. use the linearization in part (c) to estimate the value of the savings account after 8.1 years. (round your answer to three decimal places.) thousand

the value of a savings account after t years is given by: v(t)=131(1.15^t) thousand dollars. calculator checkpoint: v(2)=173.2475 a. calculate the value of the savings account after 8 years. (round your answer to three decimal places.) b. calculate the rate of change of the savings account after 8 years. (round your answer to three decimal places.) thousand dollars per year c. write the linearization of v after 8 years. (round all numerical values to three decimal places.) v_l(t)= thousand dollars d. use the linearization in part (c) to estimate the value of the savings account after 8.1 years. (round your answer to three decimal places.) thousand

Answer

Explanation:

Step1: Find value at (t = 8)

Substitute (t = 8) into (V(t)=131(1.15^{t})). [V(8)=131\times1.15^{8}] [V(8)=131\times3.05902286] [V(8)\approx 400.732]

Step2: Find derivative of (V(t))

The derivative of (y = a\cdot b^{t}) is (y^\prime=a\cdot b^{t}\ln(b)). For (V(t)=131(1.15^{t})), (V^\prime(t)=131\times1.15^{t}\ln(1.15)). Substitute (t = 8) into (V^\prime(t)): [V^\prime(8)=131\times1.15^{8}\ln(1.15)] [V^\prime(8)=131\times3.05902286\times0.13976194] [V^\prime(8)\approx 55.903]

Step3: Find linear - ization

The linearization of a function (y = V(t)) at (t = a) is given by (V_{L}(t)=V(a)+V^\prime(a)(t - a)). Here (a = 8), (V(8)\approx400.732) and (V^\prime(8)\approx55.903). [V_{L}(t)=400.732+55.903(t - 8)] [V_{L}(t)=400.732+55.903t-447.224] [V_{L}(t)=55.903t - 46.492]

Step4: Estimate value at (t = 8.1)

Substitute (t = 8.1) into (V_{L}(t)): [V_{L}(8.1)=55.903\times8.1-46.492] [V_{L}(8.1)=452.8143 - 46.492] [V_{L}(8.1)\approx406.322]

Answer:

a. (400.732) b. (55.903) c. (V_{L}(t)=55.903t - 46.492) d. (406.322)