Question

Suppose that the GDP of a country (in billion USD) in year t. is given by the function GDP(t)=690+75t The population of the country (in millions of people) in year t is given by the funciton P(t)=0.03t² +0.7t+65 GDP per capita is defined as GDP pc​ (t)= P(t)/GDP(t)

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(a) Calculate the current GDP and population when t= GDP(2)= P(2)=
(b) Give an expression for the rate of change of GDP At arbitrary time t When t=2 GDP ′(t)= GDP ′ (2)= Units are ।
(c) Give an expression for the rate of change of populati At arbitrary time t When t=2 P ′(t)= P ′ (2)=
(d) Give an expression for the rate of change of GDP per At arbitrary time t GDPpc′​ (t)= When t=2 GDPpe′​ (2)= Units are

Answer 1

The GDP at t=2 is 840 billion USD, the population at t=2 is 66.26 million people, the rate of change of GDP at any time t is a constant 75 billion USD per year, and the rate of change of population at t=2 is 0.82 million people per year. The rate of change of GDP per capita requires applying the quotient rule to the GDP and population functions.

Step-by-step explanation:

When calculating the GDP of a country at time t=2 using the given function GDP(t)=690+75t, we substitute 2 into the function to get GDP(2) = 690 + 75(2) = 840 billion USD.

For the population P(t) at time t=2 using the function P(t)=0.03t² +0.7t+65, we substitute 2 into the function to obtain P(2) = 0.03(2)² + 0.7(2) + 65 = 66.26 million people.

The rate of change of GDP at an arbitrary time t is the derivative of GDP(t), which is GDP′(t) = 75, and at t=2, this rate remains constant, so GDP′(2) = 75 billion USD per year.

The rate of change of population at an arbitrary time t is given by the derivative of P(t), that is P′(t) = 0.06t + 0.7, and at t=2, the rate of change is P′(2) = 0.06(2) + 0.7 = 0.82 million people per year.

For the rate of change of GDP per capita, we would need to apply the quotient rule of differentiation to the function GDP per capita = GDP(t) / P(t), and evaluate it at t=2 to find GDPpc′(2).