The total spent on Research and Development (R&D) by the federal government in the USA during 2002-2012 can be approximated by:
S t( ) = 3.1ln[1210(t + 10) $billion; where t is the year since 2010.
a) Determine, to the nearest billion, the total spent in the year 2005 (t = −5 )
b) Determine at what rate the total spent on R&D was varying in the year 2005
what's the trouble? Just plug and chug.
(a) 3.1ln[1210(-5 + 10)] = 26.99
(b) dS/dt = 3.1/(t+10)
so at t = -5, that is just 0.62
To determine the total spent on research and development (R&D) by the federal government in the United States during the year 2005 (t = -5), we can substitute the given value of t into the equation and solve for S(t).
a) Substitute t = -5 into the equation S(t) = 3.1ln[1210(t + 10)]:
S(-5) = 3.1ln[1210(-5 + 10)]
Simplify the inside of the logarithm:
S(-5) = 3.1ln[1210(5)]
Evaluate the expression inside the logarithm:
S(-5) = 3.1ln[6050]
Calculate the natural logarithm of 6050:
S(-5) ≈ 3.1 * 8.705683 → S(-5) ≈ 26.980 billion
Therefore, to the nearest billion, the total spent on R&D in the year 2005 was approximately 27 billion dollars.
b) To determine the rate at which the total spent on R&D was varying in the year 2005, we need to find the derivative of the equation S(t) with respect to t, and then substitute t = -5.
The derivative of S(t) = 3.1ln[1210(t + 10)] with respect to t can be found using the chain rule. The derivative of ln(u) is 1/u multiplied by the derivative of u. In this case, u = 1210(t + 10):
S'(t) = 3.1 * (1/u) * du/dt
Now, find du/dt:
du/dt = 1210 * (d/dt)(t + 10)
Since d/dt of (t + 10) is 1, simplify further:
du/dt = 1210
Substitute back into the original equation:
S'(t) = 3.1 * (1/u) * 1210
Now, substitute t = -5:
S'(-5) = 3.1 * (1/1210) * 1210
Cancel out 1210:
S'(-5) = 3.1
Therefore, the rate at which the total spent on R&D was varying in the year 2005 was approximately 3.1 billion dollars per year.
a) To determine the total spent in the year 2005 (t = -5), we substitute t = -5 into the equation S(t):
S(t) = 3.1ln[1210(t + 10)]
S(-5) = 3.1ln[1210(-5 + 10)]
S(-5) = 3.1ln[1210(5)]
Now, we can calculate the approximate value of S(-5):
S(-5) ≈ 3.1ln(6050)
Using a calculator, we find that ln(6050) ≈ 8.7062.
S(-5) ≈ 3.1(8.7062)
S(-5) ≈ 26.99 billion
Therefore, the total spent in the year 2005 is approximately $27 billion.
b) The rate at which the total spent on R&D is varying can be determined by finding the derivative of the function S(t):
dS(t)/dt = d/dt[3.1ln[1210(t + 10)]
To find the rate in the year 2005, we substitute t = -5 into the derivative:
dS(-5)/dt = d/dt[3.1ln[1210(-5 + 10)]]
dS(-5)/dt = d/dt[3.1ln[1210(5)]]
Now, we can calculate the approximate value of the derivative:
dS(-5)/dt ≈ d/dt[3.1ln(6050)]
Using the chain rule for differentiation, we have:
dS(-5)/dt ≈ 3.1(1/6050)(d/dt[6050])
Since d/dt[6050] = 0, the derivative term becomes zero:
dS(-5)/dt ≈ 3.1(1/6050)(0)
dS(-5)/dt ≈ 0
Therefore, the rate at which the total spent on R&D was varying in the year 2005 is approximately 0 billion.