A popular search engine is targeting European countries where the number of online households is expected to grow at a steady rate. Data was taken over time, and it was found that the number of online houses (in millions) projected can be modeled by the following function:
N(t) = 34.68 + 23.88 ln(1.05t + 5.3), f or 0 _< t _< 2.
The values of t are in years and when t = 0 , the year is 2004. Address the following questions:
a. What was the projected number of online households at the beginning of 2005?
b. How fast was the projected number of online households increasing at the beginning of 2005
a. Well, plug in t=1 and evaluate N(1)
b. dN/dt = 23.88*1.05/(1.05t+5.3)
plug in t=1 and evaluate N'(1)
Hello Steve. Thank you for your help. Can you please explain the steps and in details, so i will know where my problem was. Thanks
To answer these questions, we need to evaluate the given function using the given values of t.
a. To find the projected number of online households at the beginning of 2005, we need to let t = 1 (since t=0 corresponds to the year 2004).
N(1) = 34.68 + 23.88 ln(1.05(1) + 5.3)
N(1) = 34.68 + 23.88 ln(1.05 + 5.3)
N(1) = 34.68 + 23.88 ln(6.35)
N(1) ≈ 34.68 + 23.88 ln(6.35)
N(1) ≈ 34.68 + 23.88 (1.848)
N(1) ≈ 34.68 + 44.170
N(1) ≈ 78.850
Therefore, the projected number of online households at the beginning of 2005 is approximately 78.850 million.
b. To find how fast the projected number of online households was increasing at the beginning of 2005, we need to find the derivative of the function N(t) with respect to t and evaluate it at t = 1.
N(t) = 34.68 + 23.88 ln(1.05t + 5.3)
Taking the derivative of N(t) with respect to t:
dN/dt = 23.88 / (1.05t + 5.3)
Now we can evaluate this derivative at t = 1:
dN/dt at t = 1 = 23.88 / (1.05(1) + 5.3)
dN/dt at t = 1 = 23.88 / (1.05 + 5.3)
dN/dt at t = 1 ≈ 23.88 / 6.35
dN/dt at t = 1 ≈ 3.758
Therefore, the projected number of online households was increasing at a rate of approximately 3.758 million households per year at the beginning of 2005.