The price in dollars of a house during a period of mild inflation is described by the formula P(t)=94000 e0.03 t, where t is the number of years after 1990. Answer the following questions:
B. In the year 2000 the value will be increasing at a rate of dollars per year. (Round your answer to the nearest dollar.)
C.How long will it take for a house to double in value? Answer: years. (Round your answer to two decimal places.)
The equation is 94000e^(0.03t)
To find the rate of increase, we need to find the derivative of the function P(t) with respect to t.
Given: P(t) = 94000e^(0.03t)
Step 1: Calculate the derivative of P(t) with respect to t.
dP/dt = d/dt (94000e^(0.03t))
= 94000 * (d/dt (e^(0.03t)))
Step 2: Apply the chain rule to find d/dt (e^(0.03t)).
The derivative of e^(kt) with respect to t is ke^(kt).
Therefore, d/dt (e^(0.03t)) = 0.03e^(0.03t).
Step 3: Substitute the derivative back into the derivative of the original function.
dP/dt = 94000 * (0.03e^(0.03t))
= 2820e^(0.03t)
B. To find the rate of increase in the year 2000 (t = 10 since it's 10 years after 1990), we substitute t = 10 into the derivative formula:
Rate of increase = dP/dt at t = 10
= 2820e^(0.03 * 10)
≈ 7969.84 dollars per year (rounded to the nearest dollar)
Therefore, the value of the house will be increasing at a rate of approximately 7969.84 dollars per year in the year 2000.
C. To find how long it will take for a house to double in value, we need to set P(t) equal to twice its original value and solve for t.
Given: P(t) = 94000e^(0.03t)
Setting P(t) = 2 * P(0) (since we want to find when it doubles):
94000e^(0.03t) = 2 * 94000
e^(0.03t) = 2
Take the natural logarithm (ln) of both sides to solve for t:
ln(e^(0.03t)) = ln(2)
0.03t = ln(2)
t = ln(2) / 0.03
Using a calculator to evaluate the expression, we find:
t ≈ 23.10 years (rounded to two decimal places)
Therefore, it will take approximately 23.10 years for a house to double in value.