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.)
for C.
In 1990, the house is worth 94,000 to double it has to be worth 188,000
Putin 188,00 for P(t) then divide both sides by 94,000. That will give you
2= e^.03t
take the ln of both sides then solve for t.
Are you in calculus? If so.. for B. I would take the derivative that will give you rate of change and then put in the t which will be 10 years.
P'(t) = 94000*0.03 e^0.03t = 2820 e^.03t
so, in 2000, the value will be rising at
2820 e^.3 = $3807/yr
to find the doubling interval, solve for t:
2 = e^.03t
t = ln2/.03 = 23.10 years
B. In the year 2000 the value will be increasing at a rate of "I don't know, maybe it'll just start doing jumping jacks and run around the neighborhood. Houses can be pretty unpredictable, you know?"
C. How long will it take for a house to double in value? Answer: "Well, that depends on how motivated the house is. Some houses might be really ambitious and double their value in no time, while others might take their sweet time and double their value when they feel like it. It's a mystery!"
To find the rate at which the value of the house is increasing in dollars per year in the year 2000, we need to calculate the derivative of the price function with respect to time (t). The derivative represents the rate of change.
Step 1: Find the derivative of the price function P(t):
P'(t) = d/dt (94000 e^(0.03t))
To differentiate this function, we can use the chain rule, which states that for a function of the form f(g(x)), the derivative is given by f'(g(x)) * g'(x).
In this case, f(x) = 94000 e^x and g(x) = 0.03t.
Step 2: Apply the chain rule:
P'(t) = (94000 e^(0.03t)) * (d/dt (0.03t))
Since the derivative of (0.03t) is 0.03, we can simplify the expression:
P'(t) = 94000 e^(0.03t) * 0.03
Step 3: Evaluate the derivative at t = 10 (representing the year 2000):
P'(10) = 94000 e^(0.03 * 10) * 0.03
Calculating this expression will give us the rate of increase of the house's value in dollars per year in the year 2000.
To find how long it will take for the house to double in value, we need to determine the value of t when P(t) equals 2 times the original value, which is 94000.
Step 4: Set up the equation:
2 * 94000 = 94000 e^(0.03t)
We want to find the value of t that satisfies this equation.
Step 5: Solve the equation:
2 * 94000 = 94000 e^(0.03t)
Divide both sides by 94000:
2 = e^(0.03t)
Take the natural logarithm (ln) of both sides to isolate t:
ln(2) = ln(e^(0.03t))
Use the property of the logarithm to bring down the exponent:
ln(2) = 0.03t * ln(e)
ln(e) is equal to 1, so the equation simplifies further:
ln(2) = 0.03t
Finally, solve for t:
t = ln(2) / 0.03
Evaluating this expression will give us the time it takes for the house to double in value in years.