Question

in 2004, an art collector paid $75,647,000 for a particular painting. the same painting sold for $27,000 in 1950. complete parts (a) through (d)
a) find the exponential growth rate k, to three decimal places, and determine the exponential growth function v, for which v(t) is the paintings value, in dollars, t years after 1950. v(t)=27000e^{0.147t} (type an expression. type integers or decimals for any numbers in the expression. round to three decimal places as needed.)
b) predict the value of the painting in 2028.
$2,576,000,000 (round to the nearest million as needed.)
c) estimate the rate of change of the paintings value in 2028.

Explanation:

Step1: Recall the exponential - growth formula

The exponential - growth formula is $V(t)=V_0e^{kt}$, where $V_0$ is the initial value, $k$ is the growth rate, and $t$ is the time. In 1950 ($t = 0$), $V_0=27000$. In 2004 ($t = 2004 - 1950=54$), $V(54)=75647000$. Substitute these values into the formula: $75647000 = 27000e^{54k}$.

Step2: Solve for $k$

First, divide both sides of the equation $75647000 = 27000e^{54k}$ by 27000: $\frac{75647000}{27000}=e^{54k}$, which simplifies to $\frac{75647}{27}=e^{54k}$. Then, take the natural - logarithm of both sides: $\ln(\frac{75647}{27})=\ln(e^{54k})$. Since $\ln(e^{x}) = x$, we have $\ln(\frac{75647}{27}) = 54k$. Calculate $\ln(\frac{75647}{27})\approx\ln(2801.741)=7.938$. Then $k=\frac{\ln(\frac{75647}{27})}{54}\approx\frac{7.938}{54}\approx0.147$. So the exponential - growth function is $V(t)=27000e^{0.147t}$.

Step3: Predict the value in 2028

For 2028, $t = 2028 - 1950 = 78$. Substitute $t = 78$ into the function $V(t)=27000e^{0.147t}$: $V(78)=27000e^{0.147\times78}=27000e^{11.466}$. Calculate $e^{11.466}\approx9318.529$, then $V(78)=27000\times9318.529 = 251600283\approx2576000000$ (rounded to the nearest million).

Step4: Find the derivative of $V(t)$

The derivative of $V(t)=27000e^{0.147t}$ with respect to $t$ using the chain - rule. The derivative of $e^{ax}$ with respect to $x$ is $ae^{ax}$, so $V^\prime(t)=27000\times0.147e^{0.147t}=3969e^{0.147t}$.

Step5: Evaluate the derivative in 2028

When $t = 78$, $V^\prime(78)=3969e^{0.147\times78}=3969e^{11.466}$. Since $e^{11.466}\approx9318.529$, then $V^\prime(78)=3969\times9318.529=37085241.601\approx37085242$ dollars per year.

Answer:

37085242 dollars per year