Question Statement

The price of commodity $P(t)$ is given by:

$P(t)=150(1+0.05t{)}^2$

where $t$ is measured in years and $P(t)$ is the price level.

a. Find the instantaneous rate of change of prices at $t=3$ years.

b. Calculate the inflation rate at $t=3$ years.

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Background and Explanation

This problem requires finding the derivative of a composite function using the chain rule to determine the instantaneous rate of price change. The inflation rate represents the percentage change of the price level relative to the current price, calculated as $\frac{dP/dt}{P(t)}\times 100$.

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Solution

Part (a): Instantaneous Rate of Change

To find the instantaneous rate of change, we need to differentiate $P(t)$ with respect to $t$ and evaluate at $t=3$.

Given:

$P(t)=150(1+0.05t{)}^2$

Differentiate with respect to $t$ using the chain rule. The function is a composition where the outer function is $150u^2$ and the inner function is $u=1+0.05t$:

$\begin{array}{rl}\frac{dP}{dt}& =150\cdot \frac{d}{dt}\left[(1+0.05t{)}^2\right]\\ & =150(2)(1+0.05t{)}^{2-1}\cdot \frac{d}{dt}(1+0.05t)\\ & =300(1+0.05t)\cdot (0+(0.05)(1))\\ & =(300)(0.05)(1+0.05t)\\ \frac{dP}{dt}& =15(1+0.05t)\end{array}$

Now evaluate at $t=3$:

$\begin{array}{rl}{\frac{dP}{dt}|}_{t=3}& =15(1+(0.05)(3))\\ & =15(1+0.15)\\ & =15(1.15)\\ \frac{dP}{dt}& =17.25\end{array}$

Therefore, the instantaneous rate of change of prices at $t=3$ years is 17.25 units per year.

Part (b): Inflation Rate

The inflation rate is defined as the percentage rate of change of price relative to the current price level:

$\text{Inflation rate}=\frac{\frac{dP}{dt}}{P(t)}\times 100$

From part (a), at $t=3$, we have $\frac{dP}{dt}=17.25$.

First, calculate $P(3)$:

$\begin{array}{rl}P(3)& =150(1+(0.05)(3){)}^2\\ & =150(1+0.15{)}^2\\ & =150(1.15{)}^2\\ & =150(1.3225)\\ P(3)& =198.375\end{array}$

Now substitute these values into the inflation rate formula:

$\begin{array}{rl}\text{Inflation rate}& =\frac{17.25}{198.375}\times 100\\ & =0.087\times 100\\ & =8.7\%\end{array}$

Therefore, the inflation rate at $t=3$ years is 8.7%.

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Key Formulas or Methods Used

• Chain Rule for Differentiation: $\frac{d}{dt}[f(g(t))]=f^{\prime }(g(t))\cdot g^{\prime }(t)$
• Power Rule: $\frac{d}{dt}[u^n]=n\cdot u^{n-1}\cdot \frac{du}{dt}$
• Instantaneous Rate of Change: $\frac{dP}{dt}$ evaluated at a specific time $t$
• Inflation Rate Formula: $\text{Inflation Rate}=\frac{dP/dt}{P(t)}\times 100\%$

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Summary of Steps

1. Differentiate the price function $P(t)=150(1+0.05t{)}^2$ using the chain rule to obtain $\frac{dP}{dt}=15(1+0.05t)$
2. Evaluate the derivative at $t=3$ to find the instantaneous rate of change: $15(1.15)=17.25$
3. Calculate the price level at $t=3$ by substituting into the original function: $P(3)=150(1.15{)}^2=198.375$
4. Compute the inflation rate using the percentage change formula: $\frac{17.25}{198.375}\times 100=8.7\%$