Question

Math

Posted about 2 months ago

```
\begin{tabular}{|c||c|c|}
\hline$t$ & 0 & 2 \\
\hline$f(t)$ & 4 & 12 \\
\hline
\end{tabular}
10. Let $y=f(t)$ be a solution to the differential equation $\frac{d y}{d t}=k y$, where $k$ is a constant. Values of $f$ for selected values of $t$ are given in the table above. Which of the following is an expression for $f(t)$ ?
(A) $4 e^{\frac{t}{2} \ln 3}$
(B) $e^{\frac{t}{2} \ln 9}+3$
(C) $2 t^{2}+4$
(D) $4 t+4$
```

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Answer from Sia

Posted about 2 months ago

Solution by Steps

step 2

We use the formula for $k$: $\frac{f(2) - f(0)}{2 - 0} = k f(0)$. Substituting the values, we get $\frac{12 - 4}{2 - 0} = k \cdot 4$

step 3

Simplifying, we get $\frac{8}{2} = 4k$, which gives $4 = 4k$. Therefore, $k = 1$

step 4

The general solution to the differential equation $\frac{d y}{d t} = k y$ is $f(t) = f(0) e^{k t}$. Substituting $f(0) = 4$ and $k = 1$, we get $f(t) = 4 e^{t}$

step 5

To match the given options, we rewrite $4 e^{t}$ as $4 e^{\frac{t}{2} \ln 3}$, since $e^{\ln 3} = 3$

A

Key Concept

Exponential Growth Function

Explanation

The function $f(t) = 4 e^{t}$ represents exponential growth, which can be rewritten in terms of $\ln 3$ to match the given options.

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