Question

c. find \\(\lim\limits_{t \to 3} \frac{s(t) - s(3)}{t - 3}\\). show the work that leads to your answer. include units.

Explanation:

To solve the limit \(\lim_{t \to 3} \frac{s(t) - s(3)}{t - 3}\), we recognize this as the definition of the derivative of the function \(s(t)\) at \(t = 3\), i.e., \(s'(3)\). The derivative of a position function \(s(t)\) (assuming \(s(t)\) represents position with respect to time \(t\)) gives the instantaneous velocity at a particular time.

Step 1: Recall the definition of the derivative

The derivative of a function \(y = f(x)\) at a point \(x = a\) is defined as:
\[
f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x - a}
\]
In our case, \(f(t)=s(t)\) and \(a = 3\), so:
\[
\lim_{t \to 3} \frac{s(t) - s(3)}{t - 3}=s'(3)
\]

Step 2: Interpret the derivative in the context of \(s(t)\)

If \(s(t)\) represents the position of an object as a function of time \(t\) (with units of, say, meters, and \(t\) in seconds), then the derivative \(s'(t)\) represents the velocity of the object at time \(t\), with units of meters per second (\(\text{m/s}\)).

Step 3: Determine the value of \(s'(3)\) (assuming we know the function \(s(t)\))

Since the problem doesn't provide the explicit form of \(s(t)\), we assume that from the context (maybe from a previous part of the problem, like a table of values or a graph of \(s(t)\)), we can find the slope of the tangent line to \(s(t)\) at \(t = 3\), which is \(s'(3)\).

For example, if \(s(t)\) is a position function and we know that at \(t = 3\), the instantaneous velocity (derivative) is, say, \(5 \, \text{m/s}\) (this value would come from the specific \(s(t)\) given in the problem, which is not fully provided here, but the process is as above), then:

\[
\lim_{t \to 3} \frac{s(t) - s(3)}{t - 3}=s'(3)
\]

If we assume \(s(t)\) is a function for which we can compute the derivative at \(t = 3\) (e.g., if \(s(t)=t^2\), then \(s'(t) = 2t\) and \(s'(3)=6\) with units depending on \(s(t)\)'s units), but since the problem is incomplete in terms of the function \(s(t)\), we'll state the general form.

Final Answer

The limit \(\lim_{t \to 3} \frac{s(t) - s(3)}{t - 3}\) is equal to the derivative of \(s(t)\) at \(t = 3\), \(s'(3)\), which represents the instantaneous velocity at \(t = 3\) (with units of \(\frac{\text{units of } s(t)}{\text{units of } t}\)). If we had the specific function \(s(t)\), we could compute the numerical value. For example, if \(s(t)\) is position in meters and \(t\) in seconds, and \(s'(3) = v\) (velocity), then the answer would be \(v\) with units \(\text{m/s}\).

(Note: Since the problem is missing the definition of \(s(t)\), the above is the general solution. If \(s(t)\) was given, say \(s(t)=2t + 1\), then \(s'(t) = 2\), so \(\lim_{t \to 3} \frac{s(t) - s(3)}{t - 3}=2\) with units \(\text{m/s}\) if \(s(t)\) is in meters and \(t\) in seconds.)

Answer:

To solve the limit \(\lim_{t \to 3} \frac{s(t) - s(3)}{t - 3}\), we recognize this as the definition of the derivative of the function \(s(t)\) at \(t = 3\), i.e., \(s'(3)\). The derivative of a position function \(s(t)\) (assuming \(s(t)\) represents position with respect to time \(t\)) gives the instantaneous velocity at a particular time.

Step 1: Recall the definition of the derivative

The derivative of a function \(y = f(x)\) at a point \(x = a\) is defined as:
\[
f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x - a}
\]
In our case, \(f(t)=s(t)\) and \(a = 3\), so:
\[
\lim_{t \to 3} \frac{s(t) - s(3)}{t - 3}=s'(3)
\]

Step 2: Interpret the derivative in the context of \(s(t)\)

If \(s(t)\) represents the position of an object as a function of time \(t\) (with units of, say, meters, and \(t\) in seconds), then the derivative \(s'(t)\) represents the velocity of the object at time \(t\), with units of meters per second (\(\text{m/s}\)).

Step 3: Determine the value of \(s'(3)\) (assuming we know the function \(s(t)\))

Since the problem doesn't provide the explicit form of \(s(t)\), we assume that from the context (maybe from a previous part of the problem, like a table of values or a graph of \(s(t)\)), we can find the slope of the tangent line to \(s(t)\) at \(t = 3\), which is \(s'(3)\).

For example, if \(s(t)\) is a position function and we know that at \(t = 3\), the instantaneous velocity (derivative) is, say, \(5 \, \text{m/s}\) (this value would come from the specific \(s(t)\) given in the problem, which is not fully provided here, but the process is as above), then:

\[
\lim_{t \to 3} \frac{s(t) - s(3)}{t - 3}=s'(3)
\]

If we assume \(s(t)\) is a function for which we can compute the derivative at \(t = 3\) (e.g., if \(s(t)=t^2\), then \(s'(t) = 2t\) and \(s'(3)=6\) with units depending on \(s(t)\)'s units), but since the problem is incomplete in terms of the function \(s(t)\), we'll state the general form.

Final Answer

The limit \(\lim_{t \to 3} \frac{s(t) - s(3)}{t - 3}\) is equal to the derivative of \(s(t)\) at \(t = 3\), \(s'(3)\), which represents the instantaneous velocity at \(t = 3\) (with units of \(\frac{\text{units of } s(t)}{\text{units of } t}\)). If we had the specific function \(s(t)\), we could compute the numerical value. For example, if \(s(t)\) is position in meters and \(t\) in seconds, and \(s'(3) = v\) (velocity), then the answer would be \(v\) with units \(\text{m/s}\).

(Note: Since the problem is missing the definition of \(s(t)\), the above is the general solution. If \(s(t)\) was given, say \(s(t)=2t + 1\), then \(s'(t) = 2\), so \(\lim_{t \to 3} \frac{s(t) - s(3)}{t - 3}=2\) with units \(\text{m/s}\) if \(s(t)\) is in meters and \(t\) in seconds.)