The size of a balloon is changing with respect to time, such that the volume is given by V(t), where V(t) is measured in cubic centimeters and t is measured in seconds. A tangent line is shown for V(t) at t = 2 seconds. Determine the best estimate for the value of the instantaneous rate of change for V(t) when t = 2.

curve on coordinate plane with axes labeled t and V of t, tangent to curve shown at t equal to 2

 V of 2 minus V of 1 all over 1
 V of the limit of t as t approaches 1
 the change in V of t all over 2
 V of 2 and 1 hundredth minus V of 1 and 99 hundredths all over 2 hundredths

To determine the **instantaneous rate of change** of the volume \( V(t) \) at \( t = 2 \), we need to find the derivative \( V'(2) \), which represents the slope of the tangent line to the curve at \( t = 2 \).

Let's break down each option provided:

1. **\( \frac{V(2) - V(1)}{1} \)**  
   This expression represents the **average rate of change** of \( V(t) \) between \( t = 1 \) and \( t = 2 \), not the instantaneous rate of change at \( t = 2 \).

2. **\( \lim_{t \to 1} V(t) \)**  
   This expression is not the correct form for the derivative at \( t = 2 \). This is more about evaluating the function as \( t \) approaches 1, which is unrelated to finding the derivative at \( t = 2 \).

3. **The change in \( V(t) \) all over 2**  
   This is vague but seems like an approximation for the average rate of change over some interval, which still wouldn't give the instantaneous rate of change.

4. **\( \frac{V(2.01) - V(1.99)}{0.02} \)**  
   This is a **difference quotient** that approximates the derivative by taking values very close to \( t = 2 \). It's the best approximation for the instantaneous rate of change at \( t = 2 \) using nearby points.

Therefore, the **best estimate** for the instantaneous rate of change of \( V(t) \) at \( t = 2 \) is given by the expression:

\[
\frac{V(2.01) - V(1.99)}{0.02}
\]

This represents a small difference in \( V(t) \) over a small interval around \( t = 2 \), which closely approximates the slope of the tangent line at \( t = 2 \).

### Would you like more details or have any further questions? Here are some related questions for deeper exploration:

1. How do we formally define the derivative as a limit?
2. What is the geometric interpretation of the derivative on a curve?
3. How can we calculate the average rate of change over larger intervals?
4. What is the difference between instantaneous and average rates of change?
5. Can we estimate derivatives using even smaller intervals for better accuracy?

**Tip:** When estimating instantaneous rates of change, using values very close to the point of interest gives a better approximation.

The size of a balloon is changing with respect to time, such that the volume is given by V(t), where V(t) is measured in cubic centimeters and t is measured in seconds. A tangent line is shown for V(t) at t = 2 seconds. Determine the best estimate for the value of the instantaneous rate of change for V(t) when t = 2.

This page explains how to estimate the instantaneous rate of change of the volume of a balloon at t = 2 seconds using the difference quotient method. Learn about key calculus concepts, including the derivative as a limit, with a step-by-step solution.

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