Find the instantaneous rate of change for the function at the given value.
 g left parenthesis t right parenthesis equals 5 minus t squared
 
at tequals
negative 5
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Part 1
The instantaneous rate of change at tequals
negative 5 
is enter your response here
.

To find the instantaneous rate of change of the function \( g(t) = 5 - t^2 \) at \( t = -5 \), we need to compute the derivative of the function and then evaluate it at \( t = -5 \).

### Step 1: Find the derivative of \( g(t) \)
The derivative of \( g(t) = 5 - t^2 \) with respect to \( t \) is:
\[
g'(t) = \frac{d}{dt}(5 - t^2)
\]
The constant 5 has a derivative of 0, and the derivative of \( -t^2 \) is \( -2t \). Therefore:
\[
g'(t) = -2t
\]

### Step 2: Evaluate the derivative at \( t = -5 \)
Now, substitute \( t = -5 \) into the derivative:
\[
g'(-5) = -2(-5) = 10
\]

### Conclusion:
The instantaneous rate of change of the function \( g(t) = 5 - t^2 \) at \( t = -5 \) is \( 10 \).

Do you need further details or have any questions?

Here are some related questions:
1. What is the instantaneous rate of change for \( g(t) = 5 - t^2 \) at \( t = 0 \)?
2. How does the instantaneous rate of change relate to the slope of the tangent line?
3. What is the geometric interpretation of an instantaneous rate of change?
4. How would you find the average rate of change over an interval instead?
5. Can you compute the second derivative of \( g(t) \), and what does it signify?

**Tip:** The instantaneous rate of change at a point on a curve corresponds to the slope of the tangent line at that point.

Find the instantaneous rate of change for the function g(t) = 5 - t^2 at t = -5.

Learn how to find the instantaneous rate of change for the function g(t) = 5 - t^2 at t = -5. This calculus problem involves finding the derivative using the power rule and evaluating it at a specific point. Suitable for Grades 11-12 students.

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