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Find the instantaneous rates of change of the given functions at the indicated points. \(f(v)=-v^{3}+1, c=1\)

Short Answer

Expert verified
The instantaneous rate of change at \(v = 1\) is \(-3\).

Step by step solution

01

Understand Instantaneous Rate of Change

The instantaneous rate of change of a function at a point is the derivative of the function evaluated at that point. It represents the slope of the tangent line to the function's graph at that point.
02

Find the Derivative

To find the instantaneous rate of change of the function, start by finding the derivative of the function. We have the function \(f(v) = -v^3 + 1\). The derivative \(f'(v)\) is given by the rule \( \frac{d}{dv}(-v^3) = -3v^2 \). Thus, \(f'(v) = -3v^2\).
03

Evaluate the Derivative at the Given Point

Now, we evaluate the derivative at the given point, \(c = 1\). Substitute \(v = 1\) into the derivative: \(f'(1) = -3(1)^2 = -3\). This provides the instantaneous rate of change of the function at \(v = 1\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Derivative of a Function
The derivative of a function is a fundamental concept in calculus. It measures how a function's output changes as its input changes. Simply put, it's the function's rate of change at any given point. To compute it, we apply specific rules of differentiation to the function's formula.

For our function, \( f(v) = -v^3 + 1 \), we use the power rule to find its derivative. The power rule states that the derivative of \( -v^3 \) is \( -3v^2 \). Therefore, the derivative, denoted as \( f'(v) \), is \( -3v^2 \).

Always remember, the derivative gives us a new function that describes the original function’s slope at any point along its curve. It's called the derivative function.
Tangent Line
The tangent line to a curve at a given point is a straight line that just 'touches' the curve at that point. It doesn't cross the curve; instead, it represents the curve's instantaneous direction or slope at that spot.

In mathematical terms, the slope of the tangent line at any point is equivalent to the derivative evaluated at that point. For our function \( f(v) = -v^3 + 1 \), the equation of the tangent line at \( v = 1 \) would involve substituting \( v = 1 \) into the derivative \( f'(v) = -3v^2 \).

This means that at \( v = 1 \), the slope of the tangent line or the instantaneous rate of change is \( -3 \). The tangent line can be thought of as a snapshot of the curve's behavior around that specific point.
Evaluating Derivatives
Evaluating derivatives involves finding the value of the derivative at a specific point on the function. This process helps us to pinpoint the exact rate of change of the function at that particular location, rather than across its entirety.

Let's dive into our example to showcase this process. We determined that the derivative of \( f(v) = -v^3 + 1 \) is \( f'(v) = -3v^2 \). Evaluating this derivative at \( v = 1 \) involves replacing \( v \) with \( 1 \) in the derivative equation.

Thus, \( f'(1) = -3(1)^2 = -3 \). What this tells us is that at \( v = 1 \), for our function, the rate of change or the slope is \( -3 \). Practically, this indicates how steeply the function is increasing or decreasing at that specific point.

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