Problem 20 - University Calculus: Early Transcendentals

Question

Find the values of the derivatives. \(\left.\frac{d s}{d t}\right|_{t=-1} \quad\) if \(\quad s=1-3 t^{2}\) if \(\quad w=z+\sqrt{z}\)

Step-by-Step Solution

Answer

The derivative \( \frac{ds}{dt} \) evaluated at \( t = -1 \) is 6.

Step 1: Identify the Function

The function provided is \( s = 1 - 3t^2 \). We are tasked with finding the derivative of \( s \) with respect to \( t \) and evaluating it at \( t = -1 \).

Step 2: Find the Derivative

To find \( \frac{ds}{dt} \), differentiate the function \( s = 1 - 3t^2 \) with respect to \( t \). Use the power rule: \( \frac{d}{dt}[t^n] = nt^{n-1} \). This gives us \( \frac{ds}{dt} = -6t \).

Step 3: Evaluate the Derivative at Specific Point

Substitute \( t = -1 \) into the derivative \( \frac{ds}{dt} = -6t \) to find \( \left.\frac{ds}{dt}\right|_{t=-1} = -6(-1) = 6 \).

Step 4: Interpret the Second Function

The expression \( w = z + \sqrt{z} \) describes a function in terms of \( z \). However, this is separate from the given expression \( s = 1 - 3t^2 \) for which the derivative was calculated. There are no further calculations related to \( s \) and \( t \) in this context.

Key Concepts

Understanding Derivatives

In calculus, derivatives are fundamental to describing how a function changes. Derivatives illustrate the "rate of change" of one quantity relative to another. For example, in physics, they may quantify how velocity changes over time.
To understand derivatives, consider them as the slope of a function at any given point. This slope tells us how steep a curve is at that point and in which direction it's moving.

• For a linear function, the derivative (slope) is constant.
• For non-linear functions, the derivative varies depending on the point.

By applying differentiation, we find the equation that describes these changes, revealing valuable insights about the behavior of the function.
In this exercise, we focus on finding the derivative of the function \( s = 1 - 3t^2 \), which characterizes a parabolic downward shape.

Applying the Power Rule

One of the simplest and most common techniques for finding derivatives, especially for polynomials, is the power rule. The power rule is a quick way to differentiate expressions where the variable is raised to a power.
The power rule states: if you have a function \( f(t) = t^n \), then its derivative \( f'(t) = n \, t^{n-1} \).
Let's apply this rule to the function \( s = 1 - 3t^2 \):

• The "1" is a constant, and its derivative is 0.
• The term "\(- 3t^2\)" uses the power rule: bring down the exponent (2) and multiply it by the constant coefficient (-3) to get -6.
• Decrease the exponent by one, resulting in \(-6t^{1} = -6t\).

The derivative \( \frac{ds}{dt} = -6t \) indicates how \( s \) changes as \( t \) changes.

Evaluating Derivatives at a Specific Point

Once you have determined the derivative of a function, evaluating it at a particular point allows you to determine specific rates of change at that point.
This process involves substituting the given value into the derivative. In our exercise, we substitute \( t = -1 \) into the expression \( \frac{ds}{dt} = -6t \).
Performing this substitution, we calculate:

• \( \frac{ds}{dt} = -6(-1) = 6 \)

This result indicates that when \( t = -1 \), the rate of change of \( s \) is 6.
Such evaluations are crucial for interpreting real-world scenarios, such as determining specific moments of maximum or minimum values in various applications. They turn abstract calculus into concrete insights.