Problem 19 - Thomas Calculus

Question

In Exercises \(19-22,\) find the values of the derivatives. $$ \left.\frac{d s}{d t}\right|_{t=-1} \quad \text { if } \quad s=1-3 t^{2} $$

Step-by-Step Solution

Answer

The derivative is 6.

Step 1: Identify the function to differentiate

The given function is \( s = 1 - 3t^2 \). We need to find the derivative of this function with respect to \( t \).

Step 2: Differentiate the function

Differentiate \( s = 1 - 3t^2 \) with respect to \( t \). The derivative of a constant is zero and the derivative of \( -3t^2 \) is \( -6t \). So, the derivative \( \frac{ds}{dt} = -6t \).

Step 3: Evaluate the derivative at \( t = -1 \)

Substitute \( t = -1 \) into the derivative \( \frac{ds}{dt} = -6t \). This gives: \( \frac{ds}{dt} = -6(-1) = 6 \).

Key Concepts

Understanding Function Differentiation

Differentiation in calculus is the process of finding the derivative of a function. It's like asking, "How does the function's output change when we change the input slightly?" When we differentiate, we're finding the rate at which something changes and this is an essential idea in calculus.
For the function given in the exercise, we need to find how the function "s" changes with respect to "t". The function here is quadratic, meaning it has a term with a variable raised to the power of two, which is

• The base function is: \[ s = 1 - 3t^2 \]

Differentiating a quadratic function involves using the power rule. The power rule is simple: bring down the exponent as a coefficient, then reduce the exponent by one.
For example, to differentiate \( 3t^2 \), we apply the power rule: move the 2 in front to multiply with the constant (3), then subtract 1 from the exponent. This results in:

• \[ 2\times(-3)\times t^{2-1} = -6t \]

Evaluating Derivatives at Specific Points

After differentiating a function, you can evaluate the derivative at a specific input value. Evaluating a derivative at a point answers the question: "What is the rate of change of the function at this exact point?"
In this problem, we're asked to find the rate of change when \( t = -1 \). After differentiating the function

• The derivative is:\[ \frac{ds}{dt} = -6t \]

Substituting \(t = -1\) into this expression means we're simply plugging this value into the formula:

• \[ \frac{ds}{dt} = -6(-1) = 6 \]

This tells us that at \(t = -1\), the rate of change of the function is 6. So, for each unit of change in \(t\) around this point, the function \(s\) increases by 6 units.

Tackling a Basic Calculus Problem

Problems like this are the building blocks of learning calculus. They reinforce key skills such as identifying the function to differentiate and applying the correct rules to find derivatives. Here’s a straightforward approach that helps break down such problems.
First, scan through the function given, identifying what type of function it is and what terms you’ll need to differentiate. In this exercise, notice that the function is a simple quadratic. Employ the power rule for each term to find the derivative overall. Second, once you have the derivative, use simple algebra to substitute any given points to evaluate the rate of change.
As you tackle more calculus problems:

• Don’t rush. Recognize the structure of each problem and the rules that apply.
• Practice regularly, as familiarizing yourself with common types of problems makes it easier for you to identify the right strategies.
• Break down the problem into smaller parts to ensure each step is understood before moving forward.