Problem 14 - Thomas Calculus

Question

In Exercises \(13-16,\) find \(d s / d t\) $$ s=t^{2}-\sec t+1 $$

Step-by-Step Solution

Answer

\( \frac{ds}{dt} = 2t - \sec t \tan t \)

Step 1: Understand the Derivative

We need to find the derivative of the function \( s(t) = t^2 - \sec t + 1 \) with respect to \( t \). This means we are looking for \( \frac{ds}{dt} \).

Step 2: Differentiate Each Term

Differentiate each term of the function separately. The first term is \( t^2 \), which differentiates to \( 2t \). The second term is \( -\sec t \), which differentiates to \( -\sec t \tan t \). The third term is a constant, \( 1 \), and its derivative is \( 0 \).

Step 3: Combine the Derivatives

Combine the derivatives from Step 2 to get \( \frac{ds}{dt} = 2t - \sec t \tan t + 0 \).

Step 4: Simplify the Expression

The simplified form of the expression for \( \frac{ds}{dt} \) is \( 2t - \sec t \tan t \).

Key Concepts

Derivative Calculation

Calculating derivatives is a fundamental skill in calculus, involving finding the rate at which a function changes with respect to a variable. In our exercise, we focus on differentiating the function \( s(t) = t^2 - \sec t + 1 \). Here are the steps involved:

• Identify the terms that need differentiation: \( t^2 \), \(-\sec t\), and the constant \( 1 \).


• Apply the power rule for \( t^2 \): the derivative is \( 2t \), as the power rule states that the derivative of \( t^n \) is \( nt^{n-1} \).


• Use the derivative for trigonometric functions: the derivative of \(-\sec t\) is \(-\sec t \tan t\), which is derived from known formulas in calculus.


• Remember, the derivative of a constant like \( 1 \) is always \( 0 \).

Combining these derivatives, the overall derivative \( \frac{ds}{dt} \) is \( 2t - \sec t \tan t \). This calculation gives insight into how the original function \( s(t) \) changes as \( t \) changes.

Trigonometric Functions

Trigonometric functions such as \( \sec t \) and \( \tan t \) play essential roles in calculus, especially in differentiation. In this exercise, we specifically dealt with the trigonometric function \(-\sec t\):

• The secant function, \( \sec t \), is the reciprocal of the cosine function, \( \sec t = \frac{1}{\cos t} \).
• The tangent function, \( \tan t \), is defined as \( \tan t = \frac{\sin t}{\cos t} \).


• For differentiation, the derivative of \( \sec t \) with respect to \( t \) is \( \sec t \tan t \). Therefore, \(-\sec t\) differentiates to \(-\sec t \tan t\).

These functions are vital in understanding more complex motion dynamics in calculus. They frequently appear in various applications including physics and engineering, where wave-like motion is modeled.

Mathematical Expressions

Understanding mathematical expressions and simplifying them is crucial in calculus. The expression we dealt with, \( s(t) = t^2 - \sec t + 1 \), is a combination of different types of terms:

• Polynomial terms like \( t^2 \) which are straightforward to differentiate using simple rules like the power rule.


• Trigonometric terms such as \(-\sec t\) require specific knowledge of trigonometric derivatives.


• Constant numbers like \( 1 \) simplify the equation by having a derivative of zero.

Simplifying the expression after differentiation is key to achieving a clear result, \( 2t - \sec t \tan t \) in this case. As you work through calculus problems, ensure each step is clear and each term is addressed properly for accurate results.