Problem 19 - Applied Calculus

Question

Differentiate the functions in Problems 1-20. Assume that \(A\) and \(B\) are constants. $$ f(t)=\frac{t^{2}}{\cos t} $$

Step-by-Step Solution

Answer

\( f'(t) = \frac{2t \cos(t) + t^2 \sin(t)}{\cos^2(t)} \)

Step 1: Apply the Quotient Rule

The function is a quotient of two functions, to differentiate it we will use the quotient rule. For a function of the form \( \frac{u}{v} \), the derivative \( f'(t) \) is given by \( \frac{u'v - uv'}{v^2} \). Let \( u = t^2 \) and \( v = \cos(t) \).

Step 2: Differentiate the Numerator

Differentiate \( u = t^2 \). The derivative is \( u' = 2t \).

Step 3: Differentiate the Denominator

Differentiate \( v = \cos(t) \). The derivative is \( v' = -\sin(t) \).

Step 4: Substitute into Quotient Rule

Substitute \( u' = 2t \), \( u = t^2 \), \( v = \cos(t) \), and \( v' = -\sin(t) \) into the quotient rule formula. This gives:\[f'(t) = \frac{(2t) \cos(t) - (t^2)(-\sin(t))}{\cos^2(t)}\]

Step 5: Simplify the Expression

Simplify the expression obtained:\[f'(t) = \frac{2t \cos(t) + t^2 \sin(t)}{\cos^2(t)}\]

Key Concepts

Quotient Rule

When faced with differentiating a function that is a fraction of two other functions, the quotient rule is your go-to strategy. Imagine splitting your function into a top part and a bottom part. In mathematical terms, this means taking a function that is structured as \( \frac{u}{v} \), where \( u \) and \( v \) are both separate functions of a variable, usually \( t \) in our discussion, and finding its derivative. In such a setup, according to the quotient rule, the derivative of the function, denoted as \( f'(t) \), is given by:\[ f'(t) = \frac{u'v - uv'}{v^2} \]

• Take the derivative of the top part (numerator), \( u \), which we write as \( u' \).
• Take the derivative of the bottom part (denominator), \( v \), known as \( v' \).
• Plug these into the formula and simplify.

It’s a powerful technique in calculus, especially when functions can’t be easily simplified further.

Trigonometric Functions

Trigonometric functions, like \( \cos(t) \) and \( \sin(t) \), occur frequently in mathematical problems involving angles and periodic phenomena. Here, \( \cos(t) \) is the baseline function we are working with in the denominator, and some important characteristics about it help us understand how it behaves:

• The derivative of \( \cos(t) \) is \(-\sin(t)\). This means that when \( \cos(t) \) increases or decreases, \( \sin(t) \) tells us the rate and direction of this change.
• Trigonometric functions are periodic, repeating every \( 2\pi \), which means after one complete cycle, the pattern repeats.

Recognizing and applying the derivatives of trigonometric functions is a crucial step in dealing with problems like these, involving both polynomial and trigonometric parts.

Derivatives

Derivatives form the core of calculus and denote how one quantity changes with respect to another. In the context of this exercise, the derivative \( f'(t) \) represents how the value of the function \( f(t) \) changes as \( t \) changes.Here’s why it’s important:

• The derivative of \( t^2 \) is \( 2t \), showcasing how the rate of change increases with \( t \).
• The derivative of \( \cos(t) \) is \(-\sin(t)\), guiding us to handle changes in trigonometric functions.
• Derivatives are essential for understanding dynamic systems in physics, engineering, and other applied sciences.

The process involves applying rules like the quotient rule, and integrating knowledge of trigonometric derivatives to achieve simplification and understanding of function behavior across different domains.