Problem 34 - Thomas Calculus

Question

Find the derivatives of the functions in Exercises \(19-38\) $$ g(t)=\left(\frac{1+\cos t}{\sin t}\right)^{-1} $$

Step-by-Step Solution

Answer

The derivative is \( g'(t) = \frac{1}{1 + \cos t} \).

Step 1: Understand the function

The given function is \( g(t) = \left(\frac{1 + \cos t}{\sin t}\right)^{-1} \). This function involves a fraction raised to the power of -1, which is equivalent to taking the reciprocal of the fraction. This means we can rewrite the function as \( g(t) = \frac{\sin t}{1 + \cos t} \).

Step 2: Apply the quotient rule

To find the derivative of \( g(t) = \frac{\sin t}{1 + \cos t} \), we apply the quotient rule. The quotient rule states that for a function \( h(x) = \frac{u(x)}{v(x)} \), the derivative \( h'(x) \) is \( \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} \). Here, \( u(t) = \sin t \) and \( v(t) = 1 + \cos t \).

Step 3: Differentiate the numerator and denominator

Differentiate the numerator \( u(t) = \sin t \) to get \( u'(t) = \cos t \). Differentiate the denominator \( v(t) = 1 + \cos t \) to get \( v'(t) = -\sin t \).

Step 4: Substitute into the quotient rule formula

Substitute \( u(t) = \sin t \), \( u'(t) = \cos t \), \( v(t) = 1 + \cos t \), and \( v'(t) = -\sin t \) into the quotient rule formula:\[g'(t) = \frac{\cos t (1 + \cos t) - \sin t (-\sin t)}{(1 + \cos t)^2}\]

Step 5: Simplify the expression

Simplify the expression:\[g'(t) = \frac{\cos t + \cos^2 t + \sin^2 t}{(1 + \cos t)^2}\]Using the identity \( \sin^2 t + \cos^2 t = 1 \), this simplifies further to:\[g'(t) = \frac{1 + \cos t}{(1 + \cos t)^2} = \frac{1}{1 + \cos t}\]

Key Concepts

Quotient Rule

The Quotient Rule is an essential tool in calculus used to find the derivative of a fraction where both the numerator and the denominator are functions of a variable. In this exercise, we use the quotient rule to differentiate the function \( g(t) = \frac{\sin t}{1 + \cos t} \). The general formula for the quotient rule is:

• Let \( h(x) = \frac{u(x)}{v(x)} \).
• The derivative, \( h'(x) \), is given by: \[h'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}\]

To apply it here:

• The numerator \( u(t) = \sin t \) and its derivative \( u'(t) = \cos t \).
• The denominator \( v(t) = 1 + \cos t \) and its derivative \( v'(t) = -\sin t \).
• Substitute these into the quotient rule formula to compute \( g'(t) \).

This approach is valuable for breaking down complex functions with fractions and ensures precision in calculations.

Trigonometric Functions

Trigonometric functions like sine (\( \sin t \)) and cosine (\( \cos t \)) are foundational in mathematics, particularly in calculus and physics. These functions describe relationships in triangles and have periodic properties that make them invaluable in modeling waves and oscillations.
For this problem:

• The sine function (\( \sin t \)) represents the opposite side over the hypotenuse in a right triangle.
• The cosine function (\( \cos t \)) represents the adjacent side over the hypotenuse.
• Both functions have simple derivatives: \( \frac{d}{dt} (\sin t) = \cos t \) and \( \frac{d}{dt} (\cos t) = -\sin t \).

The derivative properties of these functions simplify the differentiation process, making computations straightforward once you identify the components involved.

Reciprocal Function

A reciprocal function is one that essentially "flips" a given function, taking the expression \( \frac{1}{f(x)} \). In the original function, \( g(t) = \left(\frac{1 + \cos t}{\sin t}\right)^{-1} \), it implies reversing the fraction.
Understanding reciprocals can change the approach to solving problems:

• The original problem simplifies by identifying this reciprocal nature, changing the focus from complex powers to simpler fractions.
• By converting the function \( \left(\frac{1 + \cos t}{\sin t}\right)^{-1} \) to \( \frac{\sin t}{1 + \cos t} \), math operations become more manageable through standard rules like the quotient rule.
• This technique is helpful in both simplifying expressions and in making complex derivatives more approachable.

Deciphering reciprocal functions can often be the key to unlocking more intuitive solutions in calculus exercises.