Problem 107 - Precalculus

Question

In calculus, the value of \(F(b)-F(a)\) of a function \(F(x)\) at \(x=a\) and \(x=b\) plays an important role in the calculation of definite integrals. Find the exact value of \(F(b)-F(a)\). $$F(x)=\tan x+2 \cos x, a=0, b=\frac{\pi}{3}$$

Step-by-Step Solution

Answer

The exact value of \(F(b)-F(a)\) is \(\sqrt{3} - 1\).

Step 1: Evaluate Function at Upper Limit

Substitute the value of the upper limit into the function. Here, the upper limit is \(x = \frac{\pi}{3}\) and the function is \(F(x) = \tan x + 2 \cos x\). Evaluate the function: \[ F\left(\frac{\pi}{3}\right) = \tan\left(\frac{\pi}{3}\right) + 2 \cos\left(\frac{\pi}{3}\right) \]Using trigonometric values, \(\tan\left(\frac{\pi}{3}\right) = \sqrt{3}\) and \(\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}\). Substituting these, we get:\[ F\left(\frac{\pi}{3}\right) = \sqrt{3} + 2 \cdot \frac{1}{2} = \sqrt{3} + 1 \]

Step 2: Evaluate Function at Lower Limit

Substitute the value of the lower limit into the function. Here, the lower limit is \(x = 0\) and the function is still \(F(x) = \tan x + 2 \cos x\). Evaluate the function:\[ F(0) = \tan(0) + 2 \cos(0) \]Using known trigonometric values, \(\tan(0) = 0\) and \(\cos(0) = 1\). Hence,\[ F(0) = 0 + 2 \cdot 1 = 2 \]

Step 3: Calculate the Difference

Now that we have the values of the function at both the upper and lower limits, subtract the two:\[ F\left(\frac{\pi}{3}\right) - F(0) \]Substituting the values from Steps 1 and 2, we have:\[ (\sqrt{3} + 1) - 2 \]This simplifies to:\[ \sqrt{3} + 1 - 2 = \sqrt{3} - 1 \]

Key Concepts

Trigonometric functions

Trigonometric functions are crucial when dealing with angles and wave-like patterns in calculus. They allow for the transformation and evaluation of mathematical expressions involving angles.
One of the functions we'll frequently encounter is the tangent function, denoted as \( \tan x \). It represents the sine of an angle divided by its cosine, \( \tan x = \frac{\sin x}{\cos x} \).

• At \( x = 0 \), \( \tan(0) = 0 \) since \( \sin(0) = 0 \).
• At \( x = \frac{\pi}{3} \), a special triangle helps us see \( \tan(\frac{\pi}{3}) = \sqrt{3} \).

Similarly, the cosine function, \( \cos x \), describes the ratio of the adjacent side to the hypotenuse in a right triangle.

• At \( x = 0 \), \( \cos(0) = 1 \).
• At \( x = \frac{\pi}{3} \), \( \cos(\frac{\pi}{3}) = \frac{1}{2} \).

Understanding these basic trigonometric values is essential for evaluating functions accurately in calculus.

Calculus

Calculus is a field of mathematics focused on change and motion through derivatives and integrals. It helps analyze functions and their behavior over intervals.
In this context, we are particularly interested in definite integrals, which are used to calculate the accumulation of quantities like area under a curve. The expression \( F(b) - F(a) \) is central in definite integrals as it represents the net area between two limits using the antiderivative of a function.
The process involves evaluating the antiderivative function at two different points, \( a \) and \( b \). The difference between these values gives the area between those points.
For the problem at hand, it's these calculated values of the trigonometric function \( F(x) = \tan x + 2 \cos x \) at two points, \( x=0 \) and \( x=\frac{\pi}{3} \), that tell us how much one quantity surpasses another in the context of calculus.

Function evaluation

Function evaluation is all about substituting specific input values into a function to find out its corresponding output. This concept is crucial in forming the base for calculus-related calculations such as derivatives and integrals.
To evaluate a function like \( F(x) = \tan x + 2 \cos x \), you substitute given values for \( x \) into the function. First, let's consider \( x = \frac{\pi}{3} \):

• Find \( \tan(\frac{\pi}{3}) = \sqrt{3} \).
• Calculate \( 2 \cos(\frac{\pi}{3}) = 2 \times \frac{1}{2} = 1 \).
• Add them to evaluate the function: \( F(\frac{\pi}{3}) = \sqrt{3} + 1 \).

Next, substitute \( x = 0 \):

• \( \tan(0) = 0 \),
• \( 2 \cos(0) = 2 \times 1 = 2 \).
• Thus, \( F(0) = 0 + 2 = 2 \).

This step-by-step function evaluation assists you in calculating differences, a critical task in solving calculus problems like definite integrals.