Question

Use the Intermediate Value Theorem to prove that \((\cos t) t^{3}+6 \sin ^{5} t-3=0\) has a real solution between 0 and \(2 \pi\)

Short answer

The function has a real solution between 0 and \(2\pi\).

Step-by-step solution

Define the Function

Define the function \( f(t) = (\cos t) t^3 + 6 \sin^5 t - 3 \). This is the function we need to analyze to find a real solution.

Check for Continuity

The function \( f(t) = (\cos t) t^3 + 6 \sin^5 t - 3 \) is continuous over the interval \([0, 2\pi]\) because both \(\cos t\) and \(\sin t\) are continuous, and they only involve algebraic operations which preserve continuity.

Evaluate the Function at the Endpoints

Calculate \( f(0) = (\cos 0)(0)^3 + 6 \sin^5 0 - 3 = 0 - 3 = -3 \). Next, calculate \( f(2\pi) = (\cos 2\pi)(2\pi)^3 + 6 \sin^5 2\pi - 3 = 8\pi^3 - 3 \).

Check Value Change Through the Interval

Between \(0\) and \(2\pi\), \( f(t) \) changes from \(-3\) to a positive value \(8\pi^3 - 3\). This indicates that the function changes sign over the interval.

Apply the Intermediate Value Theorem

By the Intermediate Value Theorem, since \( f(t) \) is continuous on \([0, 2\pi]\) and changes sign, there is at least one \( t \in (0, 2\pi) \) such that \( f(t) = 0 \).

Key concepts

Continuous functions

Continuous functions are essential in calculus. A function is continuous at a point if you can draw it without lifting your pencil off the paper.
For a function to be continuous over an interval, it must be continuous at every point in that interval.
The mathematical definition states that a function \( f(x) \) is continuous at a point \( a \) if the following three conditions are satisfied:

• \( f(a) \) is defined.
• \( \lim_{{x \to a}} f(x) \) exists.
• \( \lim_{{x \to a}} f(x) = f(a) \).

For the function \( f(t) = (\cos t) t^3 + 6 \sin^5 t - 3 \), both \( \cos t \) and \( \sin t \) are continuous trigonometric functions.
Since algebraic operations such as addition, multiplication, and powers do not disrupt continuity, \( f(t) \) remains continuous on the interval \([0, 2\pi]\).
This continuity means there are no breaks or gaps in the graph of the function over this interval.

Trigonometric functions

Trigonometric functions relate angles of triangles to their sides. They are periodic functions and are crucial in mathematics and physics.
Two primary trigonometric functions are used here: cosine (\(\cos t\)) and sine (\(\sin t\)).

• The function \( \cos t \) gives the horizontal coordinate of a point on the unit circle.
• The function \( \sin t \) provides the vertical coordinate of a point on the unit circle.

Both functions are continuous over all real numbers.
The periodic nature of these functions means they repeat their values in regular intervals.
Here, both \(\cos t\) and \(\sin t\) operate within the interval \([0, 2\pi]\), covering one full cycle of their respective periods.
It's this regular repeating pattern that ensures functions like \( f(t) = (\cos t) t^3 + 6 \sin^5 t - 3 \) can be analyzed for their behavior, such as evaluating them over specific intervals.

Real solutions

A real solution is a value where a function equals zero and exists within the set of real numbers.
In simpler terms, it's the point where the function's graph crosses the x-axis.
For the given function \( f(t) = (\cos t) t^3 + 6 \sin^5 t - 3 \), we are asked to find where it equals zero between \(0\) and \(2\pi\).

• The Intermediate Value Theorem helps identify the existence of real solutions.
• This theorem states that if a function is continuous on a closed interval \([a, b]\), and takes on values \(f(a)\) and \(f(b)\), it must take on any value between \(f(a)\) and \(f(b)\) at some point.

In the exercise, evaluating \(f(0) = -3\) and \(f(2\pi) = 8\pi^3 - 3\) shows a sign change.
Thus, by the theorem, there must be a value \(t\) within this interval where \(f(t) = 0\).
This value is the real solution we were seeking.