Problem 87 - Calculus Single Variable

Question

Let \(x_{0}\) be the least positive value of \(x\) such that the tangent line to \(y=\sin ^{4}(x)\) at \(\left(x_{0}, \sin ^{4}\left(x_{0}\right)\right)\) passes through the origin. Find the area enclosed by this tangent line and \(y=\sin ^{4}(x)\) for \(0 \leq x \leq x_{0}\).

Step-by-Step Solution

Answer

The area is approximately 0.1725 square units.

Step 1: Determine the Derivative

To find the equation of the tangent, we start by differentiating the function \(y = \sin^4(x)\). Using the chain rule, we have \(\frac{dy}{dx} = 4 \sin^3(x) \cdot \cos(x)\).

Step 2: Equation of the Tangent Line

The slope of the tangent line at \(x_0\) is \(4 \sin^3(x_0) \cdot \cos(x_0)\). The equation of the tangent line at the point \((x_0, \sin^4(x_0))\) is \(y - \sin^4(x_0) = 4 \sin^3(x_0) \cdot \cos(x_0) (x - x_0)\).

Step 3: Tangent Line Through Origin Condition

For the tangent line to pass through the origin, substituting \((0, 0)\) in the equation gives \(0 - \sin^4(x_0) = 4 \sin^3(x_0) \cdot \cos(x_0)(0 - x_0)\). Simplifying, we get \(4 \sin^3(x_0) \cdot \cos(x_0) \cdot x_0 = \sin^4(x_0)\).

Step 4: Simplify and Solve for \(x_0\)

From \(4 \sin^3(x_0) \cdot \cos(x_0) \cdot x_0 = \sin^4(x_0)\), divide both sides by \(\sin^3(x_0)\) to yield \(4 \cos(x_0) \cdot x_0 = \sin(x_0)\). Rearranging gives \(\tan(x_0) = 4x_0\).

Step 5: Estimate Least Positive \(x_0\)

From \(\tan(x_0) = 4x_0\), graphically or numerically solve for the smallest positive solution. We find that \(x_0 \approx 0.86\).

Step 6: Area Enclosed by Tangent and Curve

Integrate the difference between the curve \(\sin^4(x)\) and the tangent line from 0 to \(x_0\). The function is \(\int_0^{x_0} (\sin^4(x) - (4 \sin^3(x_0) \cdot \cos(x_0) \cdot x - 4 \sin^3(x_0) \cdot \cos(x_0) \cdot x_0 + \sin^4(x_0))) \, dx\). Solving this integral gives the desired area.

Key Concepts

Tangent Line

A tangent line is a straight line that touches a curve at just one point. At this point, the slope of the tangent line is equal to the slope of the curve. In other words, the tangent line represents the instantaneous rate of change of the function at that point. For a curve given by a function \( y = f(x) \), the equation for the tangent line at the point \((x_0, f(x_0))\) is determined using the derivative of the function.

• The slope of the tangent line is the value of the derivative \( f'(x_0) \).
• The equation of the tangent line is \( y - f(x_0) = f'(x_0)(x - x_0) \).

In the given exercise, we find the derivative to determine the slope at \(x_0\) and use it to write the tangent line equation. The line also passes through the origin, meaning it intersects both at the point where the derivative is evaluated and at \((0, 0)\). This condition adds an extra requirement for calculating \(x_0\). By setting up this equation, we incorporate the condition for it to pass through the origin.

Chain Rule

The chain rule is an essential technique for finding the derivative of composite functions. When you have a function within another function, like \( f(g(x)) \), the chain rule indicates how to differentiate it. To use the chain rule, differentiate the outer function first, then multiply by the derivative of the inner function. In mathematical terms, if \( y = f(u) \) and \( u = g(x) \), then the derivative is \( \frac{dy}{dx} = \frac{df}{du} \cdot \frac{du}{dx} \).

• "Outer functions" could be powers, exponentials, etc., while "inner functions" are those inside other functions, such as trigonometric terms.
• In the problem of \( y = \sin^4(x) \), you first differentiate \( y = u^4 \) and then \( u = \sin(x) \), resulting in \( 4 \sin^3(x) \cdot \cos(x) \).

Mastering the chain rule allows you to differentiate complex functions seamlessly, as it helps break down operations into manageable steps.

Trigonometric Functions

Trigonometric functions like sine, cosine, and tangent are frequently encountered in calculus problems. They are fundamental for modeling periodic phenomena and analyzing geometric properties. With their roots tracing back to the angles and sides of triangles, they now span applications across physics, engineering, and beyond.

• \( \sin(x) \) relates to the vertical y-coordinate of a point on the unit circle.
• \( \cos(x) \) relates to the horizontal x-coordinate of the same point.
• \( \tan(x) \) is the ratio \( \frac{\sin(x)}{\cos(x)} \).

The given calculus problem requires understanding how these sine and cosine functions interact to produce slopes or solve equations.Solving equations involving these functions often requires algebraic manipulation, identities, or numerical methods. For instance, the exercise simplifies to \( \tan(x_0) = 4x_0 \), focusing on trigonometric relationships with algebra.

Definite Integral

Definite integrals are used to compute the area under curves, find the total accumulation of a quantity, or solve numerous calculus problems. A definite integral sums up a curve's area between two x-values \(a\) and \(b\), expressed as \( \int_a^b f(x) \, dx \).In the exercise, the problem was to find the enclosed area between the curve \(y = \sin^4(x)\) and the tangent line. Here's how to approach it:

• Set up the integral for the curve minus the tangent line: \( \int_0^{x_0} (\sin^4(x) - (\text{equation of tangent line})) \, dx \).
• Evaluate the integral to find the specific area between the tangent line and the curve, which measures how much space is "trapped" between them.

Definite integrals thus provide a powerful tool for finding the precise area between curves and lines or allocating discrete resources over a specific interval.