Problem 57 - Calculus and its applications

Question

Find an equation for the tangent line to the graph of \(y=\left(x^{3}-4 x\right)^{10}\) at the point (2,0)

Step-by-Step Solution

Answer

The tangent line is the horizontal line \( y = 0 \).

Step 1: Find the derivative of the function

Start by finding the derivative of the function \( y = (x^3 - 4x)^{10} \) using the chain rule. The chain rule states that if you have a function \( f(g(x)) \), then its derivative is \( f'(g(x)) \, g'(x) \).Let \( u = x^3 - 4x \), then \( y = u^{10} \). The derivative \( \frac{dy}{dx} \) is:\[ \frac{dy}{du} = 10u^9 \quad \text{and} \quad \frac{du}{dx} = 3x^2 - 4. \]Thus, using the chain rule:\[ \frac{dy}{dx} = 10(x^3 - 4x)^9 (3x^2 - 4). \]

Step 2: Evaluate the derivative at the given point

Substitute \( x = 2 \) into the derivative to find the slope of the tangent line at the point \( (2,0) \).First, calculate \( u = (x^3 - 4x) = (2^3 - 4 \times 2) = 8 - 8 = 0 \).Now, evaluate \( \frac{dy}{dx} \) at \( x = 2 \):\[ \frac{dy}{dx} = 10(0)^9(3 \times 2^2 - 4) = 10 \times 0 \times (12 - 4) = 0. \]

Step 3: Write the equation for the tangent line

Use the point-slope form of the equation for a line, \( y - y_1 = m(x - x_1) \), where \( m \) is the slope and \( (x_1, y_1) = (2, 0) \).Since the calculated slope \( m = 0 \), the equation simplifies to:\[ y - 0 = 0(x - 2). \]Thus, the equation of the tangent line is:\[ y = 0. \]

Key Concepts

Derivative

In calculus, the derivative is a fundamental concept that measures how a function changes as its input changes. Derivatives are important for understanding the behavior of graphs, rates of change, and for finding slopes of tangent lines. In simple terms, the derivative tells us the rate at which one quantity changes with respect to another.

When we have a function like \( y = (x^3 - 4x)^{10} \), the derivative, denoted as \( \frac{dy}{dx} \), represents how \( y \) changes as \( x \) changes. To find this, we apply differentiation rules, including the power rule, product rule, quotient rule, or chain rule.

Derivatives can be represented in several ways:

• As \( \frac{dy}{dx} \), meaning the change in \( y \) with respect to \( x \).
• As \( f'(x) \), which is another notation for \( \frac{dy}{dx} \).

The result gives us information about the function's growth, concavity, and any local extremum points.

Tangent Line

A tangent line is a straight line that touches a curve at a single point without crossing it. At the point of tangency, the tangent line has the same slope as the curve itself. This makes tangent lines incredibly useful for understanding the instantaneous rate of change of a curve at a given point.

In the problem, we need to find the tangent line to the curve \( y = (x^3 - 4x)^{10} \) at the point \((2, 0)\). The key step here is to find the slope of the tangent line, which is given by the derivative of the function at the specific point \( x = 2 \).

Once we have the slope, the equation of the tangent line can be written using the point-slope form. This is a linear equation that best describes the line touching the curve at \((2, 0)\). In our example, the derivative at \( x=2 \) is zero, indicating the tangent is horizontal.

Chain Rule

The chain rule is essential when dealing with functions that are composed of other functions. It's especially useful when differentiating complex expressions like \( (x^3 - 4x)^{10} \). Simply put, the chain rule helps in finding the derivative of a composite function.

If you have a function of a function, expressed as \( f(g(x)) \), the chain rule states the derivative is given by \( f'(g(x)) \cdot g'(x) \). Breaking this down:

• Find \( g'(x) \): differentiate the inner function.
• Find \( f'(g(x)) \): differentiate the outer function as if the inner function is variable, and afterward multiply by \( g'(x) \).

In our exercise, this means defining \( u = x^3 - 4x \), and then finding \( \frac{dy}{du} \) and \( \frac{du}{dx} \). Combine these according to the chain rule to solve for \( \frac{dy}{dx} \).

The magic of the chain rule lies in its ability to simplify the differentiation of complex expressions by ensuring each component is handled systematically.

Point-Slope Form

The point-slope form of a line equation is a convenient method used to construct the equation of a line knowing a specific point it passes through and its slope. This form is particularly useful with tangent lines, as it clearly expresses their equation with minimal information.

In general, the point-slope form is given by:

• \( y - y_1 = m(x - x_1) \)

Where \((x_1, y_1)\) is a given point on the line, and \(m\) is the slope of the line. For our exercise, the point is \((2, 0)\) and the slope \(m\) is 0, resulting from our earlier differentiation.

Once substituted, our equation simplifies to \( y - 0 = 0(x - 2) \), meaning \( y = 0 \). This shows that the tangent line is completely horizontal at the given point. Point-slope form is a straightforward and effective tool for forming the equation of a line quickly.