Problem 46 - Calculus for Biology and Medicine

Question

In Problems , find the tangent line, in standard form, to \(y=\) \(f(x)\) at the indicated point. $$ y=-x^{3}-2 x^{2}, \text { at } x=0 $$

Step-by-Step Solution

Answer

The tangent line at \( x = 0 \) is \( y = 0 \) in standard form: \( 0x + y = 0 \).

Step 1: Find the derivative

The first step is to find the derivative of the function. The derivative of a function gives us the slope of the tangent line at any point on the curve. Function: \[ y = -x^{3} - 2x^{2} \]Derivative: \[ \frac{dy}{dx} = -3x^{2} - 4x \]

Step 2: Evaluate the derivative at the point

Next, we need to determine the slope of the tangent line at the specific point indicated, which is at \( x = 0 \). By substituting \( x = 0 \) into the derivative, we find the slope at that point. Evaluate: \[ \frac{dy}{dx} \bigg|_{x=0} = -3(0)^{2} - 4(0) = 0 \]

Step 3: Find the y-coordinate at the indicated x-value

Determine the \( y \)-coordinate of the function at \( x = 0 \) to identify the point of tangency. Substitute \( x = 0 \) into the original function.Substitute: \[ y = -0^{3} - 2(0)^{2} = 0 \]Point of tangency: \((0,0)\)

Step 4: Write the equation of the tangent line

Now that we have the slope from Step 2 and the point from Step 3, we can write the equation of the tangent line using the point-slope form: \[ y - y_1 = m(x - x_1) \]With point \((x_1, y_1) = (0, 0)\) and slope \(m = 0\), the equation is:\[ y - 0 = 0(x - 0) \]This simplifies to \( y = 0 \).

Step 5: Convert to standard form

To convert the equation of the tangent line to standard form, we rearrange it to fit the format \( Ax + By = C \).Given equation: \[ y = 0 \]Standard form: \[ 0x + 1y = 0 \]

Key Concepts

Tangent Line

The tangent line is a fundamental concept in calculus, often introduced alongside the idea of a derivative. A tangent line touches a curve at precisely one point. It does not cross the curve at that point, assuming the curve is smooth there. This contact point is critical because it defines the closest straight-line approximation of the curve at that vicinity. Here's why tangent lines are essential:

• **Slopes and Directions**: The slope of a tangent line mirrors the curve's immediate direction and steepness at the contact point.
• **Approximation**: They help approximate the value of the curve in that local region, useful for predictions and simplifications.

Determining the tangent line involves finding its slope and pinpointing its contact location on the curve. In our example problem, the tangent line to the curve at point \( x = 0 \), using the derived slope, results in the straightforward equation \( y = 0 \). This process exemplifies how calculus derives practical, real-line alignments from purely mathematical ideas.

Derivative

The derivative is another pillar of calculus, often introduced as the rate at which a function changes at any given point. Understanding derivatives is essential because they're the resource we tap into when we need to determine the slope of a function—or in physical terms, its rate of change. To find the derivative of a polynomial function, such as in the example provided, you apply these rules:

• Differentiate each term individually. For example, the power rule states that for any \( ax^n \), the derivative is \( nax^{n-1} \).
• Combine these derivatives to form the complete derivative equation.

In the exercise, this results in the derivative \[ \frac{dy}{dx} = -3x^2 - 4x \], which reveals critical insights about the original function's behavior.Once the derivative is calculated, evaluating it at specific points gives us actionable information, like the slope of the tangent line when \( x = 0 \). Derivatives are tools not only for solving academic exercises but also for optimizing complex systems in real life.

Point of Tangency

The point of tangency is where a tangent line touches the curve, an intersection with a sole contact point. This point is special because it gives the exact location where the tangent line best approximates the curve. Understanding this point is crucial as it synthesizes the concepts of slope from derivatives for calculating real-world tangents.To find the point of tangency, you must evaluate the original function equation at the specified x-value. In our exercise, substituting \( x = 0 \) into the function gives \( y = 0 \), affirming the point of tangency at \( (0, 0) \).Key to note about the point of tangency:

• **Exact Intersection**: The coordinates give the exact point where the derivative provides the tangent line's slope.
• **Foundation for Tangent Equations**: It serves as a foundational step for forming equations of tangent lines, leading to direct solutions.

In practice, determining the point of tangency offers an invaluable starting point for further mathematical exploration, like analyzing curve behavior or solving similar tangent line problems in different contexts.