Problem 27 - Calculus for Biology and Medicine

Question

Find the equation of the tangent line to the curve \(y=3 x^{2}+1\) at the point \((0,1)\).

Step-by-Step Solution

Answer

The equation of the tangent line is \( y = 1 \).

Step 1: Differentiate the Function

To find the equation of the tangent line, we first need the slope of the tangent at the given point. Differentiate the function \( y = 3x^2 + 1 \) with respect to \( x \). The derivative \( y' \) of \( y = 3x^2 + 1 \) is calculated as follows: \( y' = \frac{dy}{dx} = 6x \).

Step 2: Evaluate the Derivative at the Point

Substitute \( x = 0 \) into the derivative \( y' = 6x \) to find the slope of the tangent line at the point \( (0,1) \). This gives \( y' = 6(0) = 0 \). Therefore, the slope of the tangent line is \( 0 \).

Step 3: Write the Equation of the Tangent Line

Use the point-slope form of a linear equation, \( y - y_1 = m(x - x_1) \), where \( (x_1, y_1) \) is the point of tangency, and \( m \) is the slope. Substituting \( (x_1, y_1) = (0, 1) \) and \( m = 0 \), we have \( y - 1 = 0(x - 0) \). Simplifying, the equation of the tangent line is \( y = 1 \).

Key Concepts

Differentiation

Differentiation is a process in calculus that determines how a function changes as its input changes. It's like a mathematical way of capturing the rate of change. To find the derivative of a function, you differentiate it with respect to one of its variables, usually denoted as \( x \). This process provides us with another expression, often dubbed the 'slope function,' which tells us the slope of the original function at any point.

In our example, we start with the function \( y = 3x^2 + 1 \). To differentiate this, we need to find \( y' \) or \( \frac{dy}{dx} \).

• The derivative of \( x^2 \) with respect to \( x \) is \( 2x \).
• Hence, the derivative of \( 3x^2 \) is \( 3 \times 2x = 6x \).
• The derivative of a constant like \( 1 \) is zero.

Thus, the derivative of our function is \( y' = 6x \). This derivative helps us find the slope of the tangent line at any given point on the curve.

Slope of Tangent

The slope of a tangent line to a curve represents how steep the line is at a specific point on the curve. Imagine placing a small ruler on the graph of a function at a certain point; the angle at which the ruler is tilted indicates the slope.

To determine the slope at a particular point, we simply evaluate the derivative at that point. For the curve \( y = 3x^2 + 1 \), we've already found that the derivative is \( y' = 6x \).

• At the point \((0, 1)\), we substitute \( x = 0 \) into \( 6x \).
• This gives us \( 6(0) = 0 \).

Thus, the slope of the tangent line at the point \((0, 1)\) is \( 0 \). A zero slope means the tangent line is horizontal at that point.

Point-Slope Form

Once we know the slope of the tangent line, we can write its equation using the point-slope form of a line. This form is especially handy when you know one point on the line and the slope. The general formula is: \[ y - y_1 = m(x - x_1) \]where \((x_1, y_1)\) is the specific point on the line, and \(m\) represents the slope.

In our case, the point given is \((0, 1)\) and the slope, as calculated, is \( 0 \). Substituting these into the formula, we get:

• \( y - 1 = 0(x - 0) \)
• Simplifying this equation, we find \( y = 1 \)

This result shows that the tangent line at the point \((0, 1)\) is parallel to the x-axis. It is essentially a horizontal line passing through \( y = 1 \). This form is powerful because it allows us to swiftly find the equation of a tangent line whenever we have a slope and a point.