Problem 77 - Calculus for Biology and Medicine

Question

Show that the tangent line to the curve $$ y=x^{2} $$ at the point \((1,1)\) passes through the point \((0,-1)\).

Step-by-Step Solution

Answer

The tangent line is \( y = 2x - 1 \), which passes through \((0, -1)\).

Step 1: Find the derivative

To find the equation of the tangent line to the curve at a given point, start by finding the derivative of the function. The derivative of the function \( y = x^2 \) is \( \frac{dy}{dx} = 2x \).

Step 2: Evaluate the derivative at the point

Evaluate the derivative at the given point \((1,1)\). Substitute \( x = 1 \) into the derivative: \( \frac{dy}{dx}\bigg|_{x=1} = 2(1) = 2 \). This value represents the slope of the tangent line at \((1,1)\).

Step 3: Write the equation of the tangent line

Using the point-slope form of a line, which is \( y - y_1 = m(x - x_1) \), where \( m \) is the slope, and \((x_1, y_1)\) is the point, plug in the slope \( m = 2 \) and the point \((1,1)\) to get: \( y - 1 = 2(x - 1) \). Simplify to find the line equation: \( y = 2x - 1 \).

Step 4: Verify the line passes through the given point

Check if the point \((0, -1)\) lies on the line. Substitute \( x = 0 \) into the line equation \( y = 2x - 1 \): \( y = 2(0) - 1 = -1 \). Since when \( x = 0 \) the corresponding \( y \) is \(-1\), the line passes through the point \((0, -1)\).

Key Concepts

Understanding Derivatives

Derivatives are fundamental in calculus and are crucial for understanding changes in functions. In simple terms, a derivative measures how a function changes as its input changes. It gives us the rate of change or the slope of the function at any given point. Suppose you have a curve described by the function \( y = x^2 \). The derivative of this function, denoted by \( \frac{dy}{dx} \), tells us how steep the curve is at any point along the \( x \)-axis.For the curve \( y = x^2 \), the derivative is \( \frac{dy}{dx} = 2x \). This means that for every unit increase in \( x \), \( y \) increases by \( 2x \). Thus, when we want to find the tangent line at a specific point, like \((1,1)\), we evaluate the derivative at that point to determine the slope. Here, substituting \( x = 1 \) gives a slope of \( 2 \), meaning the tangent line is rising steeply at that point.

The Role of Point-Slope Form

Once you've determined the slope from the derivative, the next step is to use this information to write the equation of the tangent line. That's where the point-slope form comes into play. The point-slope form of a line is a convenient way to find the equation of a line when you know:

• A point on the line, \((x_1, y_1)\).
• The slope of the line, \(m\).

The formula is: \( y - y_1 = m(x - x_1) \). By substituting the point \((1,1)\) and the slope \( 2 \) into the equation, you get: \( y - 1 = 2(x - 1) \). Simplifying this equation provides the equation of the tangent line: \( y = 2x - 1 \). This form is incredibly useful for quickly establishing the linear representation of a tangent to a curve at a specific point.

Curves and Their Tangents

Curves, such as the one described by the function \( y = x^2 \), represent a variety of shapes and can demonstrate different rates of change across different points. A curve is simply the graphical representation of a function's input-output relationship.The tangent line is a straight line that just "touches" the curve at a single point, without cutting across it. This point of tangency is unique because it represents the exact point where the slope of the line and the slope of the curve match. For the given problem, at the point \((1,1)\), the equation of the tangent we found is \( y = 2x - 1 \).Interestingly, this tangent line not only touches the curve at \((1,1)\), but also extends to pass through the point \((0, -1)\). This ability to predict where the tangent line goes helps reinforce the concept that tangents give linear approximations of curves' local behavior.