Problem 73 - Calculus for Biology and Medicine

Question

Find a point on the curve $$ y=2 x^{2}-\frac{1}{2} $$ whose tangent line is parallel to the line \(y=x\). Is there more than one such point? If so, find all other points with this property.

Step-by-Step Solution

Answer

The point is \( \left(\frac{1}{4}, -\frac{3}{8}\right) \); there is only one such point.

Step 1: Determine the Slope of the Given Line

We are given that the line is \( y = x \). Here, the slope \( m \) is 1, as can be observed from the form \( y = mx + b \), where \( b = 0 \).

Step 2: Find the Derivative of the Curve

The curve is \( y = 2x^2 - \frac{1}{2} \). To find the slope of the tangent, we need to differentiate this with respect to \( x \): \[ \frac{dy}{dx} = \frac{d}{dx}(2x^2 - \frac{1}{2}) = 4x. \]

Step 3: Set the Derivative Equal to the Slope of the Line

We set the derivative of the curve equal to the slope of the line, since the tangent is parallel to the line: \[ 4x = 1. \]

Step 4: Solve for the Value of x

Divide both sides by 4 to solve for \( x \): \[ x = \frac{1}{4}. \]

Step 5: Find the y-Coordinate of the Point on the Curve

Substitute \( x = \frac{1}{4} \) back into the curve equation to find \( y \): \[ y = 2\left(\frac{1}{4}\right)^2 - \frac{1}{2} = 2\left(\frac{1}{16}\right) - \frac{1}{2} = \frac{1}{8} - \frac{1}{2} = \frac{1}{8} - \frac{4}{8} = -\frac{3}{8}. \]

Step 6: Verify if There are More Points

Since the derivative \( 4x = 1 \) yields only one solution for \( x \), which is \( x = \frac{1}{4} \), there are no other points on the curve with a tangent line parallel to \( y = x \).

Key Concepts

Differentiation

Differentiation in calculus is a fundamental concept that helps us find the rate at which a function is changing at any point. Essentially, it's the mathematical way of finding the slope of a tangent line to a curve. Let's consider the function given in the exercise:

• Function: \( y = 2x^2 - \frac{1}{2} \)

To find the slope of the tangent line to this curve at any point, we perform differentiation. This is done by applying the derivative operator to the function with respect to \( x \). For the function \( y = 2x^2 - \frac{1}{2} \), the derivative, which gives us the slope, is calculated as follows:

• Using power and constant rules, derive: \( \frac{dy}{dx} = \frac{d}{dx}(2x^2 - \frac{1}{2}) = 4x \).

This result tells us that at any point \( x \) on the curve, the slope of the tangent line is \( 4x \). Differentiation thereby allows us to see how steep the curve is at specific points, crucial in solving for tangents that are parallel to other lines.

Parallel Lines

Lines that are parallel have the same slope. So when you need to find a tangent to a curve that is parallel to a given line, you look for where their slopes match. For example, in the problem we're dealing with:

• The line given is \( y = x \), meaning the slope \( m = 1 \).
• We computed the derivative of the curve's equation to be \( 4x \).

To find the points where the tangent is parallel to \( y = x \), we set the derivative \( 4x \) equal to the line's slope:

• \( 4x = 1 \)

Solving \( 4x = 1 \) gives us \( x = \frac{1}{4} \). This tells us that at \( x = \frac{1}{4} \), the slope of the tangent to the curve is the same as the slope of the line \( y = x \). With identical slopes, these two lines are parallel.

Quadratic Functions

A quadratic function is a type of polynomial which can be represented as \( ax^2 + bx + c \), where \( a, b, \) and \( c \) are constants, and \( a eq 0 \). Quadratic functions naturally form parabolas when graphed. In the exercise,

• Our function is \( y = 2x^2 - \frac{1}{2} \), where \( a = 2, b = 0, \) and \( c = -\frac{1}{2} \).

The graph of this function opens upwards because the leading coefficient \( a = 2 \) is positive. The vertex of the parabola is the lowest point, and finding lines tangent to the curve involves calculating the slope at any given point. Since the curve is symmetrical about its vertical axis, any point \( x \) and the corresponding calculated tangent line slope is relevant to the whole shape of the parabola.

• This symmetry is crucial for determining the unique point where the tangent is parallel to another line, as was found in the solution.

Ultimately, understanding quadratic functions enables you to predict the behavior of their graphs and find tangent lines effectively.