Problem 74 - Calculus for Biology and Medicine

Question

Find a point on the curve $$ y=1-3 x^{3} $$ 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

There are two points: \( \left( \frac{1}{3}, \frac{8}{9} \right) \) and \( \left( -\frac{1}{3}, \frac{10}{9} \right) \).

Step 1: Understand the Problem

We need to find a point on the curve \( y = 1 - 3x^3 \) such that the tangent line at this point is parallel to the line \( y = -x \). This means finding \( x \) where the slope of the tangent line, \( \frac{dy}{dx} \), is equal to \(-1\), because the slope of \( y = -x \) is \(-1\).

Step 2: Differentiate the Curve

Find the derivative of the curve \( y = 1 - 3x^3 \) to get the slope of the tangent line. The derivative is \( \frac{dy}{dx} = -9x^2 \).

Step 3: Set the Derivative Equal to -1

Since we want the tangent slope to be \(-1\), set the derivative equal to \(-1\): \[ -9x^2 = -1 \] Solve for \( x \).

Step 4: Solve for x

Divide both sides by \(-9\): \[ x^2 = \frac{1}{9} \]. Taking the square root of both sides gives the possible solutions: \[ x = \frac{1}{3} \quad \text{or} \quad x = -\frac{1}{3} \].

Step 5: Find the Corresponding y-Values

Substitute \( x = \frac{1}{3} \) into the original equation: \[ y = 1 - 3 \left( \frac{1}{3} \right)^3 = 1 - 3 \cdot \frac{1}{27} = 1 - \frac{1}{9} = \frac{8}{9} \].Now substitute \( x = -\frac{1}{3} \): \[ y = 1 - 3 \left( -\frac{1}{3} \right)^3 = 1 + \frac{1}{9} = \frac{10}{9} \].

Step 6: Conclusion

The points on the curve where the tangent line is parallel to \( y = -x \) are \( \left( \frac{1}{3}, \frac{8}{9} \right) \) and \( \left( -\frac{1}{3}, \frac{10}{9} \right) \). There are two such points.

Key Concepts

Tangent Line

A tangent line is a straight line that touches a curve at exactly one point without crossing. It represents the slope of the curve at that specific point. To find the tangent line to a curve at a given point, you need to calculate the derivative of the curve's equation. The tangent line then uses this slope, along with the point's coordinates, to form the equation of the line. For instance, if a curve has the equation \( y = f(x) \), and you want to find the tangent at \( x = a \), the slope of the tangent line would be \( f'(a) \), where \( f'(x) \) denotes the derivative of \( f(x) \). Once you have the slope, you can use the point-slope form for the equation of a line:

• \( y - y_1 = m(x - x_1) \),

where \( m \) is the slope and \((x_1, y_1)\) is the point of tangency.

Derivatives

Derivatives are fundamental in calculus and measure how a function changes as its input changes. They tell us the slope of the tangent line to the curve at any point. Given a function \( f(x) \), its derivative, represented as \( f'(x) \) or \( \frac{df}{dx} \), is the primary tool to understand and find tangents.
To differentiate a function, apply rules such as the power rule, product rule, or chain rule. For polynomial functions, like \( y = 1 - 3x^3 \), use the power rule, which tells us that the derivative of \( x^n \) is \( nx^{n-1} \). Applying this rule, the derivative becomes \( \frac{dy}{dx} = -9x^2 \). This expression gives the slope of the curve at any point \( x \).

• The derivative helps us find points where the tangent slope has specific properties, like being parallel to another line.

Parallel Lines

Parallel lines have identical slopes. When we say a tangent line is parallel to another line, it means both lines rise or fall at the same rate. In the problem context, finding where the tangent line to the curve \( y = 1 - 3x^3 \) is parallel to the line \( y = -x \) means setting the derivative equal to the slope of \( y = -x \).

• For \( y = -x \), the slope is \(-1\).
• We equate \( \frac{dy}{dx} = -9x^2 \) to \(-1\) to find the appropriate \( x \) values.

Doing so shows where the tangent has the same slope of \(-1\), which allows us to find the precise points on the curve where the tangent is parallel to the given line. Solving \( -9x^2 = -1 \) gives two solutions for \( x \), indicating more than one point can satisfy these conditions.