Problem 91 - Calculus Volume 3

Question

Find points on the curve at which tangent line is horizontal or vertical. $$ x=t\left(t^{2}-3\right), \quad y=3\left(t^{2}-3\right) $$

Step-by-Step Solution

Answer

Horizontal at (0, -9); vertical at (-2, 0) and (2, 0).

Step 1: Find the derivative of x with respect to t

First, calculate the derivative of the given parametric equation for \( x \). The expression for \( x \) is \( x = t(t^2 - 3) \). Apply the product rule to differentiate: \( x = t^3 - 3t \). Therefore, \( \frac{dx}{dt} = 3t^2 - 3 \).

Step 2: Find the derivative of y with respect to t

Now, calculate the derivative of the parametric equation for \( y \). The expression for \( y \) is \( y = 3(t^2 - 3) \). Therefore, \( \frac{dy}{dt} = 6t \).

Step 3: Find the slope of the tangent line

The slope of the tangent line is the ratio of the derivatives found in Steps 1 and 2: \( \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{6t}{3t^2 - 3} \).

Step 4: Find points where the tangent line is horizontal

For the tangent line to be horizontal, \( \frac{dy}{dx} = 0 \). This implies \( \frac{dy}{dt} = 0 \). Since \( \frac{dy}{dt} = 6t \), set \( 6t = 0 \). Thus, \( t = 0 \). Substitute \( t = 0 \) into the original parametric equations to find the point: \( x = 0 \), \( y = -9 \). Thus, the point is \( (0, -9) \).

Step 5: Find points where the tangent line is vertical

For the tangent line to be vertical, the denominator of \( \frac{dy}{dx} \), i.e., \( \frac{dx}{dt} = 0 \). Set \( 3t^2 - 3 = 0 \). Solve the equation: \( 3(t^2 - 1) = 0 \) which gives \( t^2 = 1 \), hence \( t = \pm 1 \). Substitute \( t = 1 \) into the original parametric equations: \( x = -2 \), \( y = 0 \). Substitute \( t = -1 \) into the equations: \( x = 2 \), \( y = 0 \). Thus, the points are \( (-2, 0) \) and \( (2, 0) \).

Key Concepts

Parametric Equations

Parametric equations are a way to describe a curve using parameters, often denoted as \( t \). Instead of expressing \( y \) solely in terms of \( x \), parametric equations define both \( x \) and \( y \) individually as functions of \( t \).
In this exercise, we have two parametric equations: \( x = t(t^2 - 3) \) and \( y = 3(t^2 - 3) \). Here, \( t \) acts as an intermediary variable that traces the position on the curve as its value changes.
This method of defining functions is especially useful for describing complex curves that cannot be easily expressed as a single function of \( x \) or \( y \).

• It allows the representation of more complex shapes.
• It helps track changes in direction along a curve.

By understanding these equations, we can find various properties of the curve, such as tangent lines, which are covered next.

Derivative Calculation

In calculus, derivatives play a crucial role in understanding change. Here, we begin by finding derivatives of the parametric equations with respect to \( t \).
The derivative of \( x \), given by \( \frac{dx}{dt} \), involves using the product rule. It calculates the rate of change of \( x \) as \( t \) changes: \[ \frac{dx}{dt} = 3t^2 - 3 \]
Similarly, for \( y \), the derivative \( \frac{dy}{dt} \) shows how \( y \) varies with \( t \): \[ \frac{dy}{dt} = 6t \]
The slope of the tangent line to the curve, \( \frac{dy}{dx} \), is then found by dividing these derivatives: \[ \frac{dy}{dx} = \frac{6t}{3t^2 - 3} \]
This ratio provides us the gradient of the tangent to the curve, which can help determine points of interest like horizontal or vertical tangents.

Horizontal Tangent

A horizontal tangent occurs when the tangent line has no slope, meaning it is flat or horizontal. Mathematically, this is when \( \frac{dy}{dx} = 0 \).
To achieve this, the numerator of our tangent slope \( \frac{6t}{3t^2 - 3} \) must be zero. This means \( 6t = 0 \), which implies \( t = 0 \).
By substituting \( t = 0 \) back into the parametric equations \( x = t(t^2 - 3) \) and \( y = 3(t^2 - 3) \), we find:

• \( x = 0 \)
• \( y = -9\)

Hence, the point on the curve where the tangent is horizontal is \( (0, -9) \). This point suggests that at \( t=0 \), the curve stops rising or falling momentarily.

Vertical Tangent

A vertical tangent appears when a tangent line turns directly upwards or downwards, parallel to the y-axis. This means the slope \( \frac{dy}{dx} \) becomes undefined because its denominator is zero.
To find where the tangent is vertical, we set the denominator \( 3t^2 - 3 = 0 \). Solving \( 3(t^2 - 1) = 0 \), we get \( t^2 = 1 \), leading to \( t = \pm 1 \).
Substituting \( t = 1 \) into the parametric equations gives:

• \( x = -2 \)
• \( y = 0 \)

Substituting \( t = -1 \) results in:

• \( x = 2 \)
• \( y = 0 \)

Thus, the curve has vertical tangents at the points \( (-2, 0) \) and \( (2, 0) \). These points represent occasions where the curve shoots upwards or downwards infinitely.