10.2 Homework: Tangent Lines & Parametric Curves

Parametric Curve Analysis

Parametric equations are a way to represent curves by expressing both x and y coordinates as functions of a third variable, usually t (the parameter). This approach is especially useful for describing complex curves and analyzing their properties, such as tangent lines and areas.

• Parametric Form: , , where is the parameter.

• Curve Representation: As varies, the point traces out a curve in the plane.

To Find The Tangent Lines to Parametric Curves

The tangent line to a parametric curve at a given point describes the instantaneous direction of the curve. It shows;

• The direction the curve is heading at that instant

• The rate of change of y with respect to x at that point

• The "velocity direction" if t represents time

The slope of the tangent line is found using derivatives of the parametric equations.

• Given: ,

• Derivative: The slope of the tangent line is

  This comes from the chain rule: dy/dx = (dy/dt) · (dt/dx) = (dy/dt)/(dx/dt)

• Tangent Line Equation:

  1. At

  2. The tangent line at is:

  3. Where

Homework Problem 1:

, ,

Find the point:

1. Point:

STEP 1: Find the derivatives.

1. Tangent Line at :

2. The slope of the tangent line:

3. Tangent line equation at t = -2: Using point-slope form:

   y - 4 = -1/8(x - 0)

   y = -x/8 + 4 or x + 8y = 32

4. Vertical Tangent:

   Vertical tangents occur when dx/dt = 0 and dy/dt ≠ 0:

   • dx/dt = 5t⁴ - 12t² = t²(5t² - 12) = 0

   • This gives: t = 0 or t² = 12/5 → t = ±√(12/5) = ±(2√15)/5

   For t = ±(2√15)/5 ≈ ±1.549:

   • At t = (2√15)/5:

     • x = t⁵ - 4t³ ≈ (1.549)⁵ - 4(1.549)³ ≈ 9.08 - 14.85 ≈ -5.77

     • y = t² = 12/5 = 2.4

   • At t = -(2√15)/5:

     • x = t⁵ - 4t³ ≈ (-1.549)⁵ - 4(-1.549)³ ≈ -9.08 + 14.85 ≈ 5.77

     • y = t² = 12/5 = 2.4

   Vertical tangent points: (±5.77, 2.4) or exactly (±(32√15/25), 12/5)

5. Horizontal Tangent: Horizontal tangents occur when dy/dt = 0 and dx/dt ≠ 0:

   dy/dt = 2t = 0 → t = 0

   Check: dx/dt|_{t=0} = 0 (this means we have a cusp or special point)

   At t = 0: x = 0, y = 0

   Since both derivatives are zero, we need to investigate further using higher derivatives or limits. This is a cusp point at the origin.

   For a proper horizontal tangent where only dy/dt = 0, we only have t = 0, giving point (0, 0).

To Find The Vertical and Horizontal Tangents

• Vertical Tangent: Set and solve for .

  Vertical tangents occur when dx/dt = 0 and dy/dt ≠ 0:

  • dx/dt = 5t⁴ - 12t² = t²(5t² - 12) = 0

  • This gives: t = 0 or t² = 12/5 → t = ±√(12/5) = ±(2√15)/5

  For t = ±(2√15)/5 ≈ ±1.549:

  • At t = (2√15)/5:

    • x = t⁵ - 4t³ ≈ (1.549)⁵ - 4(1.549)³ ≈ 9.08 - 14.85 ≈ -5.77

    • y = t² = 12/5 = 2.4

  • At t = -(2√15)/5:

    • x = t⁵ - 4t³ ≈ (-1.549)⁵ - 4(-1.549)³ ≈ -9.08 + 14.85 ≈ 5.77

    • y = t² = 12/5 = 2.4

  Vertical tangent points: (±5.77, 2.4) or exactly (±(32√15/25), 12/5)

• Horizontal Tangent: Set and solve for .

  Horizontal tangents occur when dy/dt = 0 and dx/dt ≠ 0:

  • dy/dt = 2t = 0 → t = 0

  • Check: dx/dt|_{t=0} = 0 (this means we have a cusp or special point)

  At t = 0: x = 0, y = 0 Since both derivatives are zero, we need to investigate further using higher derivatives or limits.

  Actually, examining the behavior: this is a cusp point at the origin.

  For a proper horizontal tangent where only dy/dt = 0, we only have t = 0, giving point (0, 0).

• Labeling Points: Substitute values into and to find the coordinates of these points.

To Graph & Label:

• Graph the parametric curve by plotting for a range of values.

• Draw tangent lines at points where the slope is vertical or horizontal.

• Label these points clearly on the graph.

  x(t) = t^5 - 4t^3

  y(t) = t^2

  (x(t), y(t))

• # Tangent at t=-2

  y = -x/8 + 4

• # Mark point at t=-2

  (0, 4)

• # Vertical tangent points

  (5.77, 2.4)

  (-5.77, 2.4)

• # Vertical tangent lines

  x = 5.77

  x = -5.77

• # Origin (cusp)

  (0, 0)

Enclosed Area by Parametric Curve

The area enclosed by a parametric curve and the x-axis can be found using integration. For a curve defined by and , the area is:

Given Parametric Equations

• x = f(t) = t³ + 1

• y = g(t) = 2t - t²

Step 1: Find where the curve intersects the x-axis

Set y = 0:

• 2t - t² = 0

• t(2 - t) = 0

• t = 0 or t = 2

At these values:

• When t = 0: x = 0³ + 1 = 1

• When t = 2: x = 2³ + 1 = 9

So the curve intersects the x-axis at points (1, 0) and (9, 0).

Step 2: Check if y is positive or negative between t = 0 and t = 2

y = 2t - t² = t(2 - t)

For 0 < t < 2: both t > 0 and (2 - t) > 0, so y > 0

This means the curve is above the x-axis between these intersection points.

Step 3: Calculate the area using the parametric area formula

For a parametric curve, the area between the curve and the x-axis is:

A = ∫ y dx = ∫ᵃᵇ y(t) · x'(t) dt

First, find x'(t):

• x'(t) = 3t²

Now set up the integral: A = ∫₀² (2t - t²) · (3t²) dt

A = ∫₀² (6t³ - 3t⁴) dt

Step 4: Evaluate the integral

A = [6 · t⁴/4 - 3 · t⁵/5]₀²

A = [3t⁴/2 - 3t⁵/5]₀²

A = [3(2)⁴/2 - 3(2)⁵/5] - [0]

A = [3(16)/2 - 3(32)/5]

A = [24 - 96/5]

A = [120/5 - 96/5]

A = 24/5

Answer:

The area enclosed by the parametric curve and the x-axis is 24/5 = 4.8 square units.

Desmos Graph:

x(t) = t^3 + 1

y(t) = 2t - t^2

(x(t), y(t)) {0 ≤ t ≤ 2}

• # Intersection points

  (1, 0)

  (9, 0)

• # Shade the region (approximate)

  y = 2((x-1)^(1/3)) - ((x-1)^(2/3)) {1 ≤ x ≤ 9}

• Note: The shaded region represents the area of 4.8 square units enclosed between the parametric curve and the x-axis.

Summary Table: Tangent Line Conditions for Parametric Curves

Additional info:

• Parametric equations are covered in Chapter 10 of most Calculus textbooks.

• Graphing parametric curves and analyzing tangent lines are essential skills for understanding motion and geometry in Calculus II.